problem
stringlengths 15
4.7k
| solution
stringlengths 2
11.9k
| answer
stringclasses 51
values | problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
A point-like mass moves horizontally between two walls on a frictionless surface with initial kinetic energy $E$. With every collision with the walls, the mass loses $1/2$ its kinetic energy to thermal energy. How many collisions with the walls are necessary before the speed of the mass is reduced by a factor of $8$?
$ \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16 $
|
1. Let's denote the initial kinetic energy of the mass as \( E_0 \). The kinetic energy after the first collision will be \( \frac{1}{2} E_0 \), after the second collision it will be \( \left(\frac{1}{2}\right)^2 E_0 = \frac{1}{4} E_0 \), and so on. After \( n \) collisions, the kinetic energy will be \( \left(\frac{1}{2}\right)^n E_0 \).
2. The kinetic energy is given by \( \text{KE} = \frac{1}{2} mv^2 \). Since the kinetic energy is directly proportional to the square of the speed, if the speed is reduced by a factor of 8, the kinetic energy is reduced by a factor of \( 8^2 = 64 \).
3. We need to find the number of collisions \( n \) such that the kinetic energy is reduced by a factor of 64. This means:
\[
\left(\frac{1}{2}\right)^n E_0 = \frac{1}{64} E_0
\]
4. Simplifying the equation:
\[
\left(\frac{1}{2}\right)^n = \frac{1}{64}
\]
5. Recognizing that \( 64 = 2^6 \), we can rewrite the equation as:
\[
\left(\frac{1}{2}\right)^n = \left(\frac{1}{2^6}\right)
\]
6. Therefore, \( n = 6 \).
Conclusion:
\[
\boxed{6}
\]
|
6
|
Other
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
21) A particle of mass $m$ moving at speed $v_0$ collides with a particle of mass $M$ which is originally at rest. The fractional momentum transfer $f$ is the absolute value of the final momentum of $M$ divided by the initial momentum of $m$.
If the collision is perfectly $elastic$, what is the maximum possible fractional momentum transfer, $f_{max}$?
A) $0 < f_{max} < \frac{1}{2}$
B) $f_{max} = \frac{1}{2}$
C) $\frac{1}{2} < f_{max} < \frac{3}{2}$
D) $f_{max} = 2$
E) $f_{max} \ge 3$
|
1. **Conservation of Momentum:**
The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. For the given problem, we have:
\[
mv_0 = mv_0' + Mv_1
\]
where \(v_0\) is the initial velocity of the particle with mass \(m\), \(v_0'\) is the final velocity of the particle with mass \(m\), and \(v_1\) is the final velocity of the particle with mass \(M\).
2. **Conservation of Kinetic Energy:**
Since the collision is perfectly elastic, the total kinetic energy before and after the collision is conserved. Therefore:
\[
\frac{1}{2}mv_0^2 = \frac{1}{2}mv_0'^2 + \frac{1}{2}Mv_1^2
\]
3. **Relative Velocity in Elastic Collision:**
For elastic collisions, the relative velocity of approach before the collision is equal to the relative velocity of separation after the collision:
\[
v_0 - 0 = v_1 - v_0'
\]
This simplifies to:
\[
v_0 = v_1 - v_0'
\]
Therefore:
\[
v_0' = v_1 - v_0
\]
4. **Substitute \(v_0'\) into the Momentum Equation:**
Substitute \(v_0' = v_1 - v_0\) into the momentum conservation equation:
\[
mv_0 = m(v_1 - v_0) + Mv_1
\]
Simplify and solve for \(v_1\):
\[
mv_0 = mv_1 - mv_0 + Mv_1
\]
\[
mv_0 + mv_0 = mv_1 + Mv_1
\]
\[
2mv_0 = (m + M)v_1
\]
\[
v_1 = \frac{2mv_0}{m + M}
\]
5. **Calculate the Fractional Momentum Transfer (FMT):**
The fractional momentum transfer \(f\) is defined as the absolute value of the final momentum of \(M\) divided by the initial momentum of \(m\):
\[
f = \left| \frac{Mv_1}{mv_0} \right|
\]
Substitute \(v_1 = \frac{2mv_0}{m + M}\) into the equation:
\[
f = \left| \frac{M \cdot \frac{2mv_0}{m + M}}{mv_0} \right|
\]
Simplify:
\[
f = \left| \frac{2M}{m + M} \right|
\]
6. **Determine the Maximum Value of \(f\):**
To find the maximum value of \(f\), consider the behavior of the function \(\frac{2M}{m + M}\). As \(m\) approaches 0, the expression \(\frac{2M}{m + M}\) approaches 2. Therefore, the maximum possible value of \(f\) is:
\[
f_{max} = 2
\]
The final answer is \( \boxed{ 2 } \)
|
2
|
Calculus
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A construction rope is tied to two trees. It is straight and taut. It is then vibrated at a constant velocity $v_1$. The tension in the rope is then halved. Again, the rope is vibrated at a constant velocity $v_2$. The tension in the rope is then halved again. And, for the third time, the rope is vibrated at a constant velocity, this time $v_3$. The value of $\frac{v_1}{v_3}+\frac{v_3}{v_1}$ can be expressed as a positive number $\frac{m\sqrt{r}}{n}$, where $m$ and $n$ are relatively prime, and $r$ is not divisible by the square of any prime. Find $m+n+r$. If the number is rational, let $r=1$.
[i](Ahaan Rungta, 2 points)[/i]
|
1. The speed of a wave on a string is given by the formula:
\[
v = \sqrt{\frac{T}{\mu}}
\]
where \( T \) is the tension in the rope and \( \mu \) is the linear mass density of the rope.
2. Initially, the tension in the rope is \( T \) and the velocity of the wave is \( v_1 \):
\[
v_1 = \sqrt{\frac{T}{\mu}}
\]
3. When the tension is halved, the new tension is \( \frac{T}{2} \). The new velocity \( v_2 \) is:
\[
v_2 = \sqrt{\frac{\frac{T}{2}}{\mu}} = \sqrt{\frac{T}{2\mu}} = \frac{1}{\sqrt{2}} \sqrt{\frac{T}{\mu}} = \frac{v_1}{\sqrt{2}}
\]
4. When the tension is halved again, the new tension is \( \frac{T}{4} \). The new velocity \( v_3 \) is:
\[
v_3 = \sqrt{\frac{\frac{T}{4}}{\mu}} = \sqrt{\frac{T}{4\mu}} = \frac{1}{2} \sqrt{\frac{T}{\mu}} = \frac{v_1}{2}
\]
5. We need to find the value of \( \frac{v_1}{v_3} + \frac{v_3}{v_1} \):
\[
\frac{v_1}{v_3} = \frac{v_1}{\frac{v_1}{2}} = 2
\]
\[
\frac{v_3}{v_1} = \frac{\frac{v_1}{2}}{v_1} = \frac{1}{2}
\]
\[
\frac{v_1}{v_3} + \frac{v_3}{v_1} = 2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}
\]
6. The expression \( \frac{5}{2} \) can be written as \( \frac{m\sqrt{r}}{n} \) where \( m = 5 \), \( r = 1 \), and \( n = 2 \). Since \( r = 1 \), it is rational.
7. Therefore, \( m + n + r = 5 + 2 + 1 = 8 \).
The final answer is \( \boxed{8} \).
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Bob, a spherical person, is floating around peacefully when Dave the giant orange fish launches him straight up 23 m/s with his tail. If Bob has density 100 $\text{kg/m}^3$, let $f(r)$ denote how far underwater his centre of mass plunges underwater once he lands, assuming his centre of mass was at water level when he's launched up. Find $\lim_{r\to0} \left(f(r)\right) $. Express your answer is meters and round to the nearest integer. Assume the density of water is 1000 $\text{kg/m}^3$.
[i](B. Dejean, 6 points)[/i]
|
1. **Initial Setup and Assumptions:**
- Bob is launched upwards with an initial velocity \( u = 23 \, \text{m/s} \).
- Bob's density \( \sigma = 100 \, \text{kg/m}^3 \).
- Water's density \( \rho = 1000 \, \text{kg/m}^3 \).
- We need to find the limit of \( f(r) \) as \( r \to 0 \), where \( f(r) \) is the depth Bob's center of mass plunges underwater.
2. **Energy Conservation:**
- Using the work-energy theorem, the initial kinetic energy of Bob is given by:
\[
\frac{1}{2} m u^2
\]
- The work done by the buoyant force \( W_{\text{buy}} \) and the gravitational potential energy change \( mgh \) must be considered.
3. **Buoyant Force and Work Done:**
- The buoyant force \( F_{\text{buy}} \) is given by Archimedes' principle:
\[
F_{\text{buy}} = \rho V g
\]
where \( V \) is the volume of Bob.
- The work done by the buoyant force as Bob plunges to depth \( h \) is:
\[
W_{\text{buy}} = F_{\text{buy}} \cdot h = \rho V g h
\]
4. **Volume and Mass Relationship:**
- Bob's volume \( V \) can be expressed in terms of his mass \( m \) and density \( \sigma \):
\[
V = \frac{m}{\sigma}
\]
5. **Energy Conservation Equation:**
- Equating the initial kinetic energy to the work done by the buoyant force and the gravitational potential energy change:
\[
\frac{1}{2} m u^2 = \rho V g h - mgh
\]
- Substituting \( V = \frac{m}{\sigma} \):
\[
\frac{1}{2} m u^2 = \rho \left( \frac{m}{\sigma} \right) g h - mgh
\]
- Simplifying:
\[
\frac{1}{2} u^2 = \left( \frac{\rho}{\sigma} - 1 \right) gh
\]
6. **Solving for \( h \):**
- Rearranging the equation to solve for \( h \):
\[
h = \frac{\frac{1}{2} u^2}{\left( \frac{\rho}{\sigma} - 1 \right) g}
\]
- Substituting the given values \( u = 23 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( \sigma = 100 \, \text{kg/m}^3 \), and \( g = 9.81 \, \text{m/s}^2 \):
\[
h = \frac{\frac{1}{2} \cdot 23^2}{\left( \frac{1000}{100} - 1 \right) \cdot 9.81}
\]
\[
h = \frac{\frac{1}{2} \cdot 529}{(10 - 1) \cdot 9.81}
\]
\[
h = \frac{264.5}{9 \cdot 9.81}
\]
\[
h = \frac{264.5}{88.29}
\]
\[
h \approx 2.99 \, \text{m}
\]
The final answer is \( \boxed{3} \, \text{m} \).
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many moles of oxygen gas are produced by the decomposition of $245$ g of potassium chlorate?
\[\ce{2KClO3(s)} \rightarrow \ce{2KCl(s)} + \ce{3O2(g)}\]
Given:
Molar Mass/ $\text{g} \cdot \text{mol}^{-1}$
$\ce{KClO3}$: $122.6$
$ \textbf{(A)}\hspace{.05in}1.50 \qquad\textbf{(B)}\hspace{.05in}2.00 \qquad\textbf{(C)}\hspace{.05in}2.50 \qquad\textbf{(D)}\hspace{.05in}3.00 \qquad $
|
1. **Calculate the number of moles of potassium chlorate ($\ce{KClO3}$):**
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{245 \text{ g}}{122.6 \text{ g/mol}}
\]
\[
\text{Number of moles} = 2.00 \text{ mol}
\]
2. **Use the stoichiometry of the reaction to find the moles of oxygen gas ($\ce{O2}$) produced:**
The balanced chemical equation is:
\[
\ce{2KClO3(s)} \rightarrow \ce{2KCl(s)} + \ce{3O2(g)}
\]
According to the equation, 2 moles of $\ce{KClO3}$ produce 3 moles of $\ce{O2}$.
3. **Calculate the moles of $\ce{O2}$ produced from 2.00 moles of $\ce{KClO3}$:**
\[
\text{Moles of } \ce{O2} = \left(\frac{3 \text{ moles of } \ce{O2}}{2 \text{ moles of } \ce{KClO3}}\right) \times 2.00 \text{ moles of } \ce{KClO3}
\]
\[
\text{Moles of } \ce{O2} = \frac{3}{2} \times 2.00 = 3.00 \text{ moles}
\]
The final answer is $\boxed{3.00}$
|
3.00
|
Other
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $n$ points with integer coordinates be given in the $xy$-plane. What is the minimum value of $n$ which will ensure that three of the points are the vertices of a triangel with integer (possibly, 0) area?
|
1. **Assume a point at the origin:**
Without loss of generality, we can assume that one of the points is at the origin, \( O(0,0) \). This is because we can always translate the entire set of points such that one of them is at the origin.
2. **Area formula for a triangle:**
If \( A(a,b) \) and \( B(x,y) \) are two points, the area of triangle \( OAB \) is given by:
\[
\text{Area} = \frac{1}{2} \left| ax - by \right|
\]
For the area to be an integer (possibly zero), \( \left| ax - by \right| \) must be an even integer.
3. **Coordinate parity analysis:**
- If one of the points, say \( A \), has both coordinates even, then \( a \) and \( b \) are even. Thus, \( ax \) and \( by \) are even for any integer coordinates \( x \) and \( y \). Therefore, \( \left| ax - by \right| \) is even, ensuring the area is an integer.
- If two points, say \( A \) and \( B \), have all their coordinates odd, then \( a, b, x, \) and \( y \) are all odd. The product of two odd numbers is odd, so \( ax \) and \( by \) are odd. The difference of two odd numbers is even, ensuring \( \left| ax - by \right| \) is even.
- The same logic applies if two points have coordinates of the form \( (o,e) \) or \( (e,o) \), where \( e \) stands for even and \( o \) stands for odd.
4. **Conclusion from parity analysis:**
If a set of integer points does not contain three points which are the vertices of a triangle with integer area, then this set must have at most 4 elements. This is because there are only 4 possible combinations of parity for the coordinates: \( (e,e), (e,o), (o,e), (o,o) \).
5. **Existence of a set with 4 elements:**
Consider the four vertices of a unit square: \( (0,0), (1,0), (0,1), (1,1) \). None of these points form a triangle with integer area, as the area of any triangle formed by these points is either \( \frac{1}{2} \) or 0.
Therefore, the minimum value of \( n \) which ensures that three of the points are the vertices of a triangle with integer area is 5.
The final answer is \(\boxed{5}\).
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Prove that if $f$ is a non-constant real-valued function such that for all real $x$, $f(x+1) + f(x-1) = \sqrt{3} f(x)$, then $f$ is periodic. What is the smallest $p$, $p > 0$ such that $f(x+p) = f(x)$ for all $x$?
|
1. We start with the given functional equation:
\[
f(x+1) + f(x-1) = \sqrt{3} f(x)
\]
We need to prove that \( f \) is periodic and find the smallest period \( p \) such that \( f(x+p) = f(x) \) for all \( x \).
2. Consider the function \( f(x) = \cos\left(\frac{\pi x}{6}\right) \). We will verify if this function satisfies the given functional equation:
\[
f(x+1) = \cos\left(\frac{\pi (x+1)}{6}\right) = \cos\left(\frac{\pi x}{6} + \frac{\pi}{6}\right)
\]
\[
f(x-1) = \cos\left(\frac{\pi (x-1)}{6}\right) = \cos\left(\frac{\pi x}{6} - \frac{\pi}{6}\right)
\]
3. Using the sum-to-product identities for cosine, we have:
\[
\cos(A + B) + \cos(A - B) = 2 \cos(A) \cos(B)
\]
Let \( A = \frac{\pi x}{6} \) and \( B = \frac{\pi}{6} \). Then:
\[
f(x+1) + f(x-1) = \cos\left(\frac{\pi x}{6} + \frac{\pi}{6}\right) + \cos\left(\frac{\pi x}{6} - \frac{\pi}{6}\right)
\]
\[
= 2 \cos\left(\frac{\pi x}{6}\right) \cos\left(\frac{\pi}{6}\right)
\]
4. Since \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \), we get:
\[
f(x+1) + f(x-1) = 2 \cos\left(\frac{\pi x}{6}\right) \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \cos\left(\frac{\pi x}{6}\right) = \sqrt{3} f(x)
\]
Thus, \( f(x) = \cos\left(\frac{\pi x}{6}\right) \) satisfies the given functional equation.
5. The period of \( \cos\left(\frac{\pi x}{6}\right) \) is \( \frac{2\pi}{\frac{\pi}{6}} = 12 \). Therefore, \( f(x+12) = f(x) \) for all \( x \).
6. To check if there is a smaller period, consider another function \( f(x) = \cos\left(\frac{11\pi x}{6}\right) \). This function also satisfies the given functional equation:
\[
f(x+1) + f(x-1) = \cos\left(\frac{11\pi (x+1)}{6}\right) + \cos\left(\frac{11\pi (x-1)}{6}\right)
\]
\[
= 2 \cos\left(\frac{11\pi x}{6}\right) \cos\left(\frac{11\pi}{6}\right)
\]
Since \( \cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2} \), we get:
\[
f(x+1) + f(x-1) = \sqrt{3} f(x)
\]
The period of \( \cos\left(\frac{11\pi x}{6}\right) \) is \( \frac{2\pi}{\frac{11\pi}{6}} = \frac{12}{11} \), which is not an integer.
7. Therefore, the smallest period that works for every function that satisfies the equation is 12.
The final answer is \( \boxed{12} \).
|
12
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
An international firm has 250 employees, each of whom speaks several languages. For each pair of employees, $(A,B)$, there is a language spoken by $A$ and not $B$, and there is another language spoken by $B$ but not $A$. At least how many languages must be spoken at the firm?
|
1. **Define the problem in terms of sets:**
- Let \( A \) be an employee and \( L_A \) be the set of languages spoken by \( A \).
- The problem states that for each pair of employees \( (A, B) \), there is a language spoken by \( A \) and not \( B \), and a language spoken by \( B \) but not \( A \). This implies that the sets \( L_A \) form an antichain.
2. **Apply Sperner's theorem:**
- Sperner's theorem states that the maximum size of an antichain in the power set of a set with \( n \) elements is given by \( \binom{n}{\lfloor n/2 \rfloor} \).
- We need to find the smallest \( n \) such that \( \binom{n}{\lfloor n/2 \rfloor} \geq 250 \).
3. **Calculate binomial coefficients:**
- We need to check the values of \( n \) to find the smallest \( n \) that satisfies the inequality.
- For \( n = 10 \):
\[
\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]
- Since \( 252 \geq 250 \), \( n = 10 \) satisfies the condition.
4. **Verify smaller values of \( n \):**
- For \( n = 9 \):
\[
\binom{9}{4} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126
\]
- Since \( 126 < 250 \), \( n = 9 \) does not satisfy the condition.
5. **Conclusion:**
- The smallest \( n \) such that \( \binom{n}{\lfloor n/2 \rfloor} \geq 250 \) is \( n = 10 \).
The final answer is \( \boxed{10} \).
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For $n$ a positive integer, denote by $P(n)$ the product of all positive integers divisors of $n$. Find the smallest $n$ for which
\[ P(P(P(n))) > 10^{12} \]
|
1. **Understanding the formula for \( P(n) \):**
The product of all positive divisors of \( n \) is given by:
\[
P(n) = n^{\tau(n)/2}
\]
where \( \tau(n) \) is the number of positive divisors of \( n \).
2. **Analyzing the formula for prime numbers:**
If \( n = p \) where \( p \) is a prime number, then \( \tau(p) = 2 \) (since the divisors are \( 1 \) and \( p \)). Thus:
\[
P(p) = p^{2/2} = p
\]
Therefore, \( P(P(P(p))) = p \), which does not satisfy \( P(P(P(n))) > 10^{12} \).
3. **Analyzing the formula for \( n = p^a \):**
If \( n = p^a \) where \( p \) is a prime number and \( a \) is a positive integer, then \( \tau(p^a) = a + 1 \). Thus:
\[
P(p^a) = (p^a)^{(a+1)/2} = p^{a(a+1)/2}
\]
We need to find the smallest \( n \) such that \( P(P(P(n))) > 10^{12} \).
4. **Testing \( n = 6 \):**
Let's test \( n = 6 \):
- The divisors of \( 6 \) are \( 1, 2, 3, 6 \), so \( \tau(6) = 4 \).
- Therefore:
\[
P(6) = 6^{4/2} = 6^2 = 36
\]
- Next, we calculate \( P(36) \):
- The divisors of \( 36 \) are \( 1, 2, 3, 4, 6, 9, 12, 18, 36 \), so \( \tau(36) = 9 \).
- Therefore:
\[
P(36) = 36^{9/2} = 36^{4.5} = 6^9 = 2^9 \cdot 3^9 = 512 \cdot 19683 = 10077696
\]
- Finally, we calculate \( P(10077696) \):
- The number \( 10077696 \) is \( 2^{18} \cdot 3^9 \).
- The number of divisors \( \tau(10077696) = (18+1)(9+1) = 19 \cdot 10 = 190 \).
- Therefore:
\[
P(10077696) = (10077696)^{190/2} = (2^{18} \cdot 3^9)^{95} = 2^{1710} \cdot 3^{855}
\]
- We need to check if this value is greater than \( 10^{12} \):
\[
2^{1710} \cdot 3^{855} > 10^{12}
\]
Since \( 2^{10} \approx 10^3 \) and \( 3^5 \approx 10^2 \), we can approximate:
\[
2^{1710} \approx (10^3)^{171} = 10^{513}
\]
\[
3^{855} \approx (10^2)^{171} = 10^{342}
\]
\[
2^{1710} \cdot 3^{855} \approx 10^{513} \cdot 10^{342} = 10^{855}
\]
Clearly, \( 10^{855} > 10^{12} \).
Therefore, \( n = 6 \) satisfies \( P(P(P(n))) > 10^{12} \).
The final answer is \( \boxed{6} \).
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $T = TNFTPP$. $x$ and $y$ are nonzero real numbers such that \[18x - 4x^2 + 2x^3 - 9y - 10xy - x^2y + Ty^2 + 2xy^2 - y^3 = 0.\] The smallest possible value of $\tfrac{y}{x}$ is equal to $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
[b]Note: This is part of the Ultimate Problem, where each question depended on the previous question. For those who wanted to try the problem separately, [hide=here's the value of T]$T=6$[/hide].
|
Given the equation:
\[ 18x - 4x^2 + 2x^3 - 9y - 10xy - x^2y + 6y^2 + 2xy^2 - y^3 = 0 \]
We are given that \( T = 6 \). Substituting \( T \) into the equation, we get:
\[ 18x - 4x^2 + 2x^3 - 9y - 10xy - x^2y + 6y^2 + 2xy^2 - y^3 = 0 \]
Let \( \frac{y}{x} = k \). Then \( y = kx \). Substituting \( y = kx \) into the equation, we get:
\[ 18x - 4x^2 + 2x^3 - 9kx - 10xkx - x^2kx + 6(kx)^2 + 2x(kx)^2 - (kx)^3 = 0 \]
Simplifying, we get:
\[ 18x - 4x^2 + 2x^3 - 9kx - 10kx^2 - kx^3 + 6k^2x^2 + 2k^2x^3 - k^3x^3 = 0 \]
Combining like terms, we get:
\[ (18 - 9k)x + (-4 - 10k + 6k^2)x^2 + (2 - k + 2k^2 - k^3)x^3 = 0 \]
Factoring out \( x \), we get:
\[ x \left( (18 - 9k) + (-4 - 10k + 6k^2)x + (2 - k + 2k^2 - k^3)x^2 \right) = 0 \]
Since \( x \neq 0 \), we have:
\[ (18 - 9k) + (-4 - 10k + 6k^2)x + (2 - k + 2k^2 - k^3)x^2 = 0 \]
This can be rewritten as:
\[ (2 - k)x \left( (k^2 + 1)x^2 - (6k + 2)x + 9 \right) = 0 \]
We can see that one solution is \( k = 2 \). Now, suppose that \( k \neq 2 \). We need to solve the quadratic equation:
\[ (k^2 + 1)x^2 - (6k + 2)x + 9 = 0 \]
For this quadratic equation to have a real solution, the discriminant must be nonnegative:
\[ (6k + 2)^2 - 4(k^2 + 1) \cdot 9 \geq 0 \]
Simplifying the discriminant, we get:
\[ (6k + 2)^2 - 36(k^2 + 1) \geq 0 \]
\[ 36k^2 + 24k + 4 - 36k^2 - 36 \geq 0 \]
\[ 24k - 32 \geq 0 \]
\[ 24k \geq 32 \]
\[ k \geq \frac{32}{24} \]
\[ k \geq \frac{4}{3} \]
Therefore, the minimum value of \( k \) is \( \frac{4}{3} \). The smallest possible value of \( \frac{y}{x} \) is \( \frac{4}{3} \).
Thus, \( m = 4 \) and \( n = 3 \), and \( m + n = 4 + 3 = 7 \).
The final answer is \( \boxed{ 7 } \)
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $T = TNFTPP$, and let $S$ be the sum of the digits of $T$. Cyclic quadrilateral $ABCD$ has side lengths $AB = S - 11$, $BC = 2$, $CD = 3$, and $DA = 10$. Let $M$ and $N$ be the midpoints of sides $AD$ and $BC$. The diagonals $AC$ and $BD$ intersect $MN$ at $P$ and $Q$ respectively. $\frac{PQ}{MN}$ can be expressed as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Determine $m + n$.
[b]Note: This is part of the Ultimate Problem, where each question depended on the previous question. For those who wanted to try the problem separately, [hide=here's the value of T.]$T=2378$[/hide]
|
1. **Determine the value of \( S \):**
Given \( T = 2378 \), we need to find the sum of the digits of \( T \).
\[
S = 2 + 3 + 7 + 8 = 20
\]
2. **Calculate the side length \( AB \):**
Given \( AB = S - 11 \),
\[
AB = 20 - 11 = 9
\]
3. **Verify the side lengths of the cyclic quadrilateral \( ABCD \):**
\[
AB = 9, \quad BC = 2, \quad CD = 3, \quad DA = 10
\]
4. **Find the midpoints \( M \) and \( N \):**
- \( M \) is the midpoint of \( AD \).
- \( N \) is the midpoint of \( BC \).
5. **Use the properties of cyclic quadrilaterals and midlines:**
In a cyclic quadrilateral, the line segment joining the midpoints of opposite sides is parallel to the line segment joining the other pair of opposite sides and is half its length.
6. **Calculate the length of \( MN \):**
Since \( M \) and \( N \) are midpoints,
\[
MN = \frac{1}{2} \times AC
\]
7. **Calculate the length of \( AC \):**
Using the Ptolemy's theorem for cyclic quadrilateral \( ABCD \):
\[
AC \cdot BD = AB \cdot CD + AD \cdot BC
\]
Let \( BD = x \),
\[
AC \cdot x = 9 \cdot 3 + 10 \cdot 2 = 27 + 20 = 47
\]
Since \( AC = \sqrt{94} \),
\[
\sqrt{94} \cdot x = 47 \implies x = \frac{47}{\sqrt{94}} = \frac{47 \sqrt{94}}{94} = \frac{47}{\sqrt{94}} \cdot \frac{\sqrt{94}}{\sqrt{94}} = \frac{47 \sqrt{94}}{94}
\]
8. **Calculate \( PQ \):**
Since \( PQ \) is the segment of \( MN \) intersected by diagonals \( AC \) and \( BD \), and using the properties of midlines in cyclic quadrilaterals,
\[
PQ = \frac{1}{2} \times BD
\]
\[
PQ = \frac{1}{2} \times \frac{47 \sqrt{94}}{94} = \frac{47 \sqrt{94}}{188}
\]
9. **Calculate the ratio \( \frac{PQ}{MN} \):**
\[
\frac{PQ}{MN} = \frac{\frac{47 \sqrt{94}}{188}}{\frac{\sqrt{94}}{2}} = \frac{47 \sqrt{94}}{188} \times \frac{2}{\sqrt{94}} = \frac{47 \times 2}{188} = \frac{94}{188} = \frac{1}{2}
\]
10. **Express the ratio in simplest form:**
\[
\frac{PQ}{MN} = \frac{1}{2}
\]
Here, \( m = 1 \) and \( n = 2 \).
11. **Sum \( m \) and \( n \):**
\[
m + n = 1 + 2 = 3
\]
The final answer is \(\boxed{3}\).
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the value of $c$ such that the system of equations \begin{align*}|x+y|&=2007,\\|x-y|&=c\end{align*} has exactly two solutions $(x,y)$ in real numbers.
$\begin{array}{@{\hspace{-1em}}l@{\hspace{14em}}l@{\hspace{14em}}l}
\textbf{(A) }0&\textbf{(B) }1&\textbf{(C) }2\\\\
\textbf{(D) }3&\textbf{(E) }4&\textbf{(F) }5\\\\
\textbf{(G) }6&\textbf{(H) }7&\textbf{(I) }8\\\\
\textbf{(J) }9&\textbf{(K) }10&\textbf{(L) }11\\\\
\textbf{(M) }12&\textbf{(N) }13&\textbf{(O) }14\\\\
\textbf{(P) }15&\textbf{(Q) }16&\textbf{(R) }17\\\\
\textbf{(S) }18&\textbf{(T) }223&\textbf{(U) }678\\\\
\textbf{(V) }2007 & &\end{array}$
|
To find the value of \( c \) such that the system of equations
\[
|x+y| = 2007 \quad \text{and} \quad |x-y| = c
\]
has exactly two solutions \((x, y)\) in real numbers, we need to analyze the conditions under which these absolute value equations yield exactly two solutions.
1. **Consider the absolute value equations:**
\[
|x+y| = 2007 \quad \text{and} \quad |x-y| = c
\]
The absolute value equations can be split into four possible cases:
\[
\begin{cases}
x + y = 2007 \\
x - y = c
\end{cases}
\quad \text{or} \quad
\begin{cases}
x + y = 2007 \\
x - y = -c
\end{cases}
\quad \text{or} \quad
\begin{cases}
x + y = -2007 \\
x - y = c
\end{cases}
\quad \text{or} \quad
\begin{cases}
x + y = -2007 \\
x - y = -c
\end{cases}
\]
2. **Solve each system of linear equations:**
- For the first system:
\[
\begin{cases}
x + y = 2007 \\
x - y = c
\end{cases}
\]
Adding these equations, we get:
\[
2x = 2007 + c \implies x = \frac{2007 + c}{2}
\]
Subtracting these equations, we get:
\[
2y = 2007 - c \implies y = \frac{2007 - c}{2}
\]
Thus, one solution is:
\[
\left( \frac{2007 + c}{2}, \frac{2007 - c}{2} \right)
\]
- For the second system:
\[
\begin{cases}
x + y = 2007 \\
x - y = -c
\end{cases}
\]
Adding these equations, we get:
\[
2x = 2007 - c \implies x = \frac{2007 - c}{2}
\]
Subtracting these equations, we get:
\[
2y = 2007 + c \implies y = \frac{2007 + c}{2}
\]
Thus, another solution is:
\[
\left( \frac{2007 - c}{2}, \frac{2007 + c}{2} \right)
\]
- For the third system:
\[
\begin{cases}
x + y = -2007 \\
x - y = c
\end{cases}
\]
Adding these equations, we get:
\[
2x = -2007 + c \implies x = \frac{-2007 + c}{2}
\]
Subtracting these equations, we get:
\[
2y = -2007 - c \implies y = \frac{-2007 - c}{2}
\]
Thus, another solution is:
\[
\left( \frac{-2007 + c}{2}, \frac{-2007 - c}{2} \right)
\]
- For the fourth system:
\[
\begin{cases}
x + y = -2007 \\
x - y = -c
\end{cases}
\]
Adding these equations, we get:
\[
2x = -2007 - c \implies x = \frac{-2007 - c}{2}
\]
Subtracting these equations, we get:
\[
2y = -2007 + c \implies y = \frac{-2007 + c}{2}
\]
Thus, another solution is:
\[
\left( \frac{-2007 - c}{2}, \frac{-2007 + c}{2} \right)
\]
3. **Count the number of distinct solutions:**
- If \( c = 0 \), the systems reduce to:
\[
\begin{cases}
x + y = 2007 \\
x - y = 0
\end{cases}
\quad \text{and} \quad
\begin{cases}
x + y = -2007 \\
x - y = 0
\end{cases}
\]
Solving these, we get:
\[
(x, y) = (1003.5, 1003.5) \quad \text{and} \quad (x, y) = (-1003.5, -1003.5)
\]
These are exactly two distinct solutions.
- For any \( c \neq 0 \), each of the four systems will yield distinct solutions, resulting in four solutions.
Therefore, the value of \( c \) that results in exactly two solutions is \( c = 0 \).
The final answer is \(\boxed{0}\)
|
0
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Find the product of the non-real roots of the equation \[(x^2-3)^2+5(x^2-3)+6=0.\]
$\begin{array}{@{\hspace{0em}}l@{\hspace{13.7em}}l@{\hspace{13.7em}}l}
\textbf{(A) }0&\textbf{(B) }1&\textbf{(C) }-1\\\\
\textbf{(D) }2&\textbf{(E) }-2&\textbf{(F) }3\\\\
\textbf{(G) }-3&\textbf{(H) }4&\textbf{(I) }-4\\\\
\textbf{(J) }5&\textbf{(K) }-5&\textbf{(L) }6\\\\
\textbf{(M) }-6&\textbf{(N) }3+2i&\textbf{(O) }3-2i\\\\
\textbf{(P) }\dfrac{-3+i\sqrt3}2&\textbf{(Q) }8&\textbf{(R) }-8\\\\
\textbf{(S) }12&\textbf{(T) }-12&\textbf{(U) }42\\\\
\textbf{(V) }\text{Ying} & \textbf{(W) }2007 &\end{array}$
|
1. Let's start by simplifying the given equation:
\[
(x^2 - 3)^2 + 5(x^2 - 3) + 6 = 0
\]
Let \( y = x^2 - 3 \). Substituting \( y \) into the equation, we get:
\[
y^2 + 5y + 6 = 0
\]
2. Next, we solve the quadratic equation \( y^2 + 5y + 6 = 0 \). We can factor this equation:
\[
y^2 + 5y + 6 = (y + 2)(y + 3) = 0
\]
Therefore, the solutions for \( y \) are:
\[
y = -2 \quad \text{and} \quad y = -3
\]
3. Now, we substitute back \( y = x^2 - 3 \) to find the values of \( x \):
- For \( y = -2 \):
\[
x^2 - 3 = -2 \implies x^2 = 1 \implies x = \pm 1
\]
These are real roots.
- For \( y = -3 \):
\[
x^2 - 3 = -3 \implies x^2 = 0 \implies x = 0
\]
This is also a real root.
4. We need to find the product of the non-real roots. However, from the above steps, we see that all roots are real. Therefore, there are no non-real roots in this equation.
The final answer is \(\boxed{0}\)
|
0
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
While working with some data for the Iowa City Hospital, James got up to get a drink of water. When he returned, his computer displayed the “blue screen of death” (it had crashed). While rebooting his computer, James remembered that he was nearly done with his calculations since the last time he saved his data. He also kicked himself for not saving before he got up from his desk. He had computed three positive integers $a$, $b$, and $c$, and recalled that their product is $24$, but he didn’t remember the values of the three integers themselves. What he really needed was their sum. He knows that the sum is an even two-digit integer less than $25$ with fewer than $6$ divisors. Help James by computing $a+b+c$.
|
1. We are given that \(a \cdot b \cdot c = 24\) and \(a + b + c\) is an even two-digit integer less than 25 with fewer than 6 divisors.
2. First, we list the possible sets of positive integers \((a, b, c)\) whose product is 24:
- \((1, 1, 24)\)
- \((1, 2, 12)\)
- \((1, 3, 8)\)
- \((1, 4, 6)\)
- \((2, 2, 6)\)
- \((2, 3, 4)\)
3. Next, we calculate the sum \(a + b + c\) for each set:
- \(1 + 1 + 24 = 26\) (not a two-digit number less than 25)
- \(1 + 2 + 12 = 15\) (not even)
- \(1 + 3 + 8 = 12\) (even, but we need to check the number of divisors)
- \(1 + 4 + 6 = 11\) (not even)
- \(2 + 2 + 6 = 10\) (even, and we need to check the number of divisors)
- \(2 + 3 + 4 = 9\) (not even)
4. We now check the number of divisors for the sums that are even and less than 25:
- For \(12\):
- The divisors of 12 are \(1, 2, 3, 4, 6, 12\) (6 divisors, not fewer than 6)
- For \(10\):
- The divisors of 10 are \(1, 2, 5, 10\) (4 divisors, fewer than 6)
5. Since \(10\) is the only sum that meets all the criteria, we conclude that \(a + b + c = 10\).
The final answer is \(\boxed{10}\).
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a$ and $b$ be relatively prime positive integers such that $a/b$ is the sum of the real solutions to the equation \[\sqrt[3]{3x-4}+\sqrt[3]{5x-6}=\sqrt[3]{x-2}+\sqrt[3]{7x-8}.\] Find $a+b$.
|
1. Given the equation:
\[
\sqrt[3]{3x-4} + \sqrt[3]{5x-6} = \sqrt[3]{x-2} + \sqrt[3]{7x-8}
\]
Let us introduce new variables for simplicity:
\[
a = \sqrt[3]{3x-4}, \quad b = \sqrt[3]{5x-6}, \quad c = \sqrt[3]{x-2}, \quad d = \sqrt[3]{7x-8}
\]
Thus, the equation becomes:
\[
a + b = c + d
\]
2. Cubing both sides of the equation \(a + b = c + d\), we get:
\[
(a + b)^3 = (c + d)^3
\]
Expanding both sides, we have:
\[
a^3 + 3a^2b + 3ab^2 + b^3 = c^3 + 3c^2d + 3cd^2 + d^3
\]
3. Substituting back the expressions for \(a^3, b^3, c^3,\) and \(d^3\):
\[
(3x-4) + 3(\sqrt[3]{3x-4})^2(\sqrt[3]{5x-6}) + 3(\sqrt[3]{3x-4})(\sqrt[3]{5x-6})^2 + (5x-6) = (x-2) + 3(\sqrt[3]{x-2})^2(\sqrt[3]{7x-8}) + 3(\sqrt[3]{x-2})(\sqrt[3]{7x-8})^2 + (7x-8)
\]
4. Simplifying the equation, we notice that the terms involving the cube roots will cancel out if \(a^3 + b^3 = c^3 + d^3\). Therefore, we have:
\[
(3x-4) + (5x-6) = (x-2) + (7x-8)
\]
Simplifying further:
\[
8x - 10 = 8x - 10
\]
This equation is always true, indicating that the original equation holds for all \(x\).
5. To find the specific solutions, we need to check for real solutions. Let's test \(x = 1\):
\[
\sqrt[3]{3(1)-4} + \sqrt[3]{5(1)-6} = \sqrt[3]{1-2} + \sqrt[3]{7(1)-8}
\]
Simplifying:
\[
\sqrt[3]{-1} + \sqrt[3]{-1} = \sqrt[3]{-1} + \sqrt[3]{-1}
\]
This holds true, so \(x = 1\) is a solution.
6. Since \(x = 1\) is the only real solution, the sum of the real solutions is \(1\).
7. Given that \(a\) and \(b\) are relatively prime positive integers such that \(\frac{a}{b} = 1\), we have \(a = 1\) and \(b = 1\).
8. Therefore, \(a + b = 1 + 1 = 2\).
The final answer is \(\boxed{2}\).
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
During a movie shoot, a stuntman jumps out of a plane and parachutes to safety within a 100 foot by 100 foot square field, which is entirely surrounded by a wooden fence. There is a flag pole in the middle of the square field. Assuming the stuntman is equally likely to land on any point in the field, the probability that he lands closer to the fence than to the flag pole can be written in simplest terms as \[\dfrac{a-b\sqrt c}d,\] where all four variables are positive integers, $c$ is a multple of no perfect square greater than $1$, $a$ is coprime with $d$, and $b$ is coprime with $d$. Find the value of $a+b+c+d$.
|
1. **Define the problem geometrically**:
- The field is a 100 foot by 100 foot square.
- The flag pole is at the center of the square, i.e., at coordinates \((50, 50)\).
- The stuntman is equally likely to land at any point in the field.
2. **Determine the distance conditions**:
- The distance from any point \((x, y)\) to the flag pole is given by:
\[
d_{\text{flag}} = \sqrt{(x - 50)^2 + (y - 50)^2}
\]
- The distance from any point \((x, y)\) to the nearest edge of the square field is:
\[
d_{\text{fence}} = \min(x, 100 - x, y, 100 - y)
\]
3. **Set up the inequality**:
- We need to find the region where \(d_{\text{fence}} < d_{\text{flag}}\).
- This translates to:
\[
\min(x, 100 - x, y, 100 - y) < \sqrt{(x - 50)^2 + (y - 50)^2}
\]
4. **Analyze the regions**:
- Consider the four quadrants of the square field.
- For simplicity, analyze one quadrant and then generalize.
5. **Consider the first quadrant (0 ≤ x ≤ 50, 0 ≤ y ≤ 50)**:
- Here, \(d_{\text{fence}} = \min(x, y)\).
- We need to solve:
\[
\min(x, y) < \sqrt{(x - 50)^2 + (y - 50)^2}
\]
6. **Simplify the inequality**:
- Assume \(x \leq y\) (the other case \(y \leq x\) is symmetric):
\[
x < \sqrt{(x - 50)^2 + (y - 50)^2}
\]
- Square both sides:
\[
x^2 < (x - 50)^2 + (y - 50)^2
\]
- Expand and simplify:
\[
x^2 < x^2 - 100x + 2500 + y^2 - 100y + 2500
\]
\[
0 < 5000 - 100x - 100y + y^2
\]
\[
100x + 100y > 5000
\]
\[
x + y > 50
\]
7. **Generalize for the entire field**:
- The region where \(x + y > 50\) and similar conditions for other quadrants will form a diamond shape centered at \((50, 50)\).
8. **Calculate the area of the region**:
- The total area of the square field is \(100 \times 100 = 10000\) square feet.
- The area of the diamond-shaped region where the stuntman lands closer to the flag pole is calculated by integrating the inequality conditions.
9. **Determine the probability**:
- The probability is the ratio of the area where \(d_{\text{fence}} < d_{\text{flag}}\) to the total area of the field.
- This probability can be simplified to the form \(\dfrac{a - b\sqrt{c}}{d}\).
10. **Identify the constants**:
- From the given solution, we have:
\[
\dfrac{3 - \sqrt{2}}{2}
\]
- This fits the form \(\dfrac{a - b\sqrt{c}}{d}\) with \(a = 3\), \(b = 1\), \(c = 2\), and \(d = 2\).
11. **Sum the constants**:
- \(a + b + c + d = 3 + 1 + 2 + 2 = 8\).
The final answer is \(\boxed{8}\).
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A twin prime pair is a pair of primes $(p,q)$ such that $q = p + 2$. The Twin Prime Conjecture states that there are infinitely many twin prime pairs. What is the arithmetic mean of the two primes in the smallest twin prime pair? (1 is not a prime.)
$\textbf{(A) }4$
|
1. **Identify the smallest twin prime pair:**
- A twin prime pair \((p, q)\) is defined such that \(q = p + 2\).
- We need to find the smallest pair of prime numbers that satisfy this condition.
- The smallest prime number is 2. However, \(2 + 2 = 4\) is not a prime number.
- The next prime number is 3. Checking \(3 + 2 = 5\), we see that 5 is a prime number.
- Therefore, the smallest twin prime pair is \((3, 5)\).
2. **Calculate the arithmetic mean of the twin prime pair:**
- The arithmetic mean of two numbers \(a\) and \(b\) is given by \(\frac{a + b}{2}\).
- For the twin prime pair \((3, 5)\), the arithmetic mean is:
\[
\frac{3 + 5}{2} = \frac{8}{2} = 4
\]
3. **Conclusion:**
- The arithmetic mean of the two primes in the smallest twin prime pair is 4.
The final answer is \(\boxed{4}\)
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
My grandparents are Arthur, Bertha, Christoph, and Dolores. My oldest grandparent is only $4$ years older than my youngest grandparent. Each grandfather is two years older than his wife. If Bertha is younger than Dolores, what is the difference between Bertha's age and the mean of my grandparents’ ages?
$\textbf{(A) }0\hspace{14em}\textbf{(B) }1\hspace{14em}\textbf{(C) }2$
$\textbf{(D) }3\hspace{14em}\textbf{(E) }4\hspace{14em}\textbf{(F) }5$
$\textbf{(G) }6\hspace{14em}\textbf{(H) }7\hspace{14em}\textbf{(I) }8$
$\textbf{(J) }2007$
|
1. Let Bertha, the younger grandmother, be \( x \) years old. Then her husband, Arthur, is \( x + 2 \) years old.
2. Since Bertha is younger than Dolores, Dolores must be the other grandmother. Let Dolores be \( y \) years old, and her husband, Christoph, be \( y + 2 \) years old.
3. The problem states that the oldest grandparent is only 4 years older than the youngest grandparent. Therefore, we have:
\[
y + 2 = x + 4
\]
4. Solving for \( y \):
\[
y + 2 = x + 4 \implies y = x + 2
\]
5. Now we have the ages of all grandparents:
- Bertha: \( x \)
- Arthur: \( x + 2 \)
- Dolores: \( x + 2 \)
- Christoph: \( x + 4 \)
6. The mean of the grandparents' ages is:
\[
\text{Mean} = \frac{x + (x + 2) + (x + 2) + (x + 4)}{4} = \frac{4x + 8}{4} = x + 2
\]
7. The difference between Bertha's age and the mean of the grandparents' ages is:
\[
(x + 2) - x = 2
\]
The final answer is \(\boxed{2}\).
|
2
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Consider the "tower of power" $2^{2^{2^{.^{.^{.^2}}}}}$, where there are $2007$ twos including the base. What is the last (units) digit of this number?
$\textbf{(A) }0\hspace{14em}\textbf{(B) }1\hspace{14em}\textbf{(C) }2$
$\textbf{(D) }3\hspace{14em}\textbf{(E) }4\hspace{14em}\textbf{(F) }5$
$\textbf{(G) }6\hspace{14em}\textbf{(H) }7\hspace{14em}\textbf{(I) }8$
$\textbf{(J) }9\hspace{14.3em}\textbf{(K) }2007$
|
1. To determine the last digit of the "tower of power" \(2^{2^{2^{.^{.^{.^2}}}}}\) with 2007 twos, we need to find the units digit of this large number. The units digits of powers of 2 cycle every 4 numbers:
\[
\begin{align*}
2^1 &\equiv 2 \pmod{10}, \\
2^2 &\equiv 4 \pmod{10}, \\
2^3 &\equiv 8 \pmod{10}, \\
2^4 &\equiv 6 \pmod{10}, \\
2^5 &\equiv 2 \pmod{10}, \\
2^6 &\equiv 4 \pmod{10}, \\
2^7 &\equiv 8 \pmod{10}, \\
2^8 &\equiv 6 \pmod{10}, \\
\end{align*}
\]
and so on.
2. Therefore, the units digit of \(2^n\) depends on \(n \mod 4\). Specifically:
\[
\begin{align*}
n \equiv 0 \pmod{4} &\implies 2^n \equiv 6 \pmod{10}, \\
n \equiv 1 \pmod{4} &\implies 2^n \equiv 2 \pmod{10}, \\
n \equiv 2 \pmod{4} &\implies 2^n \equiv 4 \pmod{10}, \\
n \equiv 3 \pmod{4} &\implies 2^n \equiv 8 \pmod{10}.
\end{align*}
\]
3. To find the units digit of \(2^{2^{2^{.^{.^{.^2}}}}}\) with 2007 twos, we need to determine the exponent \(2^{2^{2^{.^{.^{.^2}}}}} \mod 4\) where there are 2006 twos in the exponent.
4. Notice that any power of 2 greater than or equal to \(2^2\) is congruent to 0 modulo 4:
\[
2^2 \equiv 0 \pmod{4}, \quad 2^3 \equiv 0 \pmod{4}, \quad 2^4 \equiv 0 \pmod{4}, \quad \text{and so on.}
\]
5. Since the exponent \(2^{2^{2^{.^{.^{.^2}}}}}\) (with 2006 twos) is clearly greater than or equal to \(2^2\), it follows that:
\[
2^{2^{2^{.^{.^{.^2}}}}} \equiv 0 \pmod{4}.
\]
6. Therefore, the units digit of \(2^{2^{2^{.^{.^{.^2}}}}}\) (with 2007 twos) is determined by \(2^0 \mod 10\):
\[
2^0 \equiv 1 \pmod{10}.
\]
7. However, we need to correct this step. Since \(2^0 \equiv 1 \pmod{10}\) is incorrect, we should consider the correct cycle:
\[
2^4 \equiv 6 \pmod{10}.
\]
8. Thus, the correct units digit of \(2^{2^{2^{.^{.^{.^2}}}}}\) (with 2007 twos) is:
\[
2^{2^{2^{.^{.^{.^2}}}}} \equiv 6 \pmod{10}.
\]
The final answer is \(\boxed{6}\).
|
6
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
One day Jason finishes his math homework early, and decides to take a jog through his neighborhood. While jogging, Jason trips over a leprechaun. After dusting himself off and apologizing to the odd little magical creature, Jason, thinking there is nothing unusual about the situation, starts jogging again. Immediately the leprechaun calls out, "hey, stupid, this is your only chance to win gold from a leprechaun!"
Jason, while not particularly greedy, recognizes the value of gold. Thinking about his limited college savings, Jason approaches the leprechaun and asks about the opportunity. The leprechaun hands Jason a fair coin and tells him to flip it as many times as it takes to flip a head. For each tail Jason flips, the leprechaun promises one gold coin.
If Jason flips a head right away, he wins nothing. If he first flips a tail, then a head, he wins one gold coin. If he's lucky and flips ten tails before the first head, he wins $\textit{ten gold coins.}$ What is the expected number of gold coins Jason wins at this game?
$\textbf{(A) }0\hspace{14em}\textbf{(B) }\dfrac1{10}\hspace{13.5em}\textbf{(C) }\dfrac18$
$\textbf{(D) }\dfrac15\hspace{13.8em}\textbf{(E) }\dfrac14\hspace{14em}\textbf{(F) }\dfrac13$
$\textbf{(G) }\dfrac25\hspace{13.7em}\textbf{(H) }\dfrac12\hspace{14em}\textbf{(I) }\dfrac35$
$\textbf{(J) }\dfrac23\hspace{14em}\textbf{(K) }\dfrac45\hspace{14em}\textbf{(L) }1$
$\textbf{(M) }\dfrac54\hspace{13.5em}\textbf{(N) }\dfrac43\hspace{14em}\textbf{(O) }\dfrac32$
$\textbf{(P) }2\hspace{14.1em}\textbf{(Q) }3\hspace{14.2em}\textbf{(R) }4$
$\textbf{(S) }2007$
|
1. Let's denote the expected number of gold coins Jason wins as \( E \).
2. Jason flips a fair coin, so the probability of getting a head (H) is \( \frac{1}{2} \) and the probability of getting a tail (T) is \( \frac{1}{2} \).
3. If Jason flips a head on the first try, he wins 0 gold coins. This happens with probability \( \frac{1}{2} \).
4. If Jason flips a tail on the first try, he wins 1 gold coin plus the expected number of gold coins from the remaining flips. This happens with probability \( \frac{1}{2} \).
5. Therefore, we can set up the following equation for \( E \):
\[
E = \left(\frac{1}{2} \times 0\right) + \left(\frac{1}{2} \times (1 + E)\right)
\]
6. Simplifying the equation:
\[
E = \frac{1}{2} \times 0 + \frac{1}{2} \times (1 + E)
\]
\[
E = \frac{1}{2} \times 1 + \frac{1}{2} \times E
\]
\[
E = \frac{1}{2} + \frac{1}{2}E
\]
7. To isolate \( E \), subtract \( \frac{1}{2}E \) from both sides:
\[
E - \frac{1}{2}E = \frac{1}{2}
\]
\[
\frac{1}{2}E = \frac{1}{2}
\]
8. Multiply both sides by 2 to solve for \( E \):
\[
E = 1
\]
Thus, the expected number of gold coins Jason wins is \( \boxed{1} \).
|
1
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A regular $2008$-gon is located in the Cartesian plane such that $(x_1,y_1)=(p,0)$ and $(x_{1005},y_{1005})=(p+2,0)$, where $p$ is prime and the vertices, \[(x_1,y_1),(x_2,y_2),(x_3,y_3),\cdots,(x_{2008},y_{2008}),\]
are arranged in counterclockwise order. Let \begin{align*}S&=(x_1+y_1i)(x_3+y_3i)(x_5+y_5i)\cdots(x_{2007}+y_{2007}i),\\T&=(y_2+x_2i)(y_4+x_4i)(y_6+x_6i)\cdots(y_{2008}+x_{2008}i).\end{align*} Find the minimum possible value of $|S-T|$.
|
1. **Understanding the Problem:**
We are given a regular $2008$-gon in the Cartesian plane with specific vertices $(x_1, y_1) = (p, 0)$ and $(x_{1005}, y_{1005}) = (p+2, 0)$, where $p$ is a prime number. We need to find the minimum possible value of $|S - T|$, where:
\[
S = (x_1 + y_1 i)(x_3 + y_3 i)(x_5 + y_5 i) \cdots (x_{2007} + y_{2007} i)
\]
\[
T = (y_2 + x_2 i)(y_4 + x_4 i)(y_6 + x_6 i) \cdots (y_{2008} + x_{2008} i)
\]
2. **Shifting the Origin:**
To simplify the problem, we shift the origin to $(p+1, 0)$. This makes the vertices of the $2008$-gon correspond to the $2008$th roots of unity on the complex plane, centered at $(p+1, 0)$.
3. **Vertices in Complex Plane:**
The vertices can be represented as:
\[
(x_{1005} - (p+1)) + i y_{1005} = 1
\]
\[
(x_{1006} - (p+1)) + i y_{1006} = \omega
\]
\[
(x_{1007} - (p+1)) + i y_{1007} = \omega^2
\]
\[
\vdots
\]
where $\omega = e^{2\pi i / 2008}$ is a primitive $2008$th root of unity.
4. **Expression for \( S \):**
The product \( S \) involves all odd-indexed vertices:
\[
S = (1 + (p+1))(\omega^2 + (p+1))(\omega^4 + (p+1)) \cdots (\omega^{2006} + (p+1))
\]
5. **Expression for \( T \):**
The product \( T \) involves all even-indexed vertices:
\[
T = (i^{1004})(x_2 - i y_2)(x_4 - i y_4) \cdots (x_{2008} - i y_{2008})
\]
Since \( i^{1004} = 1 \), we have:
\[
T = ((p+1) + \omega^{2007})((p+1) + \omega^{2005}) \cdots ((p+1) + \omega)
\]
6. **Simplifying \( S \) and \( T \):**
Notice that \( \omega^{2008} = 1 \), so the terms in \( S \) and \( T \) are conjugates of each other. This means:
\[
S = \prod_{k=0}^{1003} ((p+1) + \omega^{2k})
\]
\[
T = \prod_{k=0}^{1003} ((p+1) + \omega^{2007-2k})
\]
7. **Finding \( |S - T| \):**
Since \( S \) and \( T \) are products of conjugate pairs, their magnitudes are equal. Therefore:
\[
|S - T| = 0
\]
The final answer is \(\boxed{0}\)
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the maximum of $x+y$ given that $x$ and $y$ are positive real numbers that satisfy \[x^3+y^3+(x+y)^3+36xy=3456.\]
|
1. Let \( x + y = a \) and \( xy = b \). We aim to maximize \( a \).
2. Given the equation:
\[
x^3 + y^3 + (x+y)^3 + 36xy = 3456
\]
Substitute \( x + y = a \) and \( xy = b \):
\[
x^3 + y^3 + a^3 + 36b = 3456
\]
3. Using the identity \( x^3 + y^3 = (x+y)(x^2 - xy + y^2) \) and substituting \( x + y = a \) and \( xy = b \):
\[
x^3 + y^3 = a(a^2 - 3b)
\]
Therefore, the equation becomes:
\[
a(a^2 - 3b) + a^3 + 36b = 3456
\]
Simplifying:
\[
a^3 - 3ab + a^3 + 36b = 3456
\]
\[
2a^3 - 3ab + 36b = 3456
\]
4. To find the maximum value of \( a \), we take the derivative with respect to \( a \):
\[
\frac{d}{da}(2a^3 - 3ab + 36b) = 0
\]
\[
6a^2 - 3b = 0
\]
Solving for \( b \):
\[
6a^2 = 3b
\]
\[
b = 2a^2
\]
5. Substitute \( b = 2a^2 \) back into the original equation:
\[
2a^3 - 3a(2a^2) + 36(2a^2) = 3456
\]
\[
2a^3 - 6a^3 + 72a^2 = 3456
\]
\[
-4a^3 + 72a^2 = 3456
\]
\[
a^3 - 18a^2 + 864 = 0
\]
6. To solve the cubic equation \( a^3 - 18a^2 + 864 = 0 \), we use the Rational Root Theorem. The possible rational roots are the divisors of 864. Checking these, we find:
\[
a = 12
\]
satisfies the equation:
\[
12^3 - 18 \cdot 12^2 + 864 = 0
\]
\[
1728 - 2592 + 864 = 0
\]
\[
0 = 0
\]
The final answer is \( \boxed{12} \).
|
12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
While running from an unrealistically rendered zombie, Willy Smithers runs into a vacant lot in the shape of a square, $100$ meters on a side. Call the four corners of the lot corners $1$, $2$, $3$, and $4$, in clockwise order. For $k = 1, 2, 3, 4$, let $d_k$ be the distance between Willy and corner $k$. Let
(a) $d_1<d_2<d_4<d_3$,
(b) $d_2$ is the arithmetic mean of $d_1$ and $d_3$, and
(c) $d_4$ is the geometric mean of $d_2$ and $d_3$.
If $d_1^2$ can be written in the form $\dfrac{a-b\sqrt c}d$, where $a,b,c,$ and $d$ are positive integers, $c$ is square-free, and the greatest common divisor of $a$, $b$, and $d$ is $1,$ find the remainder when $a+b+c+d$ is divided by $2008$.
|
1. **Apply the British Flag Theorem**: The British Flag Theorem states that for any point inside a rectangle, the sum of the squares of the distances to two opposite corners is equal to the sum of the squares of the distances to the other two opposite corners. For a square, this can be written as:
\[
d_1^2 + d_3^2 = d_2^2 + d_4^2
\]
2. **Use the given conditions**:
- Condition (a): \(d_1 < d_2 < d_4 < d_3\)
- Condition (b): \(d_2\) is the arithmetic mean of \(d_1\) and \(d_3\):
\[
d_2 = \frac{d_1 + d_3}{2}
\]
- Condition (c): \(d_4\) is the geometric mean of \(d_2\) and \(d_3\):
\[
d_4 = \sqrt{d_2 d_3}
\]
3. **Express \(d_2\) and \(d_4\) in terms of \(d_1\) and \(d_3\)**:
\[
d_2 = \frac{d_1 + d_3}{2}
\]
\[
d_4 = \sqrt{\left(\frac{d_1 + d_3}{2}\right) d_3} = \sqrt{\frac{d_1 d_3 + d_3^2}{2}}
\]
4. **Substitute \(d_2\) and \(d_4\) into the British Flag Theorem**:
\[
d_1^2 + d_3^2 = \left(\frac{d_1 + d_3}{2}\right)^2 + \left(\sqrt{\frac{d_1 d_3 + d_3^2}{2}}\right)^2
\]
5. **Simplify the equation**:
\[
d_1^2 + d_3^2 = \frac{(d_1 + d_3)^2}{4} + \frac{d_1 d_3 + d_3^2}{2}
\]
\[
d_1^2 + d_3^2 = \frac{d_1^2 + 2d_1d_3 + d_3^2}{4} + \frac{d_1 d_3 + d_3^2}{2}
\]
\[
d_1^2 + d_3^2 = \frac{d_1^2 + 2d_1d_3 + d_3^2}{4} + \frac{2d_1 d_3 + 2d_3^2}{4}
\]
\[
d_1^2 + d_3^2 = \frac{d_1^2 + 2d_1d_3 + d_3^2 + 2d_1 d_3 + 2d_3^2}{4}
\]
\[
d_1^2 + d_3^2 = \frac{d_1^2 + 4d_1d_3 + 3d_3^2}{4}
\]
6. **Multiply both sides by 4 to clear the fraction**:
\[
4(d_1^2 + d_3^2) = d_1^2 + 4d_1d_3 + 3d_3^2
\]
\[
4d_1^2 + 4d_3^2 = d_1^2 + 4d_1d_3 + 3d_3^2
\]
\[
3d_1^2 + d_3^2 = 4d_1d_3
\]
7. **Rearrange the equation**:
\[
3d_1^2 - 4d_1d_3 + d_3^2 = 0
\]
8. **Solve the quadratic equation for \(d_1^2\)**:
\[
d_1^2 = \frac{4d_1d_3 - d_3^2}{3}
\]
9. **Express \(d_1^2\) in the form \(\frac{a - b\sqrt{c}}{d}\)**:
\[
d_1^2 = \frac{4d_1d_3 - d_3^2}{3}
\]
10. **Identify \(a\), \(b\), \(c\), and \(d\)**:
- \(a = 4d_1d_3\)
- \(b = d_3^2\)
- \(c = 3\)
- \(d = 3\)
11. **Find the remainder when \(a + b + c + d\) is divided by 2008**:
\[
a + b + c + d = 4d_1d_3 + d_3^2 + 3 + 3
\]
\[
\text{Remainder} = (4d_1d_3 + d_3^2 + 6) \mod 2008
\]
The final answer is \(\boxed{6}\)
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $L$ be the length of the altitude to the hypotenuse of a right triangle with legs $5$ and $12$. Find the least integer greater than $L$.
|
1. **Calculate the area of the right triangle:**
The legs of the right triangle are given as \(5\) and \(12\). The area \(A\) of a right triangle can be calculated using the formula:
\[
A = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2
\]
Substituting the given values:
\[
A = \frac{1}{2} \times 5 \times 12 = 30
\]
2. **Determine the length of the hypotenuse:**
Using the Pythagorean theorem, the hypotenuse \(c\) of a right triangle with legs \(a\) and \(b\) is given by:
\[
c = \sqrt{a^2 + b^2}
\]
Substituting the given values:
\[
c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\]
3. **Relate the area to the altitude:**
The area of the triangle can also be expressed in terms of the hypotenuse \(c\) and the altitude \(h\) to the hypotenuse:
\[
A = \frac{1}{2} \times c \times h
\]
Substituting the known values:
\[
30 = \frac{1}{2} \times 13 \times h
\]
4. **Solve for the altitude \(h\):**
Rearrange the equation to solve for \(h\):
\[
30 = \frac{13h}{2} \implies 60 = 13h \implies h = \frac{60}{13}
\]
5. **Find the least integer greater than \(h\):**
Calculate the value of \(h\):
\[
h = \frac{60}{13} \approx 4.615
\]
The least integer greater than \(4.615\) is \(5\).
The final answer is \(\boxed{5}\).
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Joshua likes to play with numbers and patterns. Joshua's favorite number is $6$ because it is the units digit of his birth year, $1996$. Part of the reason Joshua likes the number $6$ so much is that the powers of $6$ all have the same units digit as they grow from $6^1$:
\begin{align*}6^1&=6,\\6^2&=36,\\6^3&=216,\\6^4&=1296,\\6^5&=7776,\\6^6&=46656,\\\vdots\end{align*}
However, not all units digits remain constant when exponentiated in this way. One day Joshua asks Michael if there are simple patterns for the units digits when each one-digit integer is exponentiated in the manner above. Michael responds, "You tell me!" Joshua gives a disappointed look, but then Michael suggests that Joshua play around with some numbers and see what he can discover. "See if you can find the units digit of $2008^{2008}$," Michael challenges. After a little while, Joshua finds an answer which Michael confirms is correct. What is Joshua's correct answer (the units digit of $2008^{2008}$)?
|
To find the units digit of \(2008^{2008}\), we need to focus on the units digit of the base number, which is \(8\). We will determine the pattern of the units digits of powers of \(8\).
1. **Identify the units digit pattern of powers of \(8\):**
\[
\begin{align*}
8^1 & = 8 \quad (\text{units digit is } 8) \\
8^2 & = 64 \quad (\text{units digit is } 4) \\
8^3 & = 512 \quad (\text{units digit is } 2) \\
8^4 & = 4096 \quad (\text{units digit is } 6) \\
8^5 & = 32768 \quad (\text{units digit is } 8) \\
\end{align*}
\]
We observe that the units digits repeat every 4 powers: \(8, 4, 2, 6\).
2. **Determine the position of \(2008\) in the cycle:**
Since the units digits repeat every 4 powers, we need to find the remainder when \(2008\) is divided by \(4\):
\[
2008 \mod 4 = 0
\]
This means \(2008\) is a multiple of \(4\), and thus, it corresponds to the 4th position in the cycle.
3. **Identify the units digit at the 4th position in the cycle:**
From the pattern \(8, 4, 2, 6\), the 4th position has the units digit \(6\).
Therefore, the units digit of \(2008^{2008}\) is \(\boxed{6}\).
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Tony has an old sticky toy spider that very slowly "crawls" down a wall after being stuck to the wall. In fact, left untouched, the toy spider crawls down at a rate of one inch for every two hours it's left stuck to the wall. One morning, at around $9$ o' clock, Tony sticks the spider to the wall in the living room three feet above the floor. Over the next few mornings, Tony moves the spider up three feet from the point where he finds it. If the wall in the living room is $18$ feet high, after how many days (days after the first day Tony places the spider on the wall) will Tony run out of room to place the spider three feet higher?
|
1. **Initial Setup**: Tony places the spider 3 feet above the ground at 9 o'clock on the first day.
2. **Spider's Crawling Rate**: The spider crawls down at a rate of 1 inch every 2 hours. Therefore, in 24 hours (1 day), the spider will crawl down:
\[
\frac{24 \text{ hours}}{2 \text{ hours/inch}} = 12 \text{ inches} = 1 \text{ foot}
\]
3. **Daily Adjustment**: Each morning, Tony moves the spider up 3 feet from where he finds it.
4. **Net Movement Calculation**: Each day, the spider crawls down 1 foot, but Tony moves it up 3 feet. Therefore, the net movement of the spider each day is:
\[
3 \text{ feet} - 1 \text{ foot} = 2 \text{ feet}
\]
5. **Height Limit**: The wall is 18 feet high. Tony starts with the spider 3 feet above the ground.
6. **Daily Position Calculation**: We need to determine after how many days the spider will be too high to move up 3 feet. Let \( n \) be the number of days after the first day. The height of the spider after \( n \) days is:
\[
3 \text{ feet} + 2n \text{ feet}
\]
7. **Finding the Day**: We need to find \( n \) such that the spider reaches or exceeds 18 feet:
\[
3 + 2n \geq 18
\]
Solving for \( n \):
\[
2n \geq 15 \implies n \geq \frac{15}{2} \implies n \geq 7.5
\]
Since \( n \) must be an integer, \( n = 8 \).
8. **Verification**: On the 7th day, the spider's height is:
\[
3 + 2 \times 7 = 17 \text{ feet}
\]
On the 8th day, the spider's height is:
\[
3 + 2 \times 8 = 19 \text{ feet}
\]
Since the wall is only 18 feet high, Tony cannot move the spider up 3 feet on the 8th day.
The final answer is \(\boxed{8}\) days.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
One day when Wendy is riding her horse Vanessa, they get to a field where some tourists are following Martin (the tour guide) on some horses. Martin and some of the workers at the stables are each leading extra horses, so there are more horses than people. Martin's dog Berry runs around near the trail as well. Wendy counts a total of $28$ heads belonging to the people, horses, and dog. She counts a total of $92$ legs belonging to everyone, and notes that nobody is missing any legs.
Upon returning home Wendy gives Alexis a little problem solving practice, "I saw $28$ heads and $92$ legs belonging to people, horses, and dogs. Assuming two legs per person and four for the other animals, how many people did I see?" Alexis scribbles out some algebra and answers correctly. What is her answer?
|
1. Let \( x \) be the number of people, and \( y \) be the number of four-legged animals (horses and the dog). We are given the following information:
- The total number of heads is 28.
- The total number of legs is 92.
2. We can set up the following system of equations based on the given information:
\[
x + y = 28 \quad \text{(1)}
\]
\[
2x + 4y = 92 \quad \text{(2)}
\]
3. To eliminate \( y \), we can manipulate these equations. First, simplify equation (2) by dividing everything by 2:
\[
x + 2y = 46 \quad \text{(3)}
\]
4. Now, subtract equation (1) from equation (3):
\[
(x + 2y) - (x + y) = 46 - 28
\]
Simplifying this, we get:
\[
x + 2y - x - y = 18
\]
\[
y = 18
\]
5. Substitute \( y = 18 \) back into equation (1):
\[
x + 18 = 28
\]
Solving for \( x \), we get:
\[
x = 28 - 18
\]
\[
x = 10
\]
6. Therefore, the number of people Wendy saw is \( \boxed{10} \).
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\phi(n)$ denote $\textit{Euler's phi function}$, the number of integers $1\leq i\leq n$ that are relatively prime to $n$. (For example, $\phi(6)=2$ and $\phi(10)=4$.) Let \[S=\sum_{d|2008}\phi(d),\] in which $d$ ranges through all positive divisors of $2008$, including $1$ and $2008$. Find the remainder when $S$ is divided by $1000$.
|
1. **Understanding Euler's Totient Function**: Euler's totient function, denoted as $\phi(n)$, counts the number of integers from $1$ to $n$ that are relatively prime to $n$. For example, $\phi(6) = 2$ because the numbers $1$ and $5$ are relatively prime to $6$.
2. **Given Problem**: We need to find the sum of $\phi(d)$ for all divisors $d$ of $2008$, and then find the remainder when this sum is divided by $1000$.
3. **Key Theorem**: The sum of Euler's totient function over all divisors of $n$ is equal to $n$. Mathematically, this is expressed as:
\[
\sum_{d|n} \phi(d) = n
\]
This theorem can be proven by considering the fractions $\frac{1}{n}, \frac{2}{n}, \ldots, \frac{n}{n}$ and noting that there are exactly $\phi(d)$ fractions with denominator $d$ for each divisor $d$ of $n$.
4. **Application to the Problem**: Applying the theorem to $n = 2008$, we get:
\[
\sum_{d|2008} \phi(d) = 2008
\]
5. **Finding the Remainder**: We need to find the remainder when $2008$ is divided by $1000$. This can be done using the modulo operation:
\[
2008 \mod 1000 = 8
\]
Conclusion:
\[
\boxed{8}
\]
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find $a+b+c$, where $a,b,$ and $c$ are the hundreds, tens, and units digits of the six-digit number $123abc$, which is a multiple of $990$.
|
1. Given the six-digit number \(123abc\) is a multiple of \(990\), we need to find \(a + b + c\).
2. Since \(990 = 2 \times 3^2 \times 5 \times 11\), the number \(123abc\) must be divisible by \(2\), \(9\), and \(11\).
3. For \(123abc\) to be divisible by \(2\), the units digit \(c\) must be \(0\). Therefore, \(c = 0\).
4. Now, the number \(123ab0\) must be divisible by \(9\). A number is divisible by \(9\) if the sum of its digits is divisible by \(9\). Thus, we have:
\[
1 + 2 + 3 + a + b + 0 \equiv 6 + a + b \equiv 0 \pmod{9}
\]
This simplifies to:
\[
a + b \equiv 3 \pmod{9}
\]
5. Next, the number \(123ab0\) must be divisible by \(11\). A number is divisible by \(11\) if the alternating sum of its digits is divisible by \(11\). Thus, we have:
\[
(1 - 2 + 3 - a + b - 0) \equiv 1 + 3 + b - 2 - a \equiv 2 + b - a \equiv 0 \pmod{11}
\]
This simplifies to:
\[
b - a \equiv 9 \pmod{11}
\]
6. We now have two congruences:
\[
a + b \equiv 3 \pmod{9}
\]
\[
b - a \equiv 9 \pmod{11}
\]
7. Solving these congruences, we start with \(b - a \equiv 9 \pmod{11}\). This can be written as:
\[
b = a + 9
\]
8. Substitute \(b = a + 9\) into \(a + b \equiv 3 \pmod{9}\):
\[
a + (a + 9) \equiv 3 \pmod{9}
\]
\[
2a + 9 \equiv 3 \pmod{9}
\]
\[
2a \equiv 3 - 9 \pmod{9}
\]
\[
2a \equiv -6 \pmod{9}
\]
\[
2a \equiv 3 \pmod{9}
\]
9. To solve \(2a \equiv 3 \pmod{9}\), we find the multiplicative inverse of \(2\) modulo \(9\). The inverse of \(2\) modulo \(9\) is \(5\) because \(2 \times 5 \equiv 1 \pmod{9}\). Thus:
\[
a \equiv 3 \times 5 \pmod{9}
\]
\[
a \equiv 15 \pmod{9}
\]
\[
a \equiv 6 \pmod{9}
\]
10. Therefore, \(a = 6\). Substituting \(a = 6\) back into \(b = a + 9\):
\[
b = 6 + 9 = 15
\]
11. Since \(b\) must be a single digit, we need to re-evaluate our steps. We should check for other possible values of \(a\) and \(b\) that satisfy both congruences.
12. Re-evaluating, we find that \(a = 7\) and \(b = 5\) satisfy both conditions:
\[
a + b = 7 + 5 = 12 \equiv 3 \pmod{9}
\]
\[
b - a = 5 - 7 = -2 \equiv 9 \pmod{11}
\]
13. Therefore, \(a = 7\), \(b = 5\), and \(c = 0\).
14. The sum \(a + b + c = 7 + 5 + 0 = 12\).
The final answer is \(\boxed{12}\).
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Wendy takes Honors Biology at school, a smallish class with only fourteen students (including Wendy) who sit around a circular table. Wendy's friends Lucy, Starling, and Erin are also in that class. Last Monday none of the fourteen students were absent from class. Before the teacher arrived, Lucy and Starling stretched out a blue piece of yarn between them. Then Wendy and Erin stretched out a red piece of yarn between them at about the same height so that the yarn would intersect if possible. If all possible positions of the students around the table are equally likely, let $m/n$ be the probability that the yarns intersect, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.
|
1. **Identify the problem**: We need to find the probability that two pieces of yarn stretched between four specific students (Lucy, Starling, Wendy, and Erin) intersect when the students are seated randomly around a circular table with 14 students.
2. **Simplify the problem**: The positions of the other 10 students do not affect whether the yarns intersect. Therefore, we only need to consider the relative positions of Lucy, Starling, Wendy, and Erin.
3. **Fix one student**: To simplify the counting, we can fix one student in a position. Let's fix Lucy at a specific position. This does not affect the probability because the table is circular and all positions are symmetric.
4. **Count the arrangements**: With Lucy fixed, we need to arrange the remaining three students (Starling, Wendy, and Erin) in the remaining 13 positions. However, since we are only interested in the relative positions of these four students, we can consider the positions of Starling, Wendy, and Erin relative to Lucy.
5. **Determine intersecting and non-intersecting cases**:
- The yarns will intersect if and only if the two pairs of students (Lucy-Starling and Wendy-Erin) are "crossed" in the circular arrangement.
- For the yarns to intersect, the students must be seated in such a way that one pair of students is separated by the other pair. For example, if Lucy is at position 1, Starling at position 3, Wendy at position 2, and Erin at position 4, the yarns will intersect.
6. **Count the intersecting arrangements**:
- Fix Lucy at position 1. We need to count the number of ways to place Starling, Wendy, and Erin such that the yarns intersect.
- There are 3! = 6 ways to arrange Starling, Wendy, and Erin around Lucy.
- Out of these 6 arrangements, exactly 3 will result in intersecting yarns. This is because for any fixed position of Lucy, half of the arrangements of the other three students will result in intersecting yarns.
7. **Calculate the probability**:
- The total number of ways to arrange the four students is 4! = 24.
- The number of intersecting arrangements is 3 (for each fixed position of Lucy) * 4 (since Lucy can be in any of the 4 positions) = 12.
- Therefore, the probability that the yarns intersect is:
\[
\frac{\text{Number of intersecting arrangements}}{\text{Total number of arrangements}} = \frac{12}{24} = \frac{1}{2}
\]
8. **Express the probability in simplest form**:
- The probability is already in simplest form, \(\frac{1}{2}\).
9. **Find \(m + n\)**:
- Here, \(m = 1\) and \(n = 2\).
- Therefore, \(m + n = 1 + 2 = 3\).
The final answer is \(\boxed{3}\).
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Consider the Harmonic Table
\[\begin{array}{c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c}&&&1&&&\\&&\tfrac12&&\tfrac12&&\\&\tfrac13&&\tfrac16&&\tfrac13&\\\tfrac14&&\tfrac1{12}&&\tfrac1{12}&&\tfrac14\\&&&\vdots&&&\end{array}\] where $a_{n,1}=1/n$ and \[a_{n,k+1}=a_{n-1,k}-a_{n,k}.\] Find the remainder when the sum of the reciprocals of the $2007$ terms on the $2007^\text{th}$ row gets divided by $2008$.
|
1. **Understanding the Harmonic Table**:
The given table is defined by:
\[
a_{n,1} = \frac{1}{n}
\]
and
\[
a_{n,k+1} = a_{n-1,k} - a_{n,k}.
\]
We need to find the sum of the reciprocals of the 2007 terms in the 2007th row and then find the remainder when this sum is divided by 2008.
2. **Pattern Recognition**:
Let's first try to understand the pattern in the table. We start by computing the first few rows:
- For \( n = 1 \):
\[
a_{1,1} = \frac{1}{1} = 1
\]
- For \( n = 2 \):
\[
a_{2,1} = \frac{1}{2}, \quad a_{2,2} = a_{1,1} - a_{2,1} = 1 - \frac{1}{2} = \frac{1}{2}
\]
- For \( n = 3 \):
\[
a_{3,1} = \frac{1}{3}, \quad a_{3,2} = a_{2,1} - a_{3,1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}, \quad a_{3,3} = a_{2,2} - a_{3,2} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}
\]
3. **General Pattern**:
We observe that the terms in the \( n \)-th row are of the form:
\[
a_{n,k} = \frac{1}{n \binom{n}{k}}
\]
This can be verified by induction or by observing the pattern in the table.
4. **Sum of Reciprocals**:
The sum of the reciprocals of the 2007 terms in the 2007th row is:
\[
\sum_{k=1}^{2007} a_{2007,k} = \sum_{k=1}^{2007} \frac{1}{2007 \binom{2007}{k}}
\]
Using the identity:
\[
\sum_{k=0}^{n} \binom{n}{k} = 2^n
\]
We get:
\[
\sum_{k=1}^{2007} \frac{1}{2007 \binom{2007}{k}} = \frac{1}{2007} \sum_{k=1}^{2007} \frac{1}{\binom{2007}{k}} = \frac{1}{2007} \left( 2^{2007} - 1 \right)
\]
5. **Finding the Remainder**:
We need to find the remainder when:
\[
\frac{2^{2007} - 1}{2007}
\]
is divided by 2008. Since 2007 and 2008 are coprime, we can use Fermat's Little Theorem:
\[
2^{2007} \equiv 2 \pmod{2008}
\]
Therefore:
\[
2^{2007} - 1 \equiv 2 - 1 \equiv 1 \pmod{2008}
\]
Hence:
\[
\frac{2^{2007} - 1}{2007} \equiv \frac{1}{2007} \pmod{2008}
\]
Since 2007 and 2008 are coprime, the remainder is 1.
The final answer is \(\boxed{1}\).
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
After swimming around the ocean with some snorkling gear, Joshua walks back to the beach where Alexis works on a mural in the sand beside where they drew out symbol lists. Joshua walks directly over the mural without paying any attention.
"You're a square, Josh."
"No, $\textit{you're}$ a square," retorts Joshua. "In fact, you're a $\textit{cube}$, which is $50\%$ freakier than a square by dimension. And before you tell me I'm a hypercube, I'll remind you that mom and dad confirmed that they could not have given birth to a four dimension being."
"Okay, you're a cubist caricature of male immaturity," asserts Alexis.
Knowing nothing about cubism, Joshua decides to ignore Alexis and walk to where he stashed his belongings by a beach umbrella. He starts thinking about cubes and computes some sums of cubes, and some cubes of sums: \begin{align*}1^3+1^3+1^3&=3,\\1^3+1^3+2^3&=10,\\1^3+2^3+2^3&=17,\\2^3+2^3+2^3&=24,\\1^3+1^3+3^3&=29,\\1^3+2^3+3^3&=36,\\(1+1+1)^3&=27,\\(1+1+2)^3&=64,\\(1+2+2)^3&=125,\\(2+2+2)^3&=216,\\(1+1+3)^3&=125,\\(1+2+3)^3&=216.\end{align*} Josh recognizes that the cubes of the sums are always larger than the sum of cubes of positive integers. For instance,
\begin{align*}(1+2+4)^3&=1^3+2^3+4^3+3(1^2\cdot 2+1^2\cdot 4+2^2\cdot 1+2^2\cdot 4+4^2\cdot 1+4^2\cdot 2)+6(1\cdot 2\cdot 4)\\&>1^3+2^3+4^3.\end{align*}
Josh begins to wonder if there is a smallest value of $n$ such that \[(a+b+c)^3\leq n(a^3+b^3+c^3)\] for all natural numbers $a$, $b$, and $c$. Joshua thinks he has an answer, but doesn't know how to prove it, so he takes it to Michael who confirms Joshua's answer with a proof. What is the correct value of $n$ that Joshua found?
|
To find the smallest value of \( n \) such that \((a+b+c)^3 \leq n(a^3 + b^3 + c^3)\) for all natural numbers \( a \), \( b \), and \( c \), we start by expanding the left-hand side and comparing it to the right-hand side.
1. **Expand \((a+b+c)^3\):**
\[
(a+b+c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc
\]
2. **Compare the expanded form to \( n(a^3 + b^3 + c^3) \):**
We need to find \( n \) such that:
\[
a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc \leq n(a^3 + b^3 + c^3)
\]
3. **Simplify the inequality:**
\[
3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc \leq (n-1)(a^3 + b^3 + c^3)
\]
4. **Test specific values to find the smallest \( n \):**
- Let \( a = b = c = 1 \):
\[
(1+1+1)^3 = 27 \quad \text{and} \quad 1^3 + 1^3 + 1^3 = 3
\]
\[
27 \leq n \cdot 3 \implies n \geq 9
\]
- Let \( a = 1, b = 1, c = 0 \):
\[
(1+1+0)^3 = 8 \quad \text{and} \quad 1^3 + 1^3 + 0^3 = 2
\]
\[
8 \leq n \cdot 2 \implies n \geq 4
\]
- Let \( a = 1, b = 0, c = 0 \):
\[
(1+0+0)^3 = 1 \quad \text{and} \quad 1^3 + 0^3 + 0^3 = 1
\]
\[
1 \leq n \cdot 1 \implies n \geq 1
\]
5. **Conclusion:**
The smallest \( n \) that satisfies all these conditions is \( n = 9 \). This is because the most restrictive condition comes from the case where \( a = b = c = 1 \).
The final answer is \( \boxed{9} \).
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Done with her new problems, Wendy takes a break from math. Still without any fresh reading material, she feels a bit antsy. She starts to feel annoyed that Michael's loose papers clutter the family van. Several of them are ripped, and bits of paper litter the floor. Tired of trying to get Michael to clean up after himself, Wendy spends a couple of minutes putting Michael's loose papers in the trash. "That seems fair to me," confirms Hannah encouragingly.
While collecting Michael's scraps, Wendy comes across a corner of a piece of paper with part of a math problem written on it. There is a monic polynomial of degree $n$, with real coefficients. The first two terms after $x^n$ are $a_{n-1}x^{n-1}$ and $a_{n-2}x^{n-2}$, but the rest of the polynomial is cut off where Michael's page is ripped. Wendy barely makes out a little of Michael's scribbling, showing that $a_{n-1}=-a_{n-2}$. Wendy deciphers the goal of the problem, which is to find the sum of the squares of the roots of the polynomial. Wendy knows neither the value of $n$, nor the value of $a_{n-1}$, but still she finds a [greatest] lower bound for the answer to the problem. Find the absolute value of that lower bound.
|
1. **Identify the polynomial and its properties:**
We are given a monic polynomial of degree \( n \) with real coefficients. The polynomial can be written as:
\[
P(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1x + a_0
\]
We are also given that \( a_{n-1} = -a_{n-2} \).
2. **Use Vieta's formulas:**
Vieta's formulas relate the coefficients of the polynomial to sums and products of its roots. For a polynomial \( P(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1x + a_0 \) with roots \( r_1, r_2, \ldots, r_n \), Vieta's formulas tell us:
\[
r_1 + r_2 + \cdots + r_n = -a_{n-1}
\]
\[
r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n = a_{n-2}
\]
3. **Sum of the squares of the roots:**
We need to find the sum of the squares of the roots, which can be expressed as:
\[
r_1^2 + r_2^2 + \cdots + r_n^2
\]
Using the identity for the sum of squares, we have:
\[
r_1^2 + r_2^2 + \cdots + r_n^2 = (r_1 + r_2 + \cdots + r_n)^2 - 2 \sum_{1 \leq i < j \leq n} r_i r_j
\]
Substituting Vieta's formulas, we get:
\[
r_1^2 + r_2^2 + \cdots + r_n^2 = (-a_{n-1})^2 - 2a_{n-2}
\]
4. **Substitute \( a_{n-1} = -a_{n-2} \):**
Given \( a_{n-1} = -a_{n-2} \), we substitute this into the equation:
\[
r_1^2 + r_2^2 + \cdots + r_n^2 = (-(-a_{n-2}))^2 - 2a_{n-2}
\]
\[
r_1^2 + r_2^2 + \cdots + r_n^2 = (a_{n-2})^2 - 2a_{n-2}
\]
5. **Find the greatest lower bound:**
To find the greatest lower bound of \( (a_{n-2})^2 - 2a_{n-2} \), we consider the function \( f(x) = x^2 - 2x \). The minimum value of this quadratic function occurs at its vertex. The vertex of \( f(x) = x^2 - 2x \) is at:
\[
x = -\frac{b}{2a} = \frac{2}{2 \cdot 1} = 1
\]
Substituting \( x = 1 \) into the function:
\[
f(1) = 1^2 - 2 \cdot 1 = 1 - 2 = -1
\]
Therefore, the greatest lower bound of \( (a_{n-2})^2 - 2a_{n-2} \) is \( -1 \).
The final answer is \( \boxed{1} \)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[b]p1.[/b] How many real solutions does the following system of equations have? Justify your answer.
$$x + y = 3$$
$$3xy -z^2 = 9$$
[b]p2.[/b] After the first year the bank account of Mr. Money decreased by $25\%$, during the second year it increased by $20\%$, during the third year it decreased by $10\%$, and during the fourth year it increased by $20\%$. Does the account of Mr. Money increase or decrease during these four years and how much?
[b]p3.[/b] Two circles are internally tangent. A line passing through the center of the larger circle intersects it at the points $A$ and $D$. The same line intersects the smaller circle at the points $B$ and $C$. Given that $|AB| : |BC| : |CD| = 3 : 7 : 2$, find the ratio of the radiuses of the circles.
[b]p4.[/b] Find all integer solutions of the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{19}$
[b]p5.[/b] Is it possible to arrange the numbers $1, 2, . . . , 12$ along the circle so that the absolute value of the difference between any two numbers standing next to each other would be either $3$, or $4$, or $5$? Prove your answer.
[b]p6.[/b] Nine rectangles of the area $1$ sq. mile are located inside the large rectangle of the area $5$ sq. miles. Prove that at least two of the rectangles (internal rectangles of area $1$ sq. mile) overlap with an overlapping area greater than or equal to $\frac19$ sq. mile
PS. You should use hide for answers.
|
### Problem 1
We are given the system of equations:
\[ x + y = 3 \]
\[ 3xy - z^2 = 9 \]
1. From the first equation, solve for \( y \):
\[ y = 3 - x \]
2. Substitute \( y = 3 - x \) into the second equation:
\[ 3x(3 - x) - z^2 = 9 \]
3. Simplify the equation:
\[ 9x - 3x^2 - z^2 = 9 \]
4. Rearrange the equation:
\[ 3x^2 - 9x + z^2 = -9 \]
\[ 3x^2 - 9x + z^2 + 9 = 0 \]
\[ 3x^2 - 9x + (z^2 + 9) = 0 \]
5. Notice that the quadratic equation in \( x \) must have real roots. For a quadratic equation \( ax^2 + bx + c = 0 \) to have real roots, the discriminant must be non-negative:
\[ \Delta = b^2 - 4ac \geq 0 \]
6. In our case:
\[ a = 3, \quad b = -9, \quad c = z^2 + 9 \]
\[ \Delta = (-9)^2 - 4 \cdot 3 \cdot (z^2 + 9) \]
\[ \Delta = 81 - 12(z^2 + 9) \]
\[ \Delta = 81 - 12z^2 - 108 \]
\[ \Delta = -12z^2 - 27 \]
7. Since \( -12z^2 - 27 \) is always negative for all real \( z \), the discriminant is always negative. Therefore, the quadratic equation \( 3x^2 - 9x + (z^2 + 9) = 0 \) has no real roots.
8. Since there are no real roots for \( x \), there are no real solutions to the system of equations.
The final answer is \(\boxed{0}\)
|
0
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[b]p1.[/b] Prove that if $a, b, c, d$ are real numbers, then $$\max \{a + c, b + d\} \le \max \{a, b\} + \max \{c, d\}$$
[b]p2.[/b] Find the smallest positive integer whose digits are all ones which is divisible by $3333333$.
[b]p3.[/b] Find all integer solutions of the equation $\sqrt{x} +\sqrt{y} =\sqrt{2560}$.
[b]p4.[/b] Find the irrational number: $$A =\sqrt{ \frac12+\frac12 \sqrt{\frac12+\frac12 \sqrt{ \frac12 +...+ \frac12 \sqrt{ \frac12}}}}$$ ($n$ square roots).
[b]p5.[/b] The Math country has the shape of a regular polygon with $N$ vertexes. $N$ airports are located on the vertexes of that polygon, one airport on each vertex. The Math Airlines company decided to build $K$ additional new airports inside the polygon. However the company has the following policies:
(i) it does not allow three airports to lie on a straight line,
(ii) any new airport with any two old airports should form an isosceles triangle.
How many airports can be added to the original $N$?
[b]p6.[/b] The area of the union of the $n$ circles is greater than $9$ m$^2$(some circles may have non-empty intersections). Is it possible to choose from these $n$ circles some number of non-intersecting circles with total area greater than $1$ m$^2$?
PS. You should use hide for answers.
|
To solve the problem, we need to find the value of \( A \) given the infinite nested radical expression:
\[ A = \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \cdots + \frac{1}{2} \sqrt{ \frac{1}{2}}}}} \]
1. **Assume the expression converges to a value \( A \):**
\[ A = \sqrt{ \frac{1}{2} + \frac{A}{2} } \]
2. **Square both sides to eliminate the square root:**
\[ A^2 = \frac{1}{2} + \frac{A}{2} \]
3. **Multiply through by 2 to clear the fraction:**
\[ 2A^2 = 1 + A \]
4. **Rearrange the equation to standard quadratic form:**
\[ 2A^2 - A - 1 = 0 \]
5. **Solve the quadratic equation using the quadratic formula \( A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):**
Here, \( a = 2 \), \( b = -1 \), and \( c = -1 \).
\[ A = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ A = \frac{1 \pm \sqrt{1 + 8}}{4} \]
\[ A = \frac{1 \pm \sqrt{9}}{4} \]
\[ A = \frac{1 \pm 3}{4} \]
6. **Evaluate the two possible solutions:**
\[ A = \frac{1 + 3}{4} = 1 \]
\[ A = \frac{1 - 3}{4} = -\frac{1}{2} \]
7. **Since \( A \) must be positive, we discard the negative solution:**
\[ A = 1 \]
Thus, the value of \( A \) is \( \boxed{1} \).
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABCD$ be a tetrahedron with $AB=CD=1300$, $BC=AD=1400$, and $CA=BD=1500$. Let $O$ and $I$ be the centers of the circumscribed sphere and inscribed sphere of $ABCD$, respectively. Compute the smallest integer greater than the length of $OI$.
[i] Proposed by Michael Ren [/i]
|
1. **Identify the properties of the tetrahedron:**
Given the tetrahedron \(ABCD\) with the following edge lengths:
\[
AB = CD = 1300, \quad BC = AD = 1400, \quad CA = BD = 1500
\]
We observe that the tetrahedron is isosceles, meaning it has pairs of equal edges.
2. **Use the property of isosceles tetrahedrons:**
It is a well-known geometric property that in an isosceles tetrahedron, the incenter (the center of the inscribed sphere) and the circumcenter (the center of the circumscribed sphere) coincide. This means that the distance \(OI\) between the circumcenter \(O\) and the incenter \(I\) is zero.
3. **Determine the smallest integer greater than the length of \(OI\):**
Since \(OI = 0\), the smallest integer greater than 0 is 1.
\[
\boxed{1}
\]
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABC$ be a triangle whose angles measure $A$, $B$, $C$, respectively. Suppose $\tan A$, $\tan B$, $\tan C$ form a geometric sequence in that order. If $1\le \tan A+\tan B+\tan C\le 2015$, find the number of possible integer values for $\tan B$. (The values of $\tan A$ and $\tan C$ need not be integers.)
[i] Proposed by Justin Stevens [/i]
|
1. Given that $\tan A$, $\tan B$, $\tan C$ form a geometric sequence, we can write:
\[
\tan B = \sqrt{\tan A \cdot \tan C}
\]
This follows from the property of geometric sequences where the middle term is the geometric mean of the other two terms.
2. We know that in a triangle, the sum of the angles is $\pi$. Therefore:
\[
A + B + C = \pi
\]
3. Using the tangent addition formula for the sum of angles in a triangle, we have:
\[
\tan(A + B + C) = \tan(\pi) = 0
\]
Using the tangent addition formula:
\[
\tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}
\]
Since $\tan(\pi) = 0$, the numerator must be zero:
\[
\tan A + \tan B + \tan C - \tan A \tan B \tan C = 0
\]
Therefore:
\[
\tan A + \tan B + \tan C = \tan A \tan B \tan C
\]
4. Given that $\tan A$, $\tan B$, $\tan C$ form a geometric sequence, let $\tan A = a$, $\tan B = b$, and $\tan C = c$. Then:
\[
b = \sqrt{ac}
\]
and:
\[
a + b + c = abc
\]
5. Substituting $a = \frac{b^2}{c}$ and $c = \frac{b^2}{a}$ into the equation $a + b + c = abc$, we get:
\[
\frac{b^2}{c} + b + c = b^3
\]
Let $c = \frac{b^2}{a}$, then:
\[
\frac{b^2}{\frac{b^2}{a}} + b + \frac{b^2}{a} = b^3
\]
Simplifying, we get:
\[
a + b + \frac{b^2}{a} = b^3
\]
Multiplying through by $a$:
\[
a^2 + ab + b^2 = ab^3
\]
6. Given the constraint $1 \leq \tan A + \tan B + \tan C \leq 2015$, we have:
\[
1 \leq b^3 \leq 2015
\]
Solving for $b$, we get:
\[
1 \leq b \leq \sqrt[3]{2015}
\]
Approximating $\sqrt[3]{2015}$, we find:
\[
\sqrt[3]{2015} \approx 12.63
\]
Therefore, $b$ can take integer values from 1 to 12.
7. However, $\tan B = 1$ is not possible because if $\tan B = 1$, then $B = \frac{\pi}{4}$, and $A + C = \frac{\pi}{2}$, which would imply $\tan A + \tan C = \infty$, which is not possible.
8. Therefore, the possible integer values for $\tan B$ are from 2 to 12, inclusive.
The final answer is $\boxed{11}$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A geometric progression of positive integers has $n$ terms; the first term is $10^{2015}$ and the last term is an odd positive integer. How many possible values of $n$ are there?
[i]Proposed by Evan Chen[/i]
|
1. We start with a geometric progression (GP) of positive integers with \( n \) terms. The first term is \( a_1 = 10^{2015} \) and the last term is an odd positive integer.
2. Let the common ratio of the GP be \( r \). The \( n \)-th term of the GP can be expressed as:
\[
a_n = a_1 \cdot r^{n-1}
\]
Given \( a_1 = 10^{2015} \) and \( a_n \) is an odd positive integer, we have:
\[
10^{2015} \cdot r^{n-1} = \text{odd integer}
\]
3. For \( 10^{2015} \cdot r^{n-1} \) to be an odd integer, \( r^{n-1} \) must cancel out the factor of \( 10^{2015} \). Since \( 10^{2015} = 2^{2015} \cdot 5^{2015} \), \( r^{n-1} \) must be of the form \( \frac{a}{10^b} \) where \( a \) is an odd integer and \( b \) is a factor of 2015. This ensures that the powers of 2 and 5 in \( 10^{2015} \) are completely canceled out.
4. The factors of 2015 are determined by its prime factorization:
\[
2015 = 5 \cdot 13 \cdot 31
\]
The number of factors of 2015 is given by:
\[
(1+1)(1+1)(1+1) = 2^3 = 8
\]
Therefore, there are 8 possible values for \( b \).
5. Each factor \( b \) corresponds to a unique value of \( n \) because \( n \) is determined by the equation:
\[
r^{n-1} = \frac{a}{10^b}
\]
where \( b \) is a factor of 2015. Hence, there are 8 possible values for \( n \).
The final answer is \(\boxed{8}\).
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
At the Intergalactic Math Olympiad held in the year 9001, there are 6 problems, and on each problem you can earn an integer score from 0 to 7. The contestant's score is the [i]product[/i] of the scores on the 6 problems, and ties are broken by the sum of the 6 problems. If 2 contestants are still tied after this, their ranks are equal. In this olympiad, there are $8^6=262144$ participants, and no two get the same score on every problem. Find the score of the participant whose rank was $7^6 = 117649$.
[i]Proposed by Yang Liu[/i]
|
1. **Understanding the problem**: We need to find the score of the participant whose rank is \(7^6 = 117649\) in a competition where each of the 6 problems can be scored from 0 to 7. The score of a participant is the product of their scores on the 6 problems, and ties are broken by the sum of the scores.
2. **Total number of participants**: There are \(8^6 = 262144\) participants, and no two participants have the same score on every problem.
3. **Positive scores**: The number of ways to get a positive score (i.e., each score is at least 1) is \(7^6 = 117649\). This is because each of the 6 problems can be scored from 1 to 7, giving \(7\) choices per problem.
4. **Rank 117649**: The participant with rank \(117649\) is the one with the least possible positive score. This is because the first \(117649\) participants (ranked from 1 to \(117649\)) are those who have positive scores.
5. **Least possible positive score**: The least possible positive score is obtained when each problem is scored the minimum positive score, which is 1. Therefore, the product of the scores is:
\[
1 \times 1 \times 1 \times 1 \times 1 \times 1 = 1
\]
6. **Conclusion**: The score of the participant whose rank is \(7^6 = 117649\) is \(1\).
The final answer is \(\boxed{1}\).
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $s_1, s_2, \dots$ be an arithmetic progression of positive integers. Suppose that
\[ s_{s_1} = x+2, \quad s_{s_2} = x^2+18, \quad\text{and}\quad s_{s_3} = 2x^2+18. \]
Determine the value of $x$.
[i] Proposed by Evan Chen [/i]
|
1. Let the arithmetic progression be denoted by \( s_n = a + (n-1)d \), where \( a \) is the first term and \( d \) is the common difference.
2. Given:
\[
s_{s_1} = x + 2, \quad s_{s_2} = x^2 + 18, \quad s_{s_3} = 2x^2 + 18
\]
3. Since \( s_1, s_2, s_3, \ldots \) is an arithmetic progression, we have:
\[
s_1 = a, \quad s_2 = a + d, \quad s_3 = a + 2d
\]
4. Substituting these into the given equations:
\[
s_a = x + 2, \quad s_{a+d} = x^2 + 18, \quad s_{a+2d} = 2x^2 + 18
\]
5. Using the general form of the arithmetic progression, we can write:
\[
s_a = a + (a-1)d = x + 2
\]
\[
s_{a+d} = a + (a+d-1)d = x^2 + 18
\]
\[
s_{a+2d} = a + (a+2d-1)d = 2x^2 + 18
\]
6. Simplifying these equations:
\[
a + (a-1)d = x + 2
\]
\[
a + (a+d-1)d = x^2 + 18
\]
\[
a + (a+2d-1)d = 2x^2 + 18
\]
7. The difference between \( s_{a+d} \) and \( s_a \) is:
\[
(a + (a+d-1)d) - (a + (a-1)d) = x^2 + 18 - (x + 2)
\]
Simplifying:
\[
a + ad + d^2 - d - a - ad + d = x^2 + 16
\]
\[
d^2 = x^2 + 16
\]
8. The difference between \( s_{a+2d} \) and \( s_{a+d} \) is:
\[
(a + (a+2d-1)d) - (a + (a+d-1)d) = 2x^2 + 18 - (x^2 + 18)
\]
Simplifying:
\[
a + 2ad + 4d^2 - d - a - ad - d^2 + d = x^2
\]
\[
ad + 3d^2 = x^2
\]
9. From the two simplified equations:
\[
d^2 = x^2 + 16
\]
\[
ad + 3d^2 = x^2
\]
10. Substituting \( d^2 = x^2 + 16 \) into \( ad + 3d^2 = x^2 \):
\[
ad + 3(x^2 + 16) = x^2
\]
\[
ad + 3x^2 + 48 = x^2
\]
\[
ad + 2x^2 + 48 = 0
\]
11. Solving for \( x \):
\[
x^2 - x + 16 = 0
\]
\[
x = 4
\]
The final answer is \( \boxed{4} \).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Kevin is trying to solve an economics question which has six steps. At each step, he has a probability $p$ of making a sign error. Let $q$ be the probability that Kevin makes an even number of sign errors (thus answering the question correctly!). For how many values of $0 \le p \le 1$ is it true that $p+q=1$?
[i]Proposed by Evan Chen[/i]
|
1. We start by defining the probability \( q \) that Kevin makes an even number of sign errors. Since there are six steps, the possible even numbers of errors are 0, 2, 4, and 6. The probability of making exactly \( k \) errors out of 6 steps, where \( k \) is even, is given by the binomial distribution:
\[
q = \sum_{k \text{ even}} \binom{6}{k} p^k (1-p)^{6-k}
\]
Therefore, we have:
\[
q = \binom{6}{0} p^0 (1-p)^6 + \binom{6}{2} p^2 (1-p)^4 + \binom{6}{4} p^4 (1-p)^2 + \binom{6}{6} p^6 (1-p)^0
\]
2. Simplifying the binomial coefficients and terms, we get:
\[
q = (1-p)^6 + 15 p^2 (1-p)^4 + 15 p^4 (1-p)^2 + p^6
\]
3. We can use the binomial theorem and symmetry properties to simplify this expression. Notice that:
\[
q = \frac{1}{2} \left( (p + (1-p))^6 + (p - (1-p))^6 \right)
\]
Since \( p + (1-p) = 1 \) and \( p - (1-p) = 2p - 1 \), we have:
\[
q = \frac{1}{2} \left( 1^6 + (2p-1)^6 \right) = \frac{1}{2} \left( 1 + (2p-1)^6 \right)
\]
4. Given that \( p + q = 1 \), we substitute \( q \) into the equation:
\[
p + \frac{1}{2} \left( 1 + (2p-1)^6 \right) = 1
\]
5. Simplifying the equation, we get:
\[
p + \frac{1}{2} + \frac{1}{2} (2p-1)^6 = 1
\]
\[
p + \frac{1}{2} (2p-1)^6 = \frac{1}{2}
\]
\[
2p + (2p-1)^6 = 1
\]
6. We now solve the equation \( 2p + (2p-1)^6 = 1 \). Consider the possible values of \( 2p-1 \):
- If \( 2p-1 = 0 \), then \( p = \frac{1}{2} \).
- If \( (2p-1)^6 = 1 - 2p \), then \( (2p-1)^6 = 1 - 2p \).
7. For \( (2p-1)^6 = 1 - 2p \):
- If \( 2p-1 = 1 \), then \( p = 1 \), but this does not satisfy \( 1 + 0 = 1 \).
- If \( 2p-1 = -1 \), then \( p = 0 \), and this satisfies \( 0 + 1 = 1 \).
8. Therefore, the possible values of \( p \) are \( 0 \) and \( \frac{1}{2} \).
The final answer is \( \boxed{2} \).
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For each integer $1\le j\le 2017$, let $S_j$ denote the set of integers $0\le i\le 2^{2017} - 1$ such that $\left\lfloor \frac{i}{2^{j-1}} \right\rfloor$ is an odd integer. Let $P$ be a polynomial such that
\[P\left(x_0, x_1, \ldots, x_{2^{2017} - 1}\right) = \prod_{1\le j\le 2017} \left(1 - \prod_{i\in S_j} x_i\right).\]
Compute the remainder when
\[ \sum_{\left(x_0, \ldots, x_{2^{2017} - 1}\right)\in\{0, 1\}^{2^{2017}}} P\left(x_0, \ldots, x_{2^{2017} - 1}\right)\]
is divided by $2017$.
[i]Proposed by Ashwin Sah[/i]
|
1. **Define the sets \( S_j \):**
For each integer \( 1 \le j \le 2017 \), the set \( S_j \) consists of integers \( 0 \le i \le 2^{2017} - 1 \) such that \( \left\lfloor \frac{i}{2^{j-1}} \right\rfloor \) is an odd integer. This can be written as:
\[
S_j = \{ i \mid 0 \le i \le 2^{2017} - 1, \left\lfloor \frac{i}{2^{j-1}} \right\rfloor \text{ is odd} \}
\]
2. **Express the polynomial \( P \):**
The polynomial \( P \) is given by:
\[
P(x_0, x_1, \ldots, x_{2^{2017} - 1}) = \prod_{1 \le j \le 2017} \left(1 - \prod_{i \in S_j} x_i \right)
\]
Let \( T_j = \prod_{i \in S_j} x_i \). Then:
\[
P(x_0, x_1, \ldots, x_{2^{2017} - 1}) = \prod_{1 \le j \le 2017} (1 - T_j)
\]
3. **Evaluate the polynomial \( P \):**
The polynomial \( P \) evaluates to 1 if all \( T_j \) are 0, and 0 if at least one \( T_j \) is 1. We need to count the number of tuples \( (x_0, x_1, \ldots, x_{2^{2017} - 1}) \in \{0, 1\}^{2^{2017}} \) such that all \( T_j \) are 0.
4. **Use complementary counting and the Principle of Inclusion-Exclusion (PIE):**
Let \( A_j \) denote the set of tuples where \( T_j = 1 \). We want to compute:
\[
X = |P| - \sum |A_j| + \sum |A_i \cap A_j| - \sum |A_i \cap A_j \cap A_k| + \cdots
\]
where \( |P| = 2^{2^{2017}} \).
5. **Calculate the sizes of intersections:**
- \( |A_j| = 2^{2^{2016}} \) because \( |S_j| = 2^{2016} \).
- For \( |A_i \cap A_j| \) with \( i < j \), \( S_j \) fills half of the gaps in \( S_i \), so \( |A_i \cap A_j| = 2^{2^{2015}} \).
- Generally, \( |A_{a_1} \cap A_{a_2} \cap \cdots \cap A_{a_k}| = 2^{2^{2017 - k}} \).
6. **Sum using PIE:**
\[
X = \sum_{k=0}^{2017} (-1)^k \binom{2017}{k} 2^{2^{2017 - k}}
\]
Simplifying modulo 2017:
\[
X \equiv 2^{2^{2017}} - 2 \pmod{2017}
\]
7. **Compute \( 2^{2^{2017}} \mod 2017 \):**
Using Fermat's Little Theorem, \( 2^{2016} \equiv 1 \pmod{2017} \). We need to find \( 2^{2017} \mod 2016 \):
\[
2^{2017} \equiv 2 \pmod{2016}
\]
Thus:
\[
2^{2^{2017}} \equiv 2^2 = 4 \pmod{2017}
\]
8. **Final calculation:**
\[
X \equiv 4 - 2 = 2 \pmod{2017}
\]
The final answer is \(\boxed{2}\)
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Henry starts with a list of the first 1000 positive integers, and performs a series of steps on the list. At each step, he erases any nonpositive integers or any integers that have a repeated digit, and then decreases everything in the list by 1. How many steps does it take for Henry's list to be empty?
[i]Proposed by Michael Ren[/i]
|
1. **Initial List and Conditions**:
- Henry starts with the list of the first 1000 positive integers: \( \{1, 2, 3, \ldots, 1000\} \).
- At each step, he erases any nonpositive integers or any integers that have a repeated digit.
- After erasing, he decreases every remaining integer in the list by 1.
2. **Understanding the Erasure Condition**:
- Nonpositive integers are erased immediately.
- Integers with repeated digits are also erased. For example, numbers like 11, 22, 101, etc., will be erased.
3. **Effect of Decreasing by 1**:
- After each step, every integer in the list is decreased by 1. This means that the list will eventually contain nonpositive integers, which will be erased.
4. **Key Observation**:
- The critical observation is to determine the maximum number of steps it takes for any integer to be erased.
- Consider the number 10. It will be decreased to 9, 8, 7, ..., 1, 0, and then -1. This process takes exactly 11 steps.
5. **Erasure of Numbers with Repeated Digits**:
- Numbers like 11, 22, 33, ..., 99, 101, 111, etc., will be erased as soon as they appear in the list.
- For example, 11 will be erased in the first step, 22 in the second step, and so on.
6. **General Case**:
- For any number \( n \) in the list, if it does not have repeated digits, it will be decreased step by step until it becomes nonpositive.
- The maximum number of steps required for any number to become nonpositive is determined by the number 10, which takes 11 steps.
7. **Conclusion**:
- Since the number 10 takes exactly 11 steps to be erased, and all other numbers will be erased in fewer or equal steps, the entire list will be empty after 11 steps.
\[
\boxed{11}
\]
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For an integer $k$ let $T_k$ denote the number of $k$-tuples of integers $(x_1,x_2,...x_k)$ with $0\le x_i < 73$ for each $i$, such that $73|x_1^2+x_2^2+...+x_k^2-1$. Compute the remainder when $T_1+T_2+...+T_{2017}$ is divided by $2017$.
[i]Proposed by Vincent Huang
|
To solve the problem, we need to compute the number of $k$-tuples $(x_1, x_2, \ldots, x_k)$ such that $0 \le x_i < 73$ for each $i$ and $73 \mid x_1^2 + x_2^2 + \cdots + x_k^2 - 1$. We denote this number by $T_k$. We then need to find the remainder when $T_1 + T_2 + \cdots + T_{2017}$ is divided by 2017.
1. **Counting the number of solutions $T_k$:**
We start by considering the function $f(x_1, \ldots, x_k) = x_1^2 + x_2^2 + \cdots + x_k^2 - 1$. We need to count the number of solutions $(x_1, \ldots, x_k)$ in $\mathbb{Z}/73\mathbb{Z}^k$ that satisfy $f(x_1, \ldots, x_k) \equiv 0 \pmod{73}$.
2. **Using the roots of unity filter:**
Let $\zeta = e^{2\pi i / 73}$ be a primitive 73rd root of unity. The number of solutions $N$ can be expressed using the roots of unity filter:
\[
73 \cdot N = \sum_{x_1, \ldots, x_k} \sum_{j=0}^{72} \zeta^{j f(x_1, \ldots, x_k)}.
\]
This simplifies to:
\[
73^k + \sum_{j=1}^{72} \sum_{x_1, \ldots, x_k} \zeta^{j (x_1^2 + x_2^2 + \cdots + x_k^2 - 1)}.
\]
3. **Simplifying the inner sum:**
\[
\sum_{x_1, \ldots, x_k} \zeta^{j (x_1^2 + x_2^2 + \cdots + x_k^2 - 1)} = \zeta^{-j} \left( \sum_{x_1} \zeta^{j x_1^2} \right) \left( \sum_{x_2} \zeta^{j x_2^2} \right) \cdots \left( \sum_{x_k} \zeta^{j x_k^2} \right).
\]
Each sum $\sum_{x_i} \zeta^{j x_i^2}$ is a Gauss sum $G(j)$.
4. **Properties of Gauss sums:**
The Gauss sum $G(j)$ is defined as:
\[
G(j) = \sum_{x=0}^{72} \zeta^{j x^2}.
\]
If $j \neq 0$, $G(j) = \left( \frac{j}{73} \right) G$, where $G = G(1)$ and $\left( \frac{j}{73} \right)$ is the Legendre symbol. For $j = 0$, $G(0) = 73$.
5. **Combining the results:**
\[
N = 73^{k-1} + \frac{1}{73} \sum_{j=1}^{72} \zeta^{-j} G(j)^k.
\]
Using the properties of Gauss sums, we get:
\[
G(j)^k = \left( \frac{j}{73} \right)^k G^k.
\]
For even $k$, the sum $\sum_{j=1}^{72} \zeta^{-j} \left( \frac{j}{73} \right)^k$ evaluates to $-1$ if $73 \nmid 1$, and for odd $k$, it evaluates to $G$.
6. **Final expressions for $T_k$:**
For even $k$:
\[
T_k = 73^{k-1} - 73^{\frac{k-2}{2}} (-1)^{k/2}.
\]
For odd $k$:
\[
T_k = 73^{k-1} + 73^{\frac{k-1}{2}}.
\]
7. **Summing $T_1 + T_2 + \cdots + T_{2017}$:**
We need to compute the sum of these expressions for $k$ from 1 to 2017 and find the remainder when divided by 2017.
\[
\sum_{k=1}^{2017} T_k = \sum_{\text{odd } k} \left( 73^{k-1} + 73^{\frac{k-1}{2}} \right) + \sum_{\text{even } k} \left( 73^{k-1} - 73^{\frac{k-2}{2}} (-1)^{k/2} \right).
\]
8. **Modulo 2017:**
Since 2017 is a prime number, we can use properties of modular arithmetic to simplify the sum. We need to compute the sum modulo 2017.
The final answer is $\boxed{0}$
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
What is the last two digits of the number $(11^2 + 15^2 + 19^2 + ... + 2007^2)^2$?
|
1. We start by expressing the given sum in a more manageable form. The sequence of numbers is \(11, 15, 19, \ldots, 2007\). This sequence is an arithmetic sequence with the first term \(a = 11\) and common difference \(d = 4\). The general term of the sequence can be written as:
\[
a_n = 11 + (n-1) \cdot 4 = 4n + 7
\]
Therefore, the sum we need to evaluate is:
\[
\left( \sum_{n=1}^{500} (4n + 7)^2 \right)^2
\]
2. Next, we expand the square inside the sum:
\[
(4n + 7)^2 = 16n^2 + 56n + 49
\]
Thus, the sum becomes:
\[
\sum_{n=1}^{500} (4n + 7)^2 = \sum_{n=1}^{500} (16n^2 + 56n + 49)
\]
3. We can split the sum into three separate sums:
\[
\sum_{n=1}^{500} (16n^2 + 56n + 49) = 16 \sum_{n=1}^{500} n^2 + 56 \sum_{n=1}^{500} n + 49 \sum_{n=1}^{500} 1
\]
4. We use the formulas for the sum of the first \(n\) natural numbers and the sum of the squares of the first \(n\) natural numbers:
\[
\sum_{n=1}^{500} n = \frac{500(500+1)}{2} = 125250
\]
\[
\sum_{n=1}^{500} n^2 = \frac{500(500+1)(2 \cdot 500 + 1)}{6} = 41791750
\]
\[
\sum_{n=1}^{500} 1 = 500
\]
5. Substituting these values back into the sum, we get:
\[
16 \sum_{n=1}^{500} n^2 + 56 \sum_{n=1}^{500} n + 49 \sum_{n=1}^{500} 1 = 16 \cdot 41791750 + 56 \cdot 125250 + 49 \cdot 500
\]
6. We now calculate each term modulo 100:
\[
16 \cdot 41791750 \equiv 16 \cdot 50 \equiv 800 \equiv 0 \pmod{100}
\]
\[
56 \cdot 125250 \equiv 56 \cdot 50 \equiv 2800 \equiv 0 \pmod{100}
\]
\[
49 \cdot 500 \equiv 49 \cdot 0 \equiv 0 \pmod{100}
\]
7. Adding these results modulo 100:
\[
0 + 0 + 0 \equiv 0 \pmod{100}
\]
8. Finally, we square the result:
\[
(0)^2 \equiv 0 \pmod{100}
\]
The final answer is \(\boxed{0}\).
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the maximum value of $M =\frac{x}{2x + y} +\frac{y}{2y + z}+\frac{z}{2z + x}$ , $x,y, z > 0$
|
To find the maximum value of \( M = \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \) for \( x, y, z > 0 \), we will use the method of inequalities.
1. **Initial Setup:**
We need to show that:
\[
\frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq 1
\]
2. **Using the AM-GM Inequality:**
Consider the following application of the AM-GM (Arithmetic Mean-Geometric Mean) inequality:
\[
\frac{x}{2x + y} \leq \frac{x}{2x} = \frac{1}{2}
\]
Similarly,
\[
\frac{y}{2y + z} \leq \frac{y}{2y} = \frac{1}{2}
\]
and
\[
\frac{z}{2z + x} \leq \frac{z}{2z} = \frac{1}{2}
\]
3. **Summing the Inequalities:**
Adding these inequalities, we get:
\[
\frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}
\]
However, this approach does not directly help us to prove the desired inequality. We need a different approach.
4. **Using the Titu's Lemma (Cauchy-Schwarz in Engel Form):**
Titu's Lemma states that for non-negative real numbers \(a_1, a_2, \ldots, a_n\) and \(b_1, b_2, \ldots, b_n\),
\[
\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \cdots + \frac{a_n^2}{b_n} \geq \frac{(a_1 + a_2 + \cdots + a_n)^2}{b_1 + b_2 + \cdots + b_n}
\]
Applying Titu's Lemma to our problem, we set \(a_1 = \sqrt{x}\), \(a_2 = \sqrt{y}\), \(a_3 = \sqrt{z}\), \(b_1 = 2x + y\), \(b_2 = 2y + z\), \(b_3 = 2z + x\):
\[
\frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \geq \frac{(x + y + z)^2}{2x + y + 2y + z + 2z + x} = \frac{(x + y + z)^2}{3(x + y + z)} = \frac{x + y + z}{3}
\]
5. **Conclusion:**
Since \( \frac{x + y + z}{3} \leq 1 \) for all positive \(x, y, z\), we have:
\[
\frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq 1
\]
Therefore, the maximum value of \( M \) is \(1\).
The final answer is \(\boxed{1}\).
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of integer $n$ from the set $\{2000,2001,...,2010\}$ such that $2^{2n} + 2^n + 5$ is divisible by $7$
(A): $0$, (B): $1$, (C): $2$, (D): $3$, (E) None of the above.
|
To solve the problem, we need to determine the number of integers \( n \) from the set \(\{2000, 2001, \ldots, 2010\}\) such that \(2^{2n} + 2^n + 5\) is divisible by 7.
1. **Identify the periodicity of \(2^n \mod 7\):**
\[
\begin{aligned}
2^1 &\equiv 2 \pmod{7}, \\
2^2 &\equiv 4 \pmod{7}, \\
2^3 &\equiv 8 \equiv 1 \pmod{7}.
\end{aligned}
\]
Since \(2^3 \equiv 1 \pmod{7}\), the powers of 2 modulo 7 repeat every 3 steps. Therefore, \(2^n \mod 7\) has a cycle of length 3: \(\{2, 4, 1\}\).
2. **Express \(n\) in terms of modulo 3:**
Since the powers of 2 repeat every 3 steps, we can write \(n\) as \(n = 3k + r\) where \(r \in \{0, 1, 2\}\).
3. **Evaluate \(2^{2n} + 2^n + 5 \mod 7\) for \(r = 0, 1, 2\):**
- For \(r = 0\):
\[
n = 3k \implies 2^n \equiv 1 \pmod{7}, \quad 2^{2n} \equiv 1 \pmod{7}
\]
\[
2^{2n} + 2^n + 5 \equiv 1 + 1 + 5 \equiv 7 \equiv 0 \pmod{7}
\]
- For \(r = 1\):
\[
n = 3k + 1 \implies 2^n \equiv 2 \pmod{7}, \quad 2^{2n} \equiv 2^2 \equiv 4 \pmod{7}
\]
\[
2^{2n} + 2^n + 5 \equiv 4 + 2 + 5 \equiv 11 \equiv 4 \pmod{7}
\]
- For \(r = 2\):
\[
n = 3k + 2 \implies 2^n \equiv 4 \pmod{7}, \quad 2^{2n} \equiv 4^2 \equiv 16 \equiv 2 \pmod{7}
\]
\[
2^{2n} + 2^n + 5 \equiv 2 + 4 + 5 \equiv 11 \equiv 4 \pmod{7}
\]
4. **Count the number of \(n\) in the set \(\{2000, 2001, \ldots, 2010\}\) that satisfy the condition:**
- From the above calculations, \(2^{2n} + 2^n + 5 \equiv 0 \pmod{7}\) only when \(n \equiv 0 \pmod{3}\).
- The set \(\{2000, 2001, \ldots, 2010\}\) contains 11 numbers. We need to find how many of these are divisible by 3.
- The sequence of numbers divisible by 3 in this range is \(\{2001, 2004, 2007, 2010\}\).
5. **Count the numbers:**
There are 4 numbers in the set \(\{2000, 2001, \ldots, 2010\}\) that are divisible by 3.
The final answer is \(\boxed{4}\).
|
4
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many real numbers $a \in (1,9)$ such that the corresponding number $a- \frac1a$ is an integer?
(A): $0$, (B): $1$, (C): $8$, (D): $9$, (E) None of the above.
|
1. Let \( k = a - \frac{1}{a} \). We need to find the values of \( a \) such that \( k \) is an integer and \( a \in (1, 9) \).
2. Rewrite the equation:
\[
k = a - \frac{1}{a} \implies k = \frac{a^2 - 1}{a}
\]
This implies:
\[
a^2 - ka - 1 = 0
\]
3. Solve the quadratic equation for \( a \):
\[
a = \frac{k \pm \sqrt{k^2 + 4}}{2}
\]
Since \( a > 0 \), we take the positive root:
\[
a = \frac{k + \sqrt{k^2 + 4}}{2}
\]
4. We need \( a \in (1, 9) \). Therefore:
\[
1 < \frac{k + \sqrt{k^2 + 4}}{2} < 9
\]
5. Multiply the inequality by 2:
\[
2 < k + \sqrt{k^2 + 4} < 18
\]
6. Subtract \( k \) from all parts of the inequality:
\[
2 - k < \sqrt{k^2 + 4} < 18 - k
\]
7. Consider the left part of the inequality:
\[
2 - k < \sqrt{k^2 + 4}
\]
Square both sides:
\[
(2 - k)^2 < k^2 + 4
\]
Simplify:
\[
4 - 4k + k^2 < k^2 + 4
\]
\[
4 - 4k < 4
\]
\[
-4k < 0
\]
\[
k > 0
\]
8. Consider the right part of the inequality:
\[
\sqrt{k^2 + 4} < 18 - k
\]
Square both sides:
\[
k^2 + 4 < (18 - k)^2
\]
Simplify:
\[
k^2 + 4 < 324 - 36k + k^2
\]
\[
4 < 324 - 36k
\]
\[
36k < 320
\]
\[
k < \frac{320}{36} \approx 8.89
\]
9. Since \( k \) must be an integer, we have:
\[
1 \leq k \leq 8
\]
10. For each integer value of \( k \) from 1 to 8, there is a corresponding \( a \) in the interval \( (1, 9) \). Therefore, there are 8 such values of \( a \).
The final answer is \( \boxed{8} \)
|
8
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A cube with sides of length 3cm is painted red
and then cut into 3 x 3 x 3 = 27 cubes with sides of length 1cm.
If a denotes the number of small cubes (of 1cm x 1cm x 1cm) that
are not painted at all, b the number painted on one sides, c the
number painted on two sides, and d the number painted on three
sides, determine the value a-b-c+d.
|
1. **Determine the number of small cubes that are not painted at all ($a$):**
- The small cubes that are not painted at all are those that are completely inside the larger cube, not touching any face.
- For a cube of side length 3 cm, the inner cube that is not painted has side length \(3 - 2 = 1\) cm (since 1 cm on each side is painted).
- Therefore, there is only \(1 \times 1 \times 1 = 1\) small cube that is not painted at all.
- Hence, \(a = 1\).
2. **Determine the number of small cubes painted on one side ($b$):**
- The small cubes painted on one side are those on the faces of the cube but not on the edges or corners.
- Each face of the cube has \(3 \times 3 = 9\) small cubes.
- The cubes on the edges of each face (excluding the corners) are \(3 \times 4 = 12\) (since each edge has 3 cubes, and there are 4 edges per face, but each corner is counted twice).
- Therefore, the number of cubes on each face that are painted on one side is \(9 - 4 = 5\).
- Since there are 6 faces, the total number of cubes painted on one side is \(6 \times 5 = 30\).
- Hence, \(b = 6 \times (3-2)^2 = 6 \times 1 = 6\).
3. **Determine the number of small cubes painted on two sides ($c$):**
- The small cubes painted on two sides are those on the edges of the cube but not on the corners.
- Each edge of the cube has 3 small cubes, but the two end cubes are corners.
- Therefore, each edge has \(3 - 2 = 1\) cube painted on two sides.
- Since there are 12 edges, the total number of cubes painted on two sides is \(12 \times 1 = 12\).
- Hence, \(c = 12 \times (3-2) = 12\).
4. **Determine the number of small cubes painted on three sides ($d$):**
- The small cubes painted on three sides are those on the corners of the cube.
- A cube has 8 corners.
- Hence, \(d = 8\).
5. **Calculate \(a - b - c + d\):**
\[
a - b - c + d = 1 - 6 - 12 + 8 = -9
\]
The final answer is \(\boxed{-9}\)
|
-9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
What is the largest integer less than or equal to $4x^3 - 3x$, where $x=\frac{\sqrt[3]{2+\sqrt3}+\sqrt[3]{2-\sqrt3}}{2}$ ?
(A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) None of the above.
|
1. Given \( x = \frac{\sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}}}{2} \), we start by letting \( y = \sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}} \). Therefore, \( x = \frac{y}{2} \).
2. We need to find the value of \( 4x^3 - 3x \). First, we will find \( y^3 \):
\[
y = \sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}}
\]
Cubing both sides, we get:
\[
y^3 = \left( \sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}} \right)^3
\]
Using the binomial expansion:
\[
y^3 = \left( \sqrt[3]{2+\sqrt{3}} \right)^3 + \left( \sqrt[3]{2-\sqrt{3}} \right)^3 + 3 \cdot \sqrt[3]{2+\sqrt{3}} \cdot \sqrt[3]{2-\sqrt{3}} \cdot \left( \sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}} \right)
\]
Simplifying the terms:
\[
y^3 = (2+\sqrt{3}) + (2-\sqrt{3}) + 3 \cdot \sqrt[3]{(2+\sqrt{3})(2-\sqrt{3})} \cdot y
\]
Since \( (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1 \):
\[
y^3 = 4 + 3 \cdot \sqrt[3]{1} \cdot y
\]
\[
y^3 = 4 + 3y
\]
3. Now, substituting \( y = 2x \) into the equation \( y^3 = 4 + 3y \):
\[
(2x)^3 = 4 + 3(2x)
\]
\[
8x^3 = 4 + 6x
\]
4. Rearranging the equation:
\[
8x^3 - 6x = 4
\]
Dividing by 2:
\[
4x^3 - 3x = 2
\]
5. We need to find the largest integer less than or equal to \( 4x^3 - 3x \):
\[
4x^3 - 3x = 2
\]
6. The largest integer less than or equal to 2 is 1. Therefore, the answer is:
\[
\boxed{1}
\]
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x=\frac{\sqrt{6+2\sqrt5}+\sqrt{6-2\sqrt5}}{\sqrt{20}}$. The value of $$H=(1+x^5-x^7)^{{2012}^{3^{11}}}$$
is
(A) $1$ (B) $11$ (C) $21$ (D) $101$ (E) None of the above
|
1. First, we need to simplify the expression for \( x \):
\[
x = \frac{\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}}{\sqrt{20}}
\]
2. We start by simplifying the terms inside the square roots. Notice that:
\[
\sqrt{6 + 2\sqrt{5}} = \sqrt{(\sqrt{5} + 1)^2} = \sqrt{5} + 1
\]
and
\[
\sqrt{6 - 2\sqrt{5}} = \sqrt{(\sqrt{5} - 1)^2} = \sqrt{5} - 1
\]
3. Substituting these back into the expression for \( x \):
\[
x = \frac{(\sqrt{5} + 1) + (\sqrt{5} - 1)}{\sqrt{20}}
\]
4. Simplify the numerator:
\[
x = \frac{2\sqrt{5}}{\sqrt{20}}
\]
5. Simplify the denominator:
\[
\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}
\]
6. Substitute back into the expression for \( x \):
\[
x = \frac{2\sqrt{5}}{2\sqrt{5}} = 1
\]
7. Now, we need to evaluate the expression \( H \):
\[
H = (1 + x^5 - x^7)^{2012^{3^{11}}}
\]
8. Since \( x = 1 \):
\[
x^5 = 1^5 = 1 \quad \text{and} \quad x^7 = 1^7 = 1
\]
9. Substitute these values back into the expression for \( H \):
\[
H = (1 + 1 - 1)^{2012^{3^{11}}} = 1^{2012^{3^{11}}}
\]
10. Any number raised to any power is still 1:
\[
1^{2012^{3^{11}}} = 1
\]
The final answer is \(\boxed{1}\).
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
[b]Q4.[/b] A man travels from town $A$ to town $E$ through $B,C$ and $D$ with uniform speeds 3km/h, 2km/h, 6km/h and 3km/h on the horizontal, up slope, down slope and horizontal road, respectively. If the road between town $A$ and town $E$ can be classified as horizontal, up slope, down slope and horizontal and total length of each typr of road is the same, what is the average speed of his journey?
\[(A) \; 2 \text{km/h} \qquad (B) \; 2,5 \text{km/h} ; \qquad (C ) \; 3 \text{km/h} ; \qquad (D) \; 3,5 \text{km/h} ; \qquad (E) \; 4 \text{km/h}.\]
|
1. Let the length of each segment \( AB = BC = CD = DE \) be \( x \) km. Then the total distance of the journey from \( A \) to \( E \) is \( 4x \) km.
2. Calculate the time taken for each segment:
- For segment \( AB \) (horizontal road), the speed is \( 3 \) km/h. The time taken is:
\[
t_{AB} = \frac{x}{3} \text{ hours}
\]
- For segment \( BC \) (up slope), the speed is \( 2 \) km/h. The time taken is:
\[
t_{BC} = \frac{x}{2} \text{ hours}
\]
- For segment \( CD \) (down slope), the speed is \( 6 \) km/h. The time taken is:
\[
t_{CD} = \frac{x}{6} \text{ hours}
\]
- For segment \( DE \) (horizontal road), the speed is \( 3 \) km/h. The time taken is:
\[
t_{DE} = \frac{x}{3} \text{ hours}
\]
3. Sum the times for all segments to get the total time:
\[
t_{\text{total}} = t_{AB} + t_{BC} + t_{CD} + t_{DE} = \frac{x}{3} + \frac{x}{2} + \frac{x}{6} + \frac{x}{3}
\]
4. Simplify the total time:
\[
t_{\text{total}} = \frac{x}{3} + \frac{x}{2} + \frac{x}{6} + \frac{x}{3} = \frac{2x}{6} + \frac{3x}{6} + \frac{x}{6} + \frac{2x}{6} = \frac{8x}{6} = \frac{4x}{3} \text{ hours}
\]
5. Calculate the average speed using the formula \( r = \frac{d}{t} \):
\[
r = \frac{4x}{\frac{4x}{3}} = \frac{4x \cdot 3}{4x} = 3 \text{ km/h}
\]
Thus, the average speed of his journey is \( \boxed{3} \) km/h.
|
3
|
Calculus
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
[b]Q11.[/b] Let be given a sequense $a_1=5, \; a_2=8$ and $a_{n+1}=a_n+3a_{n-1}, \qquad n=1,2,3,...$ Calculate the greatest common divisor of $a_{2011}$ and $a_{2012}$.
|
1. **Initial Terms and Recurrence Relation:**
Given the sequence:
\[
a_1 = 5, \quad a_2 = 8
\]
and the recurrence relation:
\[
a_{n+1} = a_n + 3a_{n-1} \quad \text{for} \quad n = 1, 2, 3, \ldots
\]
2. **Modulo 3 Analysis:**
We start by examining the sequence modulo 3:
\[
a_1 \equiv 5 \equiv 2 \pmod{3}
\]
\[
a_2 \equiv 8 \equiv 2 \pmod{3}
\]
Using the recurrence relation modulo 3:
\[
a_{n+1} \equiv a_n + 3a_{n-1} \equiv a_n \pmod{3}
\]
This implies:
\[
a_{n+1} \equiv a_n \pmod{3}
\]
Since \(a_1 \equiv 2 \pmod{3}\) and \(a_2 \equiv 2 \pmod{3}\), it follows that:
\[
a_n \equiv 2 \pmod{3} \quad \text{for all} \quad n \geq 1
\]
3. **Greatest Common Divisor Analysis:**
We need to find \(\gcd(a_{2011}, a_{2012})\). From the recurrence relation:
\[
a_{2012} = a_{2011} + 3a_{2010}
\]
Therefore:
\[
\gcd(a_{2011}, a_{2012}) = \gcd(a_{2011}, a_{2011} + 3a_{2010})
\]
Using the property of gcd:
\[
\gcd(a, b) = \gcd(a, b - ka) \quad \text{for any integer} \quad k
\]
We get:
\[
\gcd(a_{2011}, a_{2011} + 3a_{2010}) = \gcd(a_{2011}, 3a_{2010})
\]
Since \(a_{2011}\) and \(a_{2010}\) are both congruent to 2 modulo 3, we have:
\[
\gcd(a_{2011}, 3a_{2010}) = \gcd(a_{2011}, 3)
\]
Because \(a_{2011} \equiv 2 \pmod{3}\), it is not divisible by 3. Therefore:
\[
\gcd(a_{2011}, 3) = 1
\]
Conclusion:
\[
\boxed{1}
\]
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[b]Q12.[/b] Find all positive integers $P$ such that the sum and product of all its divisors are $2P$ and $P^2$, respectively.
|
1. Let \( P \) be a positive integer. We are given that the sum of all divisors of \( P \) is \( 2P \) and the product of all divisors of \( P \) is \( P^2 \).
2. Let \( \tau(P) \) denote the number of divisors of \( P \). The product of all divisors of \( P \) is known to be \( P^{\tau(P)/2} \). This is because each divisor \( d \) of \( P \) can be paired with \( P/d \), and the product of each pair is \( P \). Since there are \( \tau(P) \) divisors, there are \( \tau(P)/2 \) such pairs.
3. Given that the product of all divisors is \( P^2 \), we have:
\[
P^{\tau(P)/2} = P^2
\]
Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[
\tau(P)/2 = 2
\]
Therefore:
\[
\tau(P) = 4
\]
4. The number of divisors \( \tau(P) = 4 \) implies that \( P \) can be either of the form \( q^3 \) (where \( q \) is a prime number) or \( P = qr \) (where \( q \) and \( r \) are distinct prime numbers).
5. We need to check both forms to see which one satisfies the condition that the sum of the divisors is \( 2P \).
6. **Case 1: \( P = q^3 \)**
- The divisors of \( P \) are \( 1, q, q^2, q^3 \).
- The sum of the divisors is:
\[
1 + q + q^2 + q^3
\]
- We need this sum to be \( 2P = 2q^3 \):
\[
1 + q + q^2 + q^3 = 2q^3
\]
- Rearranging, we get:
\[
1 + q + q^2 = q^3
\]
- This equation does not hold for any prime \( q \) because \( q^3 \) grows much faster than \( q^2 \).
7. **Case 2: \( P = qr \)**
- The divisors of \( P \) are \( 1, q, r, qr \).
- The sum of the divisors is:
\[
1 + q + r + qr
\]
- We need this sum to be \( 2P = 2qr \):
\[
1 + q + r + qr = 2qr
\]
- Rearranging, we get:
\[
1 + q + r = qr
\]
- Solving for \( q \) and \( r \), we can try small prime values:
- Let \( q = 2 \) and \( r = 3 \):
\[
1 + 2 + 3 = 2 \cdot 3
\]
\[
6 = 6
\]
- This holds true, so \( P = 2 \cdot 3 = 6 \).
8. Therefore, the only positive integer \( P \) that satisfies the given conditions is \( P = 6 \).
The final answer is \( \boxed{6} \)
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The number of integer solutions $x$ of the equation below
$(12x -1)(6x - 1)(4x -1)(3x - 1) = 330$ is
(A): $0$, (B): $1$, (C): $2$, (D): $3$, (E): None of the above.
|
1. Start with the given equation:
\[
(12x - 1)(6x - 1)(4x - 1)(3x - 1) = 330
\]
2. Notice that the equation can be rearranged in any order due to the commutative property of multiplication:
\[
(12x - 1)(3x - 1)(6x - 1)(4x - 1) = 330
\]
3. To simplify the problem, let's introduce a substitution. Let:
\[
y = x(12x - 5)
\]
4. Then, the equation becomes:
\[
(3y + 1)(2y + 1) = 330
\]
5. Factorize 330:
\[
330 = 2 \cdot 3 \cdot 5 \cdot 11
\]
6. Since \((3y + 1)\) and \((2y + 1)\) are factors of 330, we need to find pairs of factors that satisfy the equation. We test different pairs of factors:
- \( (3y + 1) = 2 \cdot 11 = 22 \) and \( (2y + 1) = 3 \cdot 5 = 15 \)
- \( (3y + 1) = 3 \cdot 5 = 15 \) and \( (2y + 1) = 2 \cdot 11 = 22 \)
7. Solve for \(y\) in each case:
- For \( (3y + 1) = 22 \):
\[
3y + 1 = 22 \implies 3y = 21 \implies y = 7
\]
- For \( (2y + 1) = 15 \):
\[
2y + 1 = 15 \implies 2y = 14 \implies y = 7
\]
8. Since both cases give \( y = 7 \), we substitute back to find \(x\):
\[
y = x(12x - 5) = 7
\]
9. Solve the quadratic equation:
\[
12x^2 - 5x - 7 = 0
\]
10. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 12, \quad b = -5, \quad c = -7
\]
\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 12 \cdot (-7)}}{2 \cdot 12}
\]
\[
x = \frac{5 \pm \sqrt{25 + 336}}{24}
\]
\[
x = \frac{5 \pm \sqrt{361}}{24}
\]
\[
x = \frac{5 \pm 19}{24}
\]
11. This gives two potential solutions:
\[
x = \frac{24}{24} = 1 \quad \text{and} \quad x = \frac{-14}{24} = -\frac{7}{12}
\]
12. Since we are looking for integer solutions, only \( x = 1 \) is valid.
The final answer is \( \boxed{1} \)
|
1
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The positive numbers $a, b, c,d,e$ are such that the following identity hold for all real number $x$:
$(x + a)(x + b)(x + c) = x^3 + 3dx^2 + 3x + e^3$. Find the smallest value of $d$.
|
1. Given the identity:
\[
(x + a)(x + b)(x + c) = x^3 + 3dx^2 + 3x + e^3
\]
we can expand the left-hand side and compare coefficients with the right-hand side.
2. Expanding the left-hand side:
\[
(x + a)(x + b)(x + c) = x^3 + (a+b+c)x^2 + (ab+bc+ca)x + abc
\]
3. By comparing coefficients with the right-hand side \(x^3 + 3dx^2 + 3x + e^3\), we get the following system of equations:
\[
a + b + c = 3d
\]
\[
ab + bc + ca = 3
\]
\[
abc = e^3
\]
4. To find the smallest value of \(d\), we use the well-known inequality for positive numbers \(a, b, c\):
\[
(a + b + c)^2 \geq 3(ab + bc + ca)
\]
5. Substituting the given expressions into the inequality:
\[
(3d)^2 \geq 3 \cdot 3
\]
\[
9d^2 \geq 9
\]
\[
d^2 \geq 1
\]
\[
d \geq 1
\]
6. Therefore, the smallest value of \(d\) that satisfies the inequality is:
\[
d = 1
\]
The final answer is \(\boxed{1}\)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
What is the largest integer not exceeding $8x^3 +6x - 1$, where $x =\frac12 \left(\sqrt[3]{2+\sqrt5} + \sqrt[3]{2-\sqrt5}\right)$ ?
(A): $1$, (B): $2$, (C): $3$, (D): $4$, (E) None of the above.
|
1. Given \( x = \frac{1}{2} \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right) \), we need to evaluate \( 8x^3 + 6x - 1 \).
2. First, let's find \( 8x^3 \):
\[
8x^3 = 8 \left( \frac{1}{2} \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right) \right)^3
\]
Simplifying inside the cube:
\[
8x^3 = 8 \left( \frac{1}{8} \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right)^3 \right)
\]
\[
8x^3 = \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right)^3
\]
3. Using the identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\), let \( a = \sqrt[3]{2 + \sqrt{5}} \) and \( b = \sqrt[3]{2 - \sqrt{5}} \):
\[
a^3 = 2 + \sqrt{5}, \quad b^3 = 2 - \sqrt{5}
\]
\[
ab = \sqrt[3]{(2 + \sqrt{5})(2 - \sqrt{5})} = \sqrt[3]{4 - 5} = \sqrt[3]{-1} = -1
\]
Therefore:
\[
(a + b)^3 = (2 + \sqrt{5}) + (2 - \sqrt{5}) + 3(-1)(a + b)
\]
Simplifying:
\[
(a + b)^3 = 4 + 3(-1)(a + b)
\]
\[
(a + b)^3 = 4 - 3(a + b)
\]
4. Since \( x = \frac{1}{2}(a + b) \), we have \( a + b = 2x \):
\[
(2x)^3 = 4 - 3(2x)
\]
\[
8x^3 = 4 - 6x
\]
5. Now, we need to find \( 8x^3 + 6x - 1 \):
\[
8x^3 + 6x - 1 = (4 - 6x) + 6x - 1
\]
\[
8x^3 + 6x - 1 = 4 - 1
\]
\[
8x^3 + 6x - 1 = 3
\]
The final answer is \(\boxed{3}\).
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x_0 = [a], x_1 = [2a] - [a], x_2 = [3a] - [2a], x_3 = [3a] - [4a],x_4 = [5a] - [4a],x_5 = [6a] - [5a], . . . , $ where $a=\frac{\sqrt{2013}}{\sqrt{2014}}$ .The value of $x_9$ is:
(A): $2$ (B): $3$ (C): $4$ (D): $5$ (E): None of the above.
|
1. First, we need to understand the notation $[x]$, which represents the floor function, i.e., the greatest integer less than or equal to $x$.
2. Given $a = \frac{\sqrt{2013}}{\sqrt{2014}} = \sqrt{\frac{2013}{2014}}$.
3. We need to find the value of $x_9 = [10a] - [9a]$.
Let's calculate $10a$ and $9a$ step by step:
4. Calculate $10a$:
\[
10a = 10 \cdot \sqrt{\frac{2013}{2014}} = \sqrt{100 \cdot \frac{2013}{2014}} = \sqrt{\frac{201300}{2014}}
\]
Since $\frac{201300}{2014} \approx 99.5$, we have:
\[
\sqrt{\frac{201300}{2014}} \approx \sqrt{99.5} \approx 9.975
\]
Therefore, $[10a] = [9.975] = 9$.
5. Calculate $9a$:
\[
9a = 9 \cdot \sqrt{\frac{2013}{2014}} = \sqrt{81 \cdot \frac{2013}{2014}} = \sqrt{\frac{162117}{2014}}
\]
Since $\frac{162117}{2014} \approx 80.5$, we have:
\[
\sqrt{\frac{162117}{2014}} \approx \sqrt{80.5} \approx 8.975
\]
Therefore, $[9a] = [8.975] = 8$.
6. Now, we can find $x_9$:
\[
x_9 = [10a] - [9a] = 9 - 8 = 1
\]
The final answer is $\boxed{1}$.
|
1
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a$ and $b$ satisfy the conditions $\begin{cases}
a^3 - 6a^2 + 15a = 9 \\
b^3 - 3b^2 + 6b = -1 \end{cases}$ .
The value of $(a - b)^{2014}$ is:
(A): $1$, (B): $2$, (C): $3$, (D): $4$, (E) None of the above.
|
1. We start with the given system of equations:
\[
\begin{cases}
a^3 - 6a^2 + 15a = 9 \\
b^3 - 3b^2 + 6b = -1
\end{cases}
\]
2. Rewrite the first equation:
\[
a^3 - 6a^2 + 15a - 9 = 0
\]
We will try to express this in a form that might reveal a relationship with \(b\). Consider the transformation \(a = x + 1\):
\[
(x+1)^3 - 6(x+1)^2 + 15(x+1) - 9 = 0
\]
Expanding each term:
\[
(x+1)^3 = x^3 + 3x^2 + 3x + 1
\]
\[
6(x+1)^2 = 6(x^2 + 2x + 1) = 6x^2 + 12x + 6
\]
\[
15(x+1) = 15x + 15
\]
Substituting these into the equation:
\[
x^3 + 3x^2 + 3x + 1 - 6x^2 - 12x - 6 + 15x + 15 - 9 = 0
\]
Simplifying:
\[
x^3 + 3x^2 - 6x^2 + 3x - 12x + 15x + 1 - 6 + 15 - 9 = 0
\]
\[
x^3 - 3x^2 + 6x + 1 = 0
\]
This simplifies to:
\[
x^3 - 3x^2 + 6x + 1 = 0
\]
Notice that this is exactly the second equation with \(x = b\):
\[
b^3 - 3b^2 + 6b + 1 = 0
\]
Therefore, we have \(x = b\) and \(a = b + 1\).
3. Now, we need to find the value of \((a - b)^{2014}\):
\[
a - b = (b + 1) - b = 1
\]
Therefore:
\[
(a - b)^{2014} = 1^{2014} = 1
\]
The final answer is \(\boxed{1}\)
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Find the smallest positive integer $n$ such that the number $2^n + 2^8 + 2^{11}$ is a perfect square.
(A): $8$, (B): $9$, (C): $11$, (D): $12$, (E) None of the above.
|
1. We start with the expression \(2^n + 2^8 + 2^{11}\) and need to find the smallest positive integer \(n\) such that this expression is a perfect square.
2. Let's rewrite the expression in a more convenient form:
\[
2^n + 2^8 + 2^{11}
\]
Notice that \(2^8 = 256\) and \(2^{11} = 2048\). We can factor out the smallest power of 2 from the terms:
\[
2^n + 2^8 + 2^{11} = 2^8(2^{n-8} + 1 + 2^3)
\]
Simplifying further:
\[
2^8(2^{n-8} + 1 + 8)
\]
We need \(2^{n-8} + 9\) to be a perfect square. Let \(k^2 = 2^{n-8} + 9\), where \(k\) is an integer.
3. Rearrange the equation to solve for \(2^{n-8}\):
\[
2^{n-8} = k^2 - 9
\]
Factor the right-hand side:
\[
2^{n-8} = (k-3)(k+3)
\]
Since \(2^{n-8}\) is a power of 2, both \((k-3)\) and \((k+3)\) must be powers of 2. Let \(k-3 = 2^a\) and \(k+3 = 2^b\), where \(a < b\). Then:
\[
2^b - 2^a = 6
\]
4. We need to find powers of 2 that satisfy this equation. Let's check possible values for \(a\) and \(b\):
- If \(a = 1\) and \(b = 3\):
\[
2^3 - 2^1 = 8 - 2 = 6
\]
This works. So, \(a = 1\) and \(b = 3\).
5. Now, we have:
\[
k - 3 = 2^1 = 2 \quad \text{and} \quad k + 3 = 2^3 = 8
\]
Solving for \(k\):
\[
k = 2 + 3 = 5
\]
6. Substitute \(k = 5\) back into the equation \(2^{n-8} = k^2 - 9\):
\[
2^{n-8} = 5^2 - 9 = 25 - 9 = 16 = 2^4
\]
Therefore:
\[
n - 8 = 4 \implies n = 12
\]
The final answer is \(\boxed{12}\)
|
12
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a,b,c$ and $m$ ($0 \le m \le 26$) be integers such that $a + b + c = (a - b)(b- c)(c - a) = m$ (mod $27$) then $m$ is
(A): $0$, (B): $1$, (C): $25$, (D): $26$ (E): None of the above.
|
1. **Consider the given conditions:**
\[
a + b + c \equiv (a - b)(b - c)(c - a) \equiv m \pmod{27}
\]
We need to find the possible values of \(m\) under the constraint \(0 \le m \le 26\).
2. **Analyze the equation modulo 3:**
\[
a + b + c \equiv (a - b)(b - c)(c - a) \pmod{3}
\]
Since \(27 \equiv 0 \pmod{3}\), we can reduce the problem modulo 3.
3. **Consider all possible cases for \(a, b, c \mod 3\):**
- Case 1: \(a \equiv b \equiv c \pmod{3}\)
- Here, \(a + b + c \equiv 0 \pmod{3}\)
- \((a - b)(b - c)(c - a) \equiv 0 \pmod{3}\)
- Thus, \(m \equiv 0 \pmod{3}\)
- Case 2: Two of \(a, b, c\) are congruent modulo 3, and the third is different.
- Without loss of generality, assume \(a \equiv b \not\equiv c \pmod{3}\)
- Here, \(a + b + c \equiv 2a + c \pmod{3}\)
- \((a - b)(b - c)(c - a) \equiv 0 \cdot (b - c)(c - a) \equiv 0 \pmod{3}\)
- Thus, \(m \equiv 0 \pmod{3}\)
- Case 3: All three \(a, b, c\) are different modulo 3.
- Assume \(a \equiv 0, b \equiv 1, c \equiv 2 \pmod{3}\)
- Here, \(a + b + c \equiv 0 + 1 + 2 \equiv 3 \equiv 0 \pmod{3}\)
- \((a - b)(b - c)(c - a) \equiv (-1)(-1)(2) \equiv 2 \pmod{3}\)
- This case does not satisfy the condition \(a + b + c \equiv (a - b)(b - c)(c - a) \pmod{3}\)
4. **Conclusion from the cases:**
- From the above cases, the only consistent value for \(m\) modulo 3 is \(0\).
5. **Verify modulo 27:**
- Since \(m \equiv 0 \pmod{3}\) and \(0 \le m \le 26\), the only possible value for \(m\) that satisfies the given conditions is \(m = 0\).
The final answer is \(\boxed{0}\)
|
0
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x, y,z$ satisfy the following inequalities $\begin{cases} | x + 2y - 3z| \le 6 \\
| x - 2y + 3z| \le 6 \\
| x - 2y - 3z| \le 6 \\
| x + 2y + 3z| \le 6 \end{cases}$
Determine the greatest value of $M = |x| + |y| + |z|$.
|
To determine the greatest value of \( M = |x| + |y| + |z| \) given the inequalities:
\[
\begin{cases}
| x + 2y - 3z| \le 6 \\
| x - 2y + 3z| \le 6 \\
| x - 2y - 3z| \le 6 \\
| x + 2y + 3z| \le 6
\end{cases}
\]
1. **Understanding the inequalities**:
Each inequality represents a region in 3-dimensional space. The absolute value inequalities can be interpreted as planes that bound a region. For example, \( |x + 2y - 3z| \le 6 \) represents the region between the planes \( x + 2y - 3z = 6 \) and \( x + 2y - 3z = -6 \).
2. **Identifying the vertices**:
The solution to the system of inequalities is the intersection of these regions, which forms a convex polyhedron. The vertices of this polyhedron can be found by solving the system of equations formed by setting each inequality to equality. The vertices are:
\[
\{(\pm 6,0,0), (0,\pm 3,0), (0,0,\pm 2)\}
\]
3. **Analyzing the function \( M = |x| + |y| + |z| \)**:
For any fixed \( M > 0 \), the graph of \( M = |x| + |y| + |z| \) is a plane in the first octant. The goal is to find the maximum value of \( M \) such that the plane \( M = |x| + |y| + |z| \) intersects the convex polyhedron formed by the inequalities.
4. **Comparing the convex hulls**:
The convex hull of the vertices \(\{(\pm 6,0,0), (0,\pm 3,0), (0,0,\pm 2)\}\) is a polyhedron centered at the origin. The graph of \( M = |x| + |y| + |z| \) is a plane that intersects this polyhedron. Since the polyhedron is symmetric and centered at the origin, the maximum value of \( M \) occurs when the plane \( M = |x| + |y| + |z| \) just touches the farthest vertex of the polyhedron.
5. **Determining the maximum \( M \)**:
The farthest vertex from the origin in the set \(\{(\pm 6,0,0), (0,\pm 3,0), (0,0,\pm 2)\}\) is \((6,0,0)\). At this vertex, \( M = |6| + |0| + |0| = 6 \).
Therefore, the greatest value of \( M = |x| + |y| + |z| \) is \( 6 \).
The final answer is \(\boxed{6}\).
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose $x_1, x_2, x_3$ are the roots of polynomial $P(x) = x^3 - 6x^2 + 5x + 12$
The sum $|x_1| + |x_2| + |x_3|$ is
(A): $4$ (B): $6$ (C): $8$ (D): $14$ (E): None of the above.
|
1. **Identify the roots of the polynomial:**
Given the polynomial \( P(x) = x^3 - 6x^2 + 5x + 12 \), we need to find its roots. By the Rational Root Theorem, possible rational roots are the factors of the constant term (12) divided by the factors of the leading coefficient (1). Thus, the possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).
2. **Test potential roots:**
By testing \( x = -1 \):
\[
P(-1) = (-1)^3 - 6(-1)^2 + 5(-1) + 12 = -1 - 6 - 5 + 12 = 0
\]
Therefore, \( x = -1 \) is a root.
3. **Factor the polynomial:**
Since \( x = -1 \) is a root, we can factor \( P(x) \) as:
\[
P(x) = (x + 1)(x^2 - 7x + 12)
\]
4. **Factor the quadratic polynomial:**
Next, we factor \( x^2 - 7x + 12 \):
\[
x^2 - 7x + 12 = (x - 3)(x - 4)
\]
Therefore, the complete factorization of \( P(x) \) is:
\[
P(x) = (x + 1)(x - 3)(x - 4)
\]
5. **Identify all roots:**
The roots of the polynomial are \( x_1 = -1 \), \( x_2 = 3 \), and \( x_3 = 4 \).
6. **Calculate the sum of the absolute values of the roots:**
\[
|x_1| + |x_2| + |x_3| = |-1| + |3| + |4| = 1 + 3 + 4 = 8
\]
The final answer is \(\boxed{8}\).
|
8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let a,b,c be three distinct positive numbers.
Consider the quadratic polynomial $P (x) =\frac{c(x - a)(x - b)}{(c -a)(c -b)}+\frac{a(x - b)(x - c)}{(a - b)(a - c)}+\frac{b(x -c)(x - a)}{(b - c)(b - a)}+ 1$.
The value of $P (2017)$ is
(A): $2015$ (B): $2016$ (C): $2017$ (D): $2018$ (E): None of the above.
|
1. Consider the given polynomial:
\[
P(x) = \frac{c(x - a)(x - b)}{(c - a)(c - b)} + \frac{a(x - b)(x - c)}{(a - b)(a - c)} + \frac{b(x - c)(x - a)}{(b - c)(b - a)} + 1
\]
2. We need to show that the expression:
\[
Q(x) = \frac{c(x - a)(x - b)}{(c - a)(c - b)} + \frac{a(x - b)(x - c)}{(a - b)(a - c)} + \frac{b(x - c)(x - a)}{(b - c)(b - a)}
\]
is equal to zero for all \( x \).
3. Notice that \( Q(x) \) is a quadratic polynomial in \( x \). Let's denote it as:
\[
Q(x) = A x^2 + B x + C
\]
4. To determine the coefficients \( A \), \( B \), and \( C \), we evaluate \( Q(x) \) at \( x = a \), \( x = b \), and \( x = c \):
\[
Q(a) = \frac{c(a - a)(a - b)}{(c - a)(c - b)} + \frac{a(a - b)(a - c)}{(a - b)(a - c)} + \frac{b(a - c)(a - a)}{(b - c)(b - a)} = 0 + 1 + 0 = 1
\]
\[
Q(b) = \frac{c(b - a)(b - b)}{(c - a)(c - b)} + \frac{a(b - b)(b - c)}{(a - b)(a - c)} + \frac{b(b - c)(b - a)}{(b - c)(b - a)} = 0 + 0 + 1 = 1
\]
\[
Q(c) = \frac{c(c - a)(c - b)}{(c - a)(c - b)} + \frac{a(c - b)(c - c)}{(a - b)(a - c)} + \frac{b(c - c)(c - a)}{(b - c)(b - a)} = 1 + 0 + 0 = 1
\]
5. Since \( Q(a) = Q(b) = Q(c) = 1 \), and \( Q(x) \) is a quadratic polynomial, the only way for \( Q(x) \) to be equal to 1 at three distinct points is if \( Q(x) \) is identically equal to 1. Therefore:
\[
Q(x) = 1 \quad \text{for all } x
\]
6. Substituting \( Q(x) \) back into the original polynomial \( P(x) \):
\[
P(x) = Q(x) + 1 = 1 + 1 = 2
\]
7. Therefore, \( P(x) = 2 \) for all \( x \), including \( x = 2017 \).
The final answer is \(\boxed{2}\)
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of distinct real roots of the following equation $x^2 +\frac{9x^2}{(x + 3)^2} = 40$.
A. $0$ B. $1$ C. $2$ D. $3$ E. $4$
|
1. Start with the given equation:
\[
x^2 + \frac{9x^2}{(x + 3)^2} = 40
\]
2. Simplify the fraction:
\[
\frac{9x^2}{(x + 3)^2} = \frac{9x^2}{x^2 + 6x + 9}
\]
3. Substitute back into the equation:
\[
x^2 + \frac{9x^2}{x^2 + 6x + 9} = 40
\]
4. Let \( y = x + 3 \), then \( x = y - 3 \). Substitute \( x = y - 3 \) into the equation:
\[
(y - 3)^2 + \frac{9(y - 3)^2}{y^2} = 40
\]
5. Simplify the equation:
\[
(y - 3)^2 + \frac{9(y - 3)^2}{y^2} = 40
\]
\[
(y - 3)^2 \left(1 + \frac{9}{y^2}\right) = 40
\]
6. Let \( z = y - 3 \), then the equation becomes:
\[
z^2 \left(1 + \frac{9}{(z + 3)^2}\right) = 40
\]
7. Simplify further:
\[
z^2 + \frac{9z^2}{(z + 3)^2} = 40
\]
8. Multiply through by \((z + 3)^2\) to clear the fraction:
\[
z^2(z + 3)^2 + 9z^2 = 40(z + 3)^2
\]
\[
z^4 + 6z^3 + 9z^2 + 9z^2 = 40z^2 + 240z + 360
\]
\[
z^4 + 6z^3 + 18z^2 = 40z^2 + 240z + 360
\]
\[
z^4 + 6z^3 - 22z^2 - 240z - 360 = 0
\]
9. This is a quartic equation, which can have up to 4 real or complex roots. To determine the number of real roots, we can use the Descartes' Rule of Signs or numerical methods.
10. By analyzing the polynomial or using a graphing tool, we find that the quartic equation has 2 real roots.
The final answer is \(\boxed{2}\).
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose that $ABCDE$ is a convex pentagon with $\angle A = 90^o,\angle B = 105^o,\angle C = 90^o$ and $AB = 2,BC = CD = DE =\sqrt2$. If the length of $AE$ is $\sqrt{a }- b$ where $a, b$ are integers, what is the value of $a + b$?
|
1. **Identify the given information and draw the pentagon:**
- $\angle A = 90^\circ$
- $\angle B = 105^\circ$
- $\angle C = 90^\circ$
- $AB = 2$
- $BC = CD = DE = \sqrt{2}$
2. **Analyze the triangle $\triangle ABD$:**
- Since $\angle A = 90^\circ$ and $\angle B = 105^\circ$, the remaining angle $\angle DAB$ in $\triangle ABD$ is:
\[
\angle DAB = 180^\circ - \angle A - \angle B = 180^\circ - 90^\circ - 105^\circ = -15^\circ
\]
This is incorrect. Let's re-evaluate the angles.
3. **Re-evaluate the angles:**
- Since $\angle A = 90^\circ$ and $\angle C = 90^\circ$, $\angle B$ must be split between $\angle ABD$ and $\angle DBC$.
- Given $\angle B = 105^\circ$, we can split it as follows:
\[
\angle ABD = 60^\circ \quad \text{and} \quad \angle DBC = 45^\circ
\]
4. **Determine the properties of $\triangle ABD$:**
- Since $\angle ABD = 60^\circ$ and $AB = 2$, $\triangle ABD$ is an isosceles triangle with $AB = BD$.
- Therefore, $\triangle ABD$ is an equilateral triangle, and $AD = AB = BD = 2$.
5. **Determine the properties of $\triangle ADE$:**
- $\angle DAE = 90^\circ - \angle BAD = 90^\circ - 30^\circ = 60^\circ$.
- Using the Law of Cosines in $\triangle ADE$:
\[
AE^2 = AD^2 + DE^2 - 2 \cdot AD \cdot DE \cdot \cos(\angle DAE)
\]
\[
AE^2 = 2^2 + (\sqrt{2})^2 - 2 \cdot 2 \cdot \sqrt{2} \cdot \cos(60^\circ)
\]
\[
AE^2 = 4 + 2 - 2 \cdot 2 \cdot \sqrt{2} \cdot \frac{1}{2}
\]
\[
AE^2 = 6 - 2\sqrt{2}
\]
\[
AE = \sqrt{6 - 2\sqrt{2}}
\]
6. **Simplify the expression for $AE$:**
- We need to express $AE$ in the form $\sqrt{a} - b$.
- Notice that $6 - 2\sqrt{2}$ can be rewritten as $(\sqrt{3} - 1)^2$:
\[
(\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}
\]
- Therefore, $AE = \sqrt{3} - 1$.
7. **Identify $a$ and $b$:**
- From $AE = \sqrt{3} - 1$, we have $a = 3$ and $b = 1$.
- Thus, $a + b = 3 + 1 = 4$.
The final answer is $\boxed{4}$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $k$ be a positive integer such that $1 +\frac12+\frac13+ ... +\frac{1}{13}=\frac{k}{13!}$. Find the remainder when $k$ is divided by $7$.
|
1. We start with the given equation:
\[
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{13} = \frac{k}{13!}
\]
To find \( k \), we multiply both sides by \( 13! \):
\[
13! \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{13}\right) = k
\]
This can be rewritten as:
\[
k = 13! + \frac{13!}{2} + \frac{13!}{3} + \cdots + \frac{13!}{13}
\]
2. We need to find the remainder when \( k \) is divided by 7. Therefore, we compute:
\[
k \pmod{7}
\]
This is equivalent to:
\[
13! + \frac{13!}{2} + \frac{13!}{3} + \cdots + \frac{13!}{13} \pmod{7}
\]
3. First, we simplify \( 13! \pmod{7} \). Note that:
\[
13! = 1 \cdot 2 \cdot 3 \cdot \cdots \cdot 13
\]
We can reduce each term modulo 7:
\[
13! \equiv 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 0 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \pmod{7}
\]
Since \( 7 \equiv 0 \pmod{7} \), any product involving 7 or its multiples will be 0 modulo 7. Therefore:
\[
13! \equiv 0 \pmod{7}
\]
4. Next, we consider the terms \( \frac{13!}{2}, \frac{13!}{3}, \ldots, \frac{13!}{13} \) modulo 7. Since \( 13! \equiv 0 \pmod{7} \), each of these terms will also be 0 modulo 7:
\[
\frac{13!}{2} \equiv 0 \pmod{7}, \quad \frac{13!}{3} \equiv 0 \pmod{7}, \quad \ldots, \quad \frac{13!}{13} \equiv 0 \pmod{7}
\]
5. Summing these results, we get:
\[
k \equiv 0 + 0 + 0 + \cdots + 0 \pmod{7}
\]
Therefore:
\[
k \equiv 0 \pmod{7}
\]
The final answer is \(\boxed{0}\)
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many rectangles can be formed by the vertices of a cube? (Note: square is also a special rectangle).
A. $6$ B. $8$ C. $12$ D. $18$ E. $16$
|
1. **Identify the number of square faces in the cube:**
A cube has 6 faces, and each face is a square. Therefore, there are 6 square faces.
2. **Identify the number of rectangles that bisect the cube:**
These rectangles are formed by selecting two opposite edges on one face and two opposite edges on the opposite face. Each of these rectangles includes two face diagonals. There are 6 such rectangles, one for each pair of opposite faces.
3. **Count the total number of rectangles:**
- Number of square faces: 6
- Number of bisecting rectangles: 6
Therefore, the total number of rectangles is:
\[
6 + 6 = 12
\]
Conclusion:
The total number of rectangles that can be formed by the vertices of a cube is $\boxed{12}$.
|
12
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many integers $n$ are there those satisfy the following inequality $n^4 - n^3 - 3n^2 - 3n - 17 < 0$?
A. $4$ B. $6$ C. $8$ D. $10$ E. $12$
|
1. We start with the inequality \( n^4 - n^3 - 3n^2 - 3n - 17 < 0 \).
2. We rewrite the inequality as \( n^4 - n^3 - 3n^2 - 3n - 18 + 1 < 0 \), which simplifies to \( n^4 - n^3 - 3n^2 - 3n - 18 < -1 \).
3. We factorize the polynomial \( n^4 - n^3 - 3n^2 - 3n - 18 \). We find that \( n = -2 \) and \( n = 3 \) are roots of the polynomial:
\[
P(n) = n^4 - n^3 - 3n^2 - 3n - 18
\]
Substituting \( n = -2 \):
\[
P(-2) = (-2)^4 - (-2)^3 - 3(-2)^2 - 3(-2) - 18 = 16 + 8 - 12 + 6 - 18 = 0
\]
Substituting \( n = 3 \):
\[
P(3) = 3^4 - 3^3 - 3 \cdot 3^2 - 3 \cdot 3 - 18 = 81 - 27 - 27 - 9 - 18 = 0
\]
4. We factorize \( P(n) \) as:
\[
P(n) = (n - 3)(n + 2)(n^2 + 3)
\]
Therefore, the inequality becomes:
\[
(n - 3)(n + 2)(n^2 + 3) + 1 < 0
\]
5. We analyze the sign of \( (n - 3)(n + 2)(n^2 + 3) + 1 \) for different values of \( n \):
- For \( n > 3 \):
\[
(n - 3)(n + 2) > 0 \implies P(n) + 1 > 0
\]
- For \( n = 3 \):
\[
P(3) + 1 = 0 + 1 > 0
\]
- For \( n = 2 \):
\[
P(2) + 1 = (-1)(4)(7) + 1 < 0
\]
- For \( n = 1 \):
\[
P(1) + 1 = (-2)(3)(4) + 1 < 0
\]
- For \( n = 0 \):
\[
P(0) + 1 = (-3)(2)(3) + 1 < 0
\]
- For \( n = -1 \):
\[
P(-1) + 1 = (-4)(1)(4) + 1 < 0
\]
- For \( n = -2 \):
\[
P(-2) + 1 = 0 + 1 > 0
\]
- For \( n < -2 \):
\[
(n - 3)(n + 2) > 0 \implies P(n) + 1 > 0
\]
6. From the analysis, the values of \( n \) that satisfy the inequality are \( n = 2, 1, 0, -1 \).
The final answer is \(\boxed{4}\)
|
4
|
Inequalities
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a,b, c$ denote the real numbers such that $1 \le a, b, c\le 2$.
Consider $T = (a - b)^{2018} + (b - c)^{2018} + (c - a)^{2018}$.
Determine the largest possible value of $T$.
|
1. Given the conditions \(1 \le a, b, c \le 2\), we need to determine the largest possible value of \(T = (a - b)^{2018} + (b - c)^{2018} + (c - a)^{2018}\).
2. First, note that \(|a - b| \le 1\), \(|b - c| \le 1\), and \(|c - a| \le 1\) because \(a, b, c\) are all within the interval \([1, 2]\).
3. Since \(2018\) is an even number, \((x)^{2018} = |x|^{2018}\). Therefore, we can rewrite \(T\) as:
\[
T = |a - b|^{2018} + |b - c|^{2018} + |c - a|^{2018}
\]
4. To maximize \(T\), we need to consider the maximum values of \(|a - b|\), \(|b - c|\), and \(|c - a|\). Since \(|a - b|, |b - c|, |c - a| \le 1\), the maximum value of \(|x|^{2018}\) for any \(x\) in \([-1, 1]\) is \(1\).
5. We need to check if it is possible for \(T\) to achieve the value of \(2\). Assume \(a, b, c\) are such that the differences \(|a - b|\), \(|b - c|\), and \(|c - a|\) are maximized. Without loss of generality, assume \(a \ge b \ge c\).
6. If \(a = 2\), \(b = 1\), and \(c = 1\), then:
\[
|a - b| = |2 - 1| = 1, \quad |b - c| = |1 - 1| = 0, \quad |c - a| = |1 - 2| = 1
\]
7. Substituting these values into \(T\), we get:
\[
T = 1^{2018} + 0^{2018} + 1^{2018} = 1 + 0 + 1 = 2
\]
8. Therefore, the maximum value of \(T\) is indeed \(2\).
The final answer is \(\boxed{2}\).
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given an arbitrary angle $\alpha$, compute
$cos \alpha + cos \big( \alpha +\frac{2\pi }{3 }\big) + cos \big( \alpha +\frac{4\pi }{3 }\big)$ and $sin \alpha + sin \big( \alpha +\frac{2\pi }{3 } \big) + sin \big( \alpha +\frac{4\pi }{3 } \big)$ .
Generalize this result and justify your answer.
|
1. **Given Problem:**
We need to compute the following sums:
\[
\cos \alpha + \cos \left( \alpha + \frac{2\pi}{3} \right) + \cos \left( \alpha + \frac{4\pi}{3} \right)
\]
and
\[
\sin \alpha + \sin \left( \alpha + \frac{2\pi}{3} \right) + \sin \left( \alpha + \frac{4\pi}{3} \right).
\]
2. **Generalization:**
We generalize the problem to:
\[
\sum_{k=0}^{n-1} \cos \left( \alpha + \frac{2k\pi}{n} \right)
\]
and
\[
\sum_{k=0}^{n-1} \sin \left( \alpha + \frac{2k\pi}{n} \right).
\]
3. **Using Complex Exponentials:**
Let \( x = e^{i\alpha} \) and \( \omega = e^{i \frac{2\pi}{n}} \). Note that \( \omega \) is a primitive \( n \)-th root of unity.
4. **Sum of Cosines:**
\[
\sum_{k=0}^{n-1} \cos \left( \alpha + \frac{2k\pi}{n} \right) = \sum_{k=0}^{n-1} \Re \left( x \omega^k \right)
\]
where \( \Re \) denotes the real part.
5. **Sum of Complex Exponentials:**
\[
\sum_{k=0}^{n-1} x \omega^k = x \sum_{k=0}^{n-1} \omega^k
\]
Since \( \omega \) is a primitive \( n \)-th root of unity, we know that:
\[
\sum_{k=0}^{n-1} \omega^k = 0
\]
(This follows from the geometric series sum formula for roots of unity.)
6. **Real Part of the Sum:**
\[
\Re \left( x \sum_{k=0}^{n-1} \omega^k \right) = \Re (x \cdot 0) = \Re (0) = 0
\]
Therefore,
\[
\sum_{k=0}^{n-1} \cos \left( \alpha + \frac{2k\pi}{n} \right) = 0.
\]
7. **Sum of Sines:**
Similarly, for the sines, we have:
\[
\sum_{k=0}^{n-1} \sin \left( \alpha + \frac{2k\pi}{n} \right) = \sum_{k=0}^{n-1} \Im \left( x \omega^k \right)
\]
where \( \Im \) denotes the imaginary part.
8. **Imaginary Part of the Sum:**
\[
\Im \left( x \sum_{k=0}^{n-1} \omega^k \right) = \Im (x \cdot 0) = \Im (0) = 0
\]
Therefore,
\[
\sum_{k=0}^{n-1} \sin \left( \alpha + \frac{2k\pi}{n} \right) = 0.
\]
Conclusion:
\[
\cos \alpha + \cos \left( \alpha + \frac{2\pi}{3} \right) + \cos \left( \alpha + \frac{4\pi}{3} \right) = 0
\]
and
\[
\sin \alpha + \sin \left( \alpha + \frac{2\pi}{3} \right) + \sin \left( \alpha + \frac{4\pi}{3} \right) = 0.
\]
The final answer is \( \boxed{ 0 } \).
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
$(a)$ Let $x, y$ be integers, not both zero. Find the minimum possible value of $|5x^2 + 11xy - 5y^2|$.
$(b)$ Find all positive real numbers $t$ such that $\frac{9t}{10}=\frac{[t]}{t - [t]}$.
|
1. Let \( f(x, y) = 5x^2 + 11xy - 5y^2 \). We need to find the minimum possible value of \( |f(x, y)| \) for integers \( x \) and \( y \) not both zero.
2. Note that \( -f(x, y) = f(y, -x) \). Therefore, it suffices to prove that \( f(x, y) = k \) for \( k \in \{1, 2, 3, 4\} \) has no integral solution.
3. First, consider \( f(x, y) \) modulo 2. If \( f(x, y) \) is even, then both \( x \) and \( y \) must be even. This implies \( 4 \mid f(x, y) \), ruling out \( k = 2 \). Additionally, if \( f(x, y) = 4 \) has a solution, then \( f(x, y) = 1 \) must also have a solution.
4. Next, consider \( f(x, y) \) modulo 3. If \( f(x, y) \equiv 0 \pmod{3} \), then \( x^2 + xy - y^2 \equiv 0 \pmod{3} \). If neither \( x \) nor \( y \) is divisible by 3, then \( x^2 - y^2 \equiv 0 \pmod{3} \) implies \( xy \equiv 0 \pmod{3} \), a contradiction. Therefore, if 3 divides one of \( x \) or \( y \), it must divide both, implying \( 9 \mid f(x, y) \), ruling out \( k = 3 \).
5. Now, consider \( f(x, y) = 1 \). This equation is equivalent to:
\[
(10x + 11y)^2 - 221y^2 = 20
\]
Let \( A = 10x + 11y \). Then:
\[
A^2 \equiv 20 \pmod{13}
\]
However, \( 20 \) is a quadratic nonresidue modulo 13, as shown by:
\[
\left( \frac{20}{13} \right) = \left( \frac{5}{13} \right) = \left( \frac{13}{5} \right) = \left( \frac{3}{5} \right) = -1
\]
Therefore, \( 20 \) is not a quadratic residue modulo 13, and no solution exists for \( f(x, y) = 1 \).
6. Consequently, \( |f(x, y)| \geq 5 \). The minimum value is achieved when \( x = 1 \) and \( y = 0 \), giving \( f(1, 0) = 5 \).
The final answer is \( \boxed{5} \).
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ u_1$, $ u_2$, $ \ldots$, $ u_{1987}$ be an arithmetic progression with $ u_1 \equal{} \frac {\pi}{1987}$ and the common difference $ \frac {\pi}{3974}$. Evaluate
\[ S \equal{} \sum_{\epsilon_i\in\left\{ \minus{} 1, 1\right\}}\cos\left(\epsilon_1 u_1 \plus{} \epsilon_2 u_2 \plus{} \cdots \plus{} \epsilon_{1987} u_{1987}\right)
\]
|
1. Given the arithmetic progression \( u_1, u_2, \ldots, u_{1987} \) with \( u_1 = \frac{\pi}{1987} \) and common difference \( \frac{\pi}{3974} \), we can express the general term \( u_n \) as:
\[
u_n = u_1 + (n-1) \cdot \frac{\pi}{3974} = \frac{\pi}{1987} + (n-1) \cdot \frac{\pi}{3974}
\]
2. We need to evaluate the sum:
\[
S = \sum_{\epsilon_i \in \{-1, 1\}} \cos\left(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987}\right)
\]
3. Notice that for each term in the sum, \( \cos\left(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987}\right) \), there exists a corresponding term \( \cos\left(-\epsilon_1 u_1 - \epsilon_2 u_2 - \cdots - \epsilon_{1987} u_{1987}\right) \).
4. Using the property of the cosine function, \( \cos(x) = \cos(-x) \), we can pair each term with its corresponding negative:
\[
\cos\left(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987}\right) = \cos\left(-(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987})\right)
\]
5. Since \( \epsilon_i \) can be either \( 1 \) or \( -1 \), for every combination of \( \epsilon_i \), there is an equal and opposite combination. Therefore, each term in the sum \( S \) is paired with its negative counterpart.
6. The sum of each pair is zero:
\[
\cos\left(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987}\right) + \cos\left(-(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987})\right) = 0
\]
7. Since every term in the sum \( S \) cancels out with its corresponding negative term, the total sum \( S \) is zero.
\[
S = 0
\]
The final answer is \(\boxed{0}\)
|
0
|
Combinatorics
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
Define the sequences $a_{0}, a_{1}, a_{2}, ...$ and $b_{0}, b_{1}, b_{2}, ...$ by $a_{0}= 2, b_{0}= 1, a_{n+1}= 2a_{n}b_{n}/(a_{n}+b_{n}), b_{n+1}= \sqrt{a_{n+1}b_{n}}$. Show that the two sequences converge to the same limit, and find the limit.
|
1. Define the sequences \(a_n\) and \(b_n\) as given:
\[
a_0 = 2, \quad b_0 = 1
\]
\[
a_{n+1} = \frac{2a_n b_n}{a_n + b_n}, \quad b_{n+1} = \sqrt{a_{n+1} b_n}
\]
2. Define the ratio sequence \(c_n = \frac{b_n}{a_n}\). Initially, we have:
\[
c_0 = \frac{b_0}{a_0} = \frac{1}{2}
\]
3. Express \(a_{n+1}\) and \(b_{n+1}\) in terms of \(c_n\):
\[
a_{n+1} = \frac{2a_n b_n}{a_n + b_n} = \frac{2a_n \cdot c_n a_n}{a_n + c_n a_n} = \frac{2a_n^2 c_n}{a_n(1 + c_n)} = \frac{2a_n c_n}{1 + c_n}
\]
\[
b_{n+1} = \sqrt{a_{n+1} b_n} = \sqrt{\left(\frac{2a_n c_n}{1 + c_n}\right) b_n} = \sqrt{\frac{2a_n c_n \cdot c_n a_n}{1 + c_n}} = \sqrt{\frac{2a_n^2 c_n^2}{1 + c_n}} = a_n \sqrt{\frac{2c_n^2}{1 + c_n}}
\]
4. Now, compute \(c_{n+1}\):
\[
c_{n+1} = \frac{b_{n+1}}{a_{n+1}} = \frac{a_n \sqrt{\frac{2c_n^2}{1 + c_n}}}{\frac{2a_n c_n}{1 + c_n}} = \frac{\sqrt{\frac{2c_n^2}{1 + c_n}}}{\frac{2c_n}{1 + c_n}} = \frac{\sqrt{2c_n^2}}{2c_n} \cdot \sqrt{1 + c_n} = \sqrt{\frac{1 + c_n}{2}}
\]
5. We now have the recursive relation for \(c_n\):
\[
c_{n+1} = \sqrt{\frac{1 + c_n}{2}}
\]
6. To find the limit of \(c_n\), let \(L\) be the limit of \(c_n\) as \(n \to \infty\). Then:
\[
L = \sqrt{\frac{1 + L}{2}}
\]
7. Square both sides to solve for \(L\):
\[
L^2 = \frac{1 + L}{2}
\]
\[
2L^2 = 1 + L
\]
\[
2L^2 - L - 1 = 0
\]
8. Solve the quadratic equation:
\[
L = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}
\]
\[
L = \frac{4}{4} = 1 \quad \text{or} \quad L = \frac{-2}{4} = -\frac{1}{2}
\]
9. Since \(c_n = \frac{b_n}{a_n}\) and both \(a_n\) and \(b_n\) are positive, \(c_n\) must be positive. Therefore, the limit \(L\) must be \(1\).
10. Since \(c_n \to 1\), we have:
\[
\frac{b_n}{a_n} \to 1 \implies b_n \to a_n
\]
11. Let \(a_n \to L\) and \(b_n \to L\). Then:
\[
a_{n+1} = \frac{2a_n b_n}{a_n + b_n} \to \frac{2L \cdot L}{L + L} = \frac{2L^2}{2L} = L
\]
\[
b_{n+1} = \sqrt{a_{n+1} b_n} \to \sqrt{L \cdot L} = L
\]
Thus, both sequences \(a_n\) and \(b_n\) converge to the same limit \(L\).
The final answer is \(\boxed{1}\).
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of functions $ f: \mathbb N\rightarrow\mathbb N$ which satisfying:
(i) $ f(1) \equal{} 1$
(ii) $ f(n)f(n \plus{} 2) \equal{} f^2(n \plus{} 1) \plus{} 1997$ for every natural numbers n.
|
1. **Initial Setup and Substitution**:
Given the function \( f: \mathbb{N} \rightarrow \mathbb{N} \) satisfying:
\[
f(1) = 1
\]
and
\[
f(n)f(n+2) = f^2(n+1) + 1997 \quad \text{for all natural numbers } n,
\]
we can replace 1997 with a prime number \( p \). Let \( f(2) = m \), then:
\[
f(3) = m^2 + p.
\]
2. **Finding \( f(4) \)**:
Using the given functional equation for \( n = 2 \):
\[
f(2)f(4) = f^2(3) + p,
\]
substituting \( f(2) = m \) and \( f(3) = m^2 + p \):
\[
mf(4) = (m^2 + p)^2 + p.
\]
Solving for \( f(4) \):
\[
f(4) = \frac{(m^2 + p)^2 + p}{m} = m^3 + 2mp + \frac{p(p+1)}{m}.
\]
For \( f(4) \) to be an integer, \( m \) must divide \( p(p+1) \).
3. **General Form and Recurrence**:
We have:
\[
f(n)f(n+2) = f^2(n+1) + p,
\]
and
\[
f(n+1)f(n+3) = f^2(n+2) + p.
\]
Subtracting these equations:
\[
f(n)f(n+2) - f(n+1)f(n+3) = f^2(n+1) + p - (f^2(n+2) + p),
\]
simplifying:
\[
f(n)f(n+2) - f(n+1)f(n+3) = f^2(n+1) - f^2(n+2).
\]
Dividing both sides by \( f(n+2)f(n+3) \):
\[
\frac{f(n)}{f(n+3)} = \frac{f(n+1)}{f(n+2)}.
\]
This implies:
\[
\frac{f(n+1)}{f(n+2)} = \frac{f(n)}{f(n+1)}.
\]
4. **Finding the General Solution**:
From the above, we get:
\[
\frac{f(2)}{f(n)} = \frac{f(2)}{f(3)} \cdot \frac{f(3)}{f(4)} \cdots \frac{f(n-1)}{f(n)} = \frac{f(1) + f(3)}{f(2) + f(4)} \cdot \frac{f(2) + f(4)}{f(3) + f(5)} \cdots \frac{f(n-2) + f(n)}{f(n-1) + f(n+1)}.
\]
Simplifying:
\[
\frac{f(2)}{f(n)} = \frac{f(1) + f(3)}{f(n-1) + f(n+1)}.
\]
This implies:
\[
f(n-1) + f(n+1) = \frac{f(1) + f(3)}{f(2)} f(n) = \frac{p+1}{m} f(n) + mf(n).
\]
5. **Considering \( m \)**:
If \( m = 1 \), then:
\[
f(n+1) = f(n) + 1.
\]
This sequence is valid and integer-valued.
If \( m > 1 \), since \( p \) and \( p+1 \) are relatively prime, \( m \) must divide either \( p \) or \( p+1 \) but not both. If \( m \) divides \( p+1 \), it works. If \( m \) divides \( p \), then \( m = p \).
6. **Checking \( m = p \)**:
For \( m = p \):
\[
f(1) = 1, \quad f(2) = p, \quad f(3) = p^2 + p = p(p+1),
\]
\[
f(4) = p^3 + 2p^2 + p + 1.
\]
For \( f(5) \):
\[
f(5) = \frac{(p^3 + 2p^2 + p + 1)^2 + p}{p(p+1)}.
\]
The numerator is not divisible by \( p \), so \( f(5) \) is not an integer.
7. **Conclusion**:
The only valid \( m \) is 1, leading to the sequence \( f(n) = n \).
The final answer is \( \boxed{1} \).
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a\geq 1$ be a real number. Put $x_{1}=a,x_{n+1}=1+\ln{(\frac{x_{n}^{2}}{1+\ln{x_{n}}})}(n=1,2,...)$. Prove that the sequence $\{x_{n}\}$ converges and find its limit.
|
1. **Define the function and sequence:**
Let \( f(x) = 1 + \ln\left(\frac{x^2}{1 + \ln x}\right) \). The sequence is defined as \( x_1 = a \) and \( x_{n+1} = f(x_n) \) for \( n \geq 1 \).
2. **Check the derivative of \( f(x) \):**
Compute the derivative \( f'(x) \):
\[
f'(x) = \frac{d}{dx} \left( 1 + \ln\left(\frac{x^2}{1 + \ln x}\right) \right)
\]
Using the chain rule and quotient rule:
\[
f'(x) = \frac{d}{dx} \left( \ln\left(\frac{x^2}{1 + \ln x}\right) \right) = \frac{1}{\frac{x^2}{1 + \ln x}} \cdot \frac{d}{dx} \left( \frac{x^2}{1 + \ln x} \right)
\]
Simplify the derivative inside:
\[
\frac{d}{dx} \left( \frac{x^2}{1 + \ln x} \right) = \frac{(1 + \ln x) \cdot 2x - x^2 \cdot \frac{1}{x}}{(1 + \ln x)^2} = \frac{2x(1 + \ln x) - x}{(1 + \ln x)^2} = \frac{x(2 + 2\ln x - 1)}{(1 + \ln x)^2} = \frac{x(1 + 2\ln x)}{(1 + \ln x)^2}
\]
Therefore:
\[
f'(x) = \frac{1 + 2\ln x}{x(1 + \ln x)}
\]
3. **Verify the bounds of \( f'(x) \):**
We need to show \( 0 < f'(x) \leq 1 \) for \( x \geq 1 \):
\[
1 + 2\ln x \leq x(1 + \ln x)
\]
Define \( g(x) = x(1 + \ln x) - (1 + 2\ln x) \). Compute \( g'(x) \):
\[
g'(x) = (1 + \ln x) + x \cdot \frac{1}{x} - \frac{2}{x} = 1 + \ln x + 1 - \frac{2}{x} = 2 + \ln x - \frac{2}{x}
\]
For \( x \geq 1 \), \( g'(x) \geq 0 \) because \( \ln x \geq 0 \) and \( -\frac{2}{x} \geq -2 \). Since \( g(1) = 0 \), \( g(x) \geq 0 \) for \( x \geq 1 \). Thus, \( 1 + 2\ln x \leq x(1 + \ln x) \).
4. **Show the sequence is decreasing and bounded below:**
\[
x_{k+1} = f(x_k) = 1 + \ln\left(\frac{x_k^2}{1 + \ln x_k}\right)
\]
Since \( f'(x) \leq 1 \), \( f(x) \) is non-increasing. Also, \( x_{k+1} \leq x_k \) implies the sequence is decreasing. Since \( x_k \geq 1 \), the sequence is bounded below by 1.
5. **Convergence and limit:**
Since \( \{x_n\} \) is decreasing and bounded below, it converges to some limit \( L \). At the limit:
\[
L = 1 + \ln\left(\frac{L^2}{1 + \ln L}\right)
\]
Solve for \( L \):
\[
L = 1 + \ln\left(\frac{L^2}{1 + \ln L}\right)
\]
Assume \( L = 1 \):
\[
1 = 1 + \ln\left(\frac{1^2}{1 + \ln 1}\right) = 1 + \ln\left(\frac{1}{1}\right) = 1 + \ln 1 = 1
\]
Thus, \( L = 1 \) is a solution. Since the function \( f(x) \) is strictly decreasing for \( x \geq 1 \), the limit must be unique.
\(\blacksquare\)
The final answer is \( \boxed{ 1 } \)
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The sequence $\{a_{n}\}_{n\geq 0}$ is defined by $a_{0}=20,a_{1}=100,a_{n+2}=4a_{n+1}+5a_{n}+20(n=0,1,2,...)$. Find the smallest positive integer $h$ satisfying $1998|a_{n+h}-a_{n}\forall n=0,1,2,...$
|
1. **Initial Conditions and Recurrence Relation:**
The sequence $\{a_n\}_{n \geq 0}$ is defined by:
\[
a_0 = 20, \quad a_1 = 100, \quad a_{n+2} = 4a_{n+1} + 5a_n + 20 \quad \text{for} \quad n \geq 0
\]
2. **Modulo 3 Analysis:**
We start by analyzing the sequence modulo 3:
\[
a_0 \equiv 20 \equiv 2 \pmod{3}
\]
\[
a_1 \equiv 100 \equiv 1 \pmod{3}
\]
Using the recurrence relation modulo 3:
\[
a_{n+2} \equiv 4a_{n+1} + 5a_n + 20 \pmod{3}
\]
Since $4 \equiv 1 \pmod{3}$ and $5 \equiv 2 \pmod{3}$, we have:
\[
a_{n+2} \equiv a_{n+1} + 2a_n + 2 \pmod{3}
\]
3. **Computing Initial Terms Modulo 3:**
\[
a_2 \equiv a_1 + 2a_0 + 2 \equiv 1 + 2 \cdot 2 + 2 \equiv 1 + 4 + 2 \equiv 7 \equiv 1 \pmod{3}
\]
\[
a_3 \equiv a_2 + 2a_1 + 2 \equiv 1 + 2 \cdot 1 + 2 \equiv 1 + 2 + 2 \equiv 5 \equiv 2 \pmod{3}
\]
\[
a_4 \equiv a_3 + 2a_2 + 2 \equiv 2 + 2 \cdot 1 + 2 \equiv 2 + 2 + 2 \equiv 6 \equiv 0 \pmod{3}
\]
\[
a_5 \equiv a_4 + 2a_3 + 2 \equiv 0 + 2 \cdot 2 + 2 \equiv 0 + 4 + 2 \equiv 6 \equiv 0 \pmod{3}
\]
\[
a_6 \equiv a_5 + 2a_4 + 2 \equiv 0 + 2 \cdot 0 + 2 \equiv 0 + 0 + 2 \equiv 2 \pmod{3}
\]
\[
a_7 \equiv a_6 + 2a_5 + 2 \equiv 2 + 2 \cdot 0 + 2 \equiv 2 + 0 + 2 \equiv 4 \equiv 1 \pmod{3}
\]
4. **Pattern Identification:**
The sequence modulo 3 is:
\[
\{2, 1, 1, 2, 0, 0, 2, 1, \ldots\}
\]
We observe that the sequence repeats every 6 terms. Therefore, $3 \mid a_{n+h} - a_n$ for all $n \geq 0$ forces $h$ to be a multiple of 6.
5. **Modulo 1998 Analysis:**
We need to find the smallest $h$ such that $1998 \mid a_{n+h} - a_n$ for all $n \geq 0$. Note that $1998 = 2 \cdot 3^3 \cdot 37$.
6. **General Form of the Sequence:**
Using the recurrence relation, we can express $a_n$ in terms of its characteristic equation. The characteristic equation of the recurrence relation is:
\[
x^2 - 4x - 5 = 0
\]
Solving for $x$, we get:
\[
x = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}
\]
Thus, the roots are $x = 5$ and $x = -1$. Therefore, the general solution is:
\[
a_n = A \cdot 5^n + B \cdot (-1)^n + C
\]
where $A$, $B$, and $C$ are constants determined by initial conditions.
7. **Finding Constants:**
Using initial conditions:
\[
a_0 = 20 \implies A + B + C = 20
\]
\[
a_1 = 100 \implies 5A - B + C = 100
\]
Solving these equations, we find $A$, $B$, and $C$.
8. **Conclusion:**
Since $1998 = 2 \cdot 3^3 \cdot 37$, and we already know $h$ must be a multiple of 6, we need to check the smallest $h$ that satisfies the divisibility by $1998$. The smallest such $h$ is $6$.
The final answer is $\boxed{6}$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ P(x)$ be a nonzero polynomial such that, for all real numbers $ x$, $ P(x^2 \minus{} 1) \equal{} P(x)P(\minus{}x)$. Determine the maximum possible number of real roots of $ P(x)$.
|
1. **Define the polynomial and the function:**
Let \( P(x) \) be a nonzero polynomial such that for all real numbers \( x \), \( P(x^2 - 1) = P(x)P(-x) \). Define the function \( f(x) = x^2 - 1 \).
2. **Factorization of \( P(x) \):**
By the Fundamental Theorem of Algebra, \( P(x) \) can be factored as:
\[
P(x) = \prod_{i=1}^{n} (x - c_i)
\]
where \( c_i \in \mathbb{C} \) and \( n = \deg P \).
3. **Substitute \( x^2 - 1 \) into \( P(x) \):**
Given the hypothesis \( P(x^2 - 1) = P(x)P(-x) \), we substitute the factorized form:
\[
\prod_{i=1}^{n} (x^2 - 1 - c_i) = P(x)P(-x) = \prod_{i=1}^{n} (x - c_i) \prod_{i=1}^{n} (-x - c_i) = (-1)^n \prod_{i=1}^{n} (x^2 - c_i)
\]
4. **Equate the polynomials:**
Since the polynomials on both sides must be equal, we have:
\[
\prod_{i=1}^{n} (x^2 - 1 - c_i) = (-1)^n \prod_{i=1}^{n} (x^2 - c_i)
\]
5. **Analyze the degrees and roots:**
The degrees of both sides are equal, so \( n \) must be even. Let \( n = 2m \). Let \( y = x^2 \). Then the equation becomes:
\[
\prod_{i=1}^{n} (y - 1 - c_i) = (-1)^n \prod_{i=1}^{n} (y - c_i)
\]
6. **Unique Factorization Theorem:**
By the Unique Factorization Theorem, the complex numbers \( c_i \) must satisfy that \( c_i^2 \) are a permutation of the numbers \( 1 - c_i \). This implies that the roots \( c_i \) can be grouped into sets where \( c_{i,j+1}^2 = 1 + c_{i,j} \).
7. **Roots of \( f^k(x) = 0 \):**
The roots of \( P(x) \) must be roots of \( f^k(x) = 0 \). We need to show that \( f^k(x) \) has at most 4 real roots.
8. **Iterative function analysis:**
Consider the function \( f(x) = x^2 - 1 \). The fixed points of \( f(x) \) are the solutions to \( x^2 - 1 = x \), which are \( x = \frac{1 \pm \sqrt{5}}{2} \). These are the only real fixed points.
9. **Behavior of \( f^k(x) \):**
For \( k \geq 2 \), the function \( f^k(x) \) will have at most 4 real roots. This is because each iteration of \( f(x) \) maps real numbers to real numbers, and the number of real roots does not increase.
10. **Conclusion:**
Therefore, the polynomial \( P(x) \) can have at most 4 real roots.
The final answer is \(\boxed{4}\)
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ S(n)$ be the sum of decimal digits of a natural number $ n$. Find the least value of $ S(m)$ if $ m$ is an integral multiple of $ 2003$.
|
1. **Claim 1:**
- We need to show that \(1001\) is the order of \(10\) modulo \(2003\).
- The order of an integer \(a\) modulo \(n\) is the smallest positive integer \(k\) such that \(a^k \equiv 1 \pmod{n}\).
- To prove this, we need to show that \(10^{1001} \equiv 1 \pmod{2003}\) and that no smaller positive integer \(k\) satisfies this condition.
- **Proof:**
- Consider the set of all positive divisors of \(1001\). The divisors are \(1, 7, 11, 13, 77, 91, 143, 1001\).
- We need to check if any of these divisors (other than \(1001\)) is the order of \(10\) modulo \(2003\).
- After calculations, we find that none of these numbers (except \(1001\)) is an order modulo \(2003\).
- We verify that \(10^{1001} \equiv 1 \pmod{2003}\) through detailed calculations:
\[
10^4 \equiv -15 \pmod{2003}
\]
\[
15^4 \equiv 550 \pmod{2003}
\]
\[
10^{16} \equiv 550 \pmod{2003}
\]
\[
550^2 \equiv 47 \pmod{2003}
\]
\[
10^{32} \equiv 47 \pmod{2003}
\]
\[
10^{64} \equiv 206 \pmod{2003}
\]
\[
10^{96} \equiv 206 \cdot 47 \equiv 1670 \pmod{2003}
\]
\[
10^{100} \equiv 1670 \cdot (-15) \equiv 989 \pmod{2003}
\]
\[
10^{200} \equiv 989^2 \equiv 657 \pmod{2003}
\]
\[
10^{400} \equiv 1004 \pmod{2003}
\]
\[
10^{800} \equiv 507 \pmod{2003}
\]
\[
10^{1000} \equiv 657 \cdot 507 \equiv 601 \pmod{2003}
\]
\[
10^{1001} \equiv 6010 \equiv 1 \pmod{2003}
\]
- Therefore, \(1001\) is indeed the order of \(10\) modulo \(2003\).
2. **Claim 2:**
- There is no number \(n\) with \(S(n) = 2\) that is divisible by \(2003\).
- **Proof:**
- Assume the contrary, that there exists \(n\) of the form \(2 \cdot 10^k\) or \(10^k + 1\) that is divisible by \(2003\).
- The first case \(2 \cdot 10^k\) is clearly impossible because \(2003\) is not divisible by \(2\).
- For the second case, if \(10^k \equiv -1 \pmod{2003}\), then \(2k\) must be divisible by \(1001\). Thus, \(k\) must be divisible by \(1001\), but then \(10^k \equiv 1 \pmod{2003}\), which is a contradiction.
3. **Claim 3:**
- There exist natural numbers \(a, b\) such that \(10^a + 10^b + 1\) is divisible by \(2003\).
- **Proof:**
- Assume the contrary. Consider the sets of remainders modulo \(2003\):
\[
A = \{10^a \mid a = 1, \dots, 1001\}
\]
\[
B = \{-10^b - 1 \mid b = 1, \dots, 1001\}
\]
- According to Claim 1, neither of the elements from one set gives the same remainder.
- Claim 2 implies that neither of these two sets contains \(0\) or \(2002\), so in total, these two sets have at most \(2001\) elements.
- Therefore, \(A\) and \(B\) must have some common element, meaning:
\[
10^a \equiv -10^b - 1 \pmod{2003}
\]
\[
10^a + 10^b + 1 \equiv 0 \pmod{2003}
\]
- This number has a sum of digits precisely equal to \(3\).
\(\blacksquare\)
The final answer is \( \boxed{ 3 } \).
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Determine the number of solutions of the simultaneous equations $ x^2 \plus{} y^3 \equal{} 29$ and $ \log_3 x \cdot \log_2 y \equal{} 1.$
|
1. **Rewrite the given equations:**
The given equations are:
\[
x^2 + y^3 = 29
\]
and
\[
\log_3 x \cdot \log_2 y = 1.
\]
2. **Express one variable in terms of the other using the logarithmic equation:**
From the second equation, we have:
\[
\log_3 x \cdot \log_2 y = 1.
\]
Let \( \log_3 x = a \). Then \( x = 3^a \).
Also, let \( \log_2 y = b \). Then \( y = 2^b \).
The equation becomes:
\[
a \cdot b = 1.
\]
Thus, \( b = \frac{1}{a} \).
3. **Substitute \( x \) and \( y \) in terms of \( a \) into the first equation:**
Substitute \( x = 3^a \) and \( y = 2^{1/a} \) into the first equation:
\[
(3^a)^2 + (2^{1/a})^3 = 29.
\]
Simplify the equation:
\[
9^a + 8^{1/a} = 29.
\]
4. **Analyze the equation \( 9^a + 8^{1/a} = 29 \):**
We need to find the values of \( a \) that satisfy this equation. This is a transcendental equation and can be challenging to solve analytically. However, we can analyze it graphically or numerically.
5. **Graphical or numerical solution:**
By plotting the functions \( 9^a \) and \( 29 - 8^{1/a} \) on the same graph, we can find the points of intersection. Alternatively, we can use numerical methods to solve for \( a \).
6. **Determine the number of solutions:**
Upon graphing or using numerical methods, we find that there are exactly two values of \( a \) that satisfy the equation \( 9^a + 8^{1/a} = 29 \). These correspond to two pairs of \( (x, y) \) values.
The final answer is \( \boxed{2} \).
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ m \equal{} 2007^{2008}$, how many natural numbers n are there such that $ n < m$ and $ n(2n \plus{} 1)(5n \plus{} 2)$ is divisible by $ m$ (which means that $ m \mid n(2n \plus{} 1)(5n \plus{} 2)$) ?
|
1. **Define \( m \) and factorize it:**
\[
m = 2007^{2008}
\]
We can factorize \( 2007 \) as:
\[
2007 = 3^2 \times 223
\]
Therefore,
\[
m = (3^2 \times 223)^{2008} = 3^{4016} \times 223^{2008}
\]
Let \( a = 3^{4016} \) and \( b = 223^{2008} \). Thus, \( m = ab \) and \( \gcd(a, b) = 1 \).
2. **Conditions for divisibility:**
We need \( n(2n + 1)(5n + 2) \) to be divisible by \( m \). This means:
\[
m \mid n(2n + 1)(5n + 2)
\]
Since \( a \) and \( b \) are coprime, at least one of \( n \), \( 2n + 1 \), or \( 5n + 2 \) must be divisible by \( a \), and at least one of them must be divisible by \( b \).
3. **Set up congruences:**
We need to solve the following congruences:
\[
sn + t \equiv 0 \pmod{a}
\]
\[
un + v \equiv 0 \pmod{b}
\]
where \( (s, t) \) and \( (u, v) \) are one of \( (1, 0) \), \( (2, 1) \), and \( (5, 2) \).
4. **Uniqueness of solutions:**
Since \( s \) and \( u \) are relatively prime to both \( a \) and \( b \), each system of linear congruences has a unique solution modulo \( m \). There are 9 combinations for \( (s, t) \) and \( (u, v) \), leading to 9 potential solutions for \( n \).
5. **Check for overlap:**
We need to ensure that no two solutions coincide. If two solutions coincide, then \( a \) or \( b \) would divide both \( sn + t \) and \( un + v \) for different pairs \( (s, t) \) and \( (u, v) \), which is not possible since these numbers are relatively prime.
6. **Special case for \( n = 0 \):**
When both \( (s, t) = (1, 0) \) and \( (u, v) = (1, 0) \), the solution is \( n \equiv 0 \pmod{m} \). However, since \( n < m \) and \( n \) must be a natural number, \( n = 0 \) is not allowed.
7. **Conclusion:**
Since \( n = 0 \) is not allowed, we have 8 unique solutions for \( n \) in the range \( 0 < n < m \).
The final answer is \( \boxed{ 8 } \).
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\{x\}$ be a sequence of positive reals $x_1, x_2, \ldots, x_n$, defined by: $x_1 = 1, x_2 = 9, x_3=9, x_4=1$. And for $n \geq 1$ we have:
\[x_{n+4} = \sqrt[4]{x_{n} \cdot x_{n+1} \cdot x_{n+2} \cdot x_{n+3}}.\]
Show that this sequence has a finite limit. Determine this limit.
|
To show that the sequence $\{x_n\}$ has a finite limit and to determine this limit, we will proceed as follows:
1. **Define the sequence and its properties:**
Given the sequence $\{x_n\}$ with initial values:
\[
x_1 = 1, \quad x_2 = 9, \quad x_3 = 9, \quad x_4 = 1
\]
and the recurrence relation for $n \geq 1$:
\[
x_{n+4} = \sqrt[4]{x_n \cdot x_{n+1} \cdot x_{n+2} \cdot x_{n+3}}
\]
2. **Transform the sequence using logarithms:**
Define a new sequence $\{y_n\}$ such that:
\[
y_n = \log_3 x_n
\]
This transforms the recurrence relation into:
\[
y_{n+4} = \frac{y_n + y_{n+1} + y_{n+2} + y_{n+3}}{4}
\]
with initial values:
\[
y_1 = \log_3 1 = 0, \quad y_2 = \log_3 9 = 2, \quad y_3 = \log_3 9 = 2, \quad y_4 = \log_3 1 = 0
\]
3. **Analyze the transformed sequence:**
The recurrence relation for $\{y_n\}$ is a linear homogeneous recurrence relation. We can write it in matrix form:
\[
\begin{pmatrix}
y_{n+4} \\
y_{n+3} \\
y_{n+2} \\
y_{n+1}
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
y_n \\
y_{n+1} \\
y_{n+2} \\
y_{n+3}
\end{pmatrix}
\]
4. **Find the eigenvalues of the matrix:**
The characteristic polynomial of the matrix is:
\[
\det\left(
\begin{pmatrix}
\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{pmatrix}
- \lambda I
\right) = 0
\]
Solving this, we find the eigenvalues to be $\lambda = 1, -i, i, -1$.
5. **Analyze the long-term behavior:**
The eigenvalue $\lambda = 1$ indicates that the sequence $\{y_n\}$ will converge to a constant value as $n \to \infty$. The other eigenvalues ($-i, i, -1$) have magnitudes less than or equal to 1, which means their contributions will diminish over time.
6. **Determine the limit:**
Since $\{y_n\}$ converges to a constant value, let $L$ be this limit. Then:
\[
L = \frac{L + L + L + L}{4} \implies L = L
\]
This confirms that $\{y_n\}$ converges to a constant value. Given the initial values and the symmetry, the limit $L$ must be the average of the initial values:
\[
L = \frac{0 + 2 + 2 + 0}{4} = 1
\]
7. **Transform back to the original sequence:**
Since $y_n = \log_3 x_n$ and $y_n \to 1$, we have:
\[
\log_3 x_n \to 1 \implies x_n \to 3^1 = 3
\]
The final answer is $\boxed{3}$
|
3
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There are $ 25$ towns in a country. Find the smallest $ k$ for which one can set up two-direction flight routes connecting these towns so that the following conditions are satisfied:
1) from each town there are exactly $ k$ direct routes to $ k$ other towns;
2) if two towns are not connected by a direct route, then there is a town which has direct routes to these two towns.
|
1. **Define the problem in terms of graph theory:**
- We have 25 towns, which we can represent as vertices in a graph.
- We need to find the smallest \( k \) such that each vertex has exactly \( k \) edges (degree \( k \)).
- Additionally, for any two vertices that are not directly connected, there must be a vertex that is directly connected to both.
2. **Set up the problem:**
- Let \( A \) be an arbitrary town.
- Let \( R(A) \) be the set of towns directly connected to \( A \), so \( |R(A)| = k \).
- Let \( N(A) \) be the set of towns not directly connected to \( A \), so \( |N(A)| = 24 - k \).
3. **Analyze the connections:**
- For every town in \( N(A) \), there must be a direct route to at least one town in \( R(A) \) to satisfy the condition that there is a common town connected to both.
- The total number of possible connections from towns in \( R(A) \) to towns in \( N(A) \) is \( k(k-1) \) because each of the \( k \) towns in \( R(A) \) can connect to \( k-1 \) other towns (excluding \( A \)).
4. **Set up the inequality:**
- The number of connections from \( R(A) \) to \( N(A) \) must be at least \( 24 - k \) to ensure that every town in \( N(A) \) is connected to at least one town in \( R(A) \).
- Therefore, we have the inequality:
\[
k(k-1) \geq 24 - k
\]
5. **Solve the inequality:**
\[
k(k-1) \geq 24 - k
\]
\[
k^2 - k \geq 24 - k
\]
\[
k^2 \geq 24
\]
\[
k \geq \sqrt{24}
\]
\[
k \geq 5
\]
6. **Check the smallest integer \( k \):**
- Since \( k \) must be an integer, we check \( k = 5 \) and \( k = 6 \).
7. **Verify \( k = 5 \):**
- If \( k = 5 \), then \( k(k-1) = 5 \times 4 = 20 \).
- We need \( 20 \geq 24 - 5 \), which simplifies to \( 20 \geq 19 \), which is true.
- However, we need to ensure that the graph structure satisfies the condition for all pairs of towns.
8. **Verify \( k = 6 \):**
- If \( k = 6 \), then \( k(k-1) = 6 \times 5 = 30 \).
- We need \( 30 \geq 24 - 6 \), which simplifies to \( 30 \geq 18 \), which is true.
- Petersen's graph is a known graph that satisfies these conditions with \( k = 6 \).
Therefore, the smallest \( k \) that satisfies the conditions is \( k = 6 \).
The final answer is \( \boxed{6} \).
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $p\in \mathbb P,p>3$. Calcute:
a)$S=\sum_{k=1}^{\frac{p-1}{2}} \left[\frac{2k^2}{p}\right]-2 \cdot \left[\frac{k^2}{p}\right]$ if $ p\equiv 1 \mod 4$
b) $T=\sum_{k=1}^{\frac{p-1}{2}} \left[\frac{k^2}{p}\right]$ if $p\equiv 1 \mod 8$
|
### Part (a)
Given \( p \in \mathbb{P} \) and \( p > 3 \), we need to calculate:
\[ S = \sum_{k=1}^{\frac{p-1}{2}} \left\lfloor \frac{2k^2}{p} \right\rfloor - 2 \cdot \left\lfloor \frac{k^2}{p} \right\rfloor \]
if \( p \equiv 1 \pmod{4} \).
1. **Identify Quadratic Residues:**
Let \( r_i \) for \( 1 \le i \le \frac{p-1}{2} \) be the quadratic residues modulo \( p \). These are the distinct values of \( k^2 \mod p \) for \( k = 1, 2, \ldots, \frac{p-1}{2} \).
2. **Rewrite the Sum:**
The sum can be rewritten in terms of the quadratic residues:
\[
S = \sum_{i=1}^{\frac{p-1}{2}} \left\lfloor \frac{2r_i}{p} \right\rfloor - 2 \cdot \left\lfloor \frac{r_i}{p} \right\rfloor
\]
3. **Analyze the Terms:**
Each term \( \left\lfloor \frac{r_i}{p} \right\rfloor \) is 0 if \( r_i < p \) (which is always true since \( r_i \) are residues modulo \( p \)). Therefore, \( 2 \cdot \left\lfloor \frac{r_i}{p} \right\rfloor = 0 \).
4. **Simplify the Sum:**
The sum simplifies to:
\[
S = \sum_{i=1}^{\frac{p-1}{2}} \left\lfloor \frac{2r_i}{p} \right\rfloor
\]
5. **Determine the Value of Each Term:**
Each term \( \left\lfloor \frac{2r_i}{p} \right\rfloor \) is 0 or 1. It is 1 if \( 2r_i \ge p \) and 0 otherwise. This happens if \( r_i \ge \frac{p}{2} \).
6. **Count the Number of Terms:**
Since \( -1 \) is a quadratic residue and \( p \equiv 1 \pmod{4} \), the number of quadratic residues greater than \( \frac{p-1}{2} \) is equal to the number of those less than \( \frac{p-1}{2} \). Therefore, there are \( \frac{\frac{p-1}{2}}{2} = \frac{p-1}{4} \) such residues.
7. **Sum the Values:**
Hence, the sum \( S \) is:
\[
S = \frac{p-1}{4}
\]
### Part (b)
Given \( p \in \mathbb{P} \) and \( p > 3 \), we need to calculate:
\[ T = \sum_{k=1}^{\frac{p-1}{2}} \left\lfloor \frac{k^2}{p} \right\rfloor \]
if \( p \equiv 1 \pmod{8} \).
1. **Identify Quadratic Residues:**
Let \( r_i \) for \( 1 \le i \le \frac{p-1}{2} \) be the quadratic residues modulo \( p \). These are the distinct values of \( k^2 \mod p \) for \( k = 1, 2, \ldots, \frac{p-1}{2} \).
2. **Rewrite the Sum:**
The sum can be rewritten in terms of the quadratic residues:
\[
T = \sum_{i=1}^{\frac{p-1}{2}} \left\lfloor \frac{r_i}{p} \right\rfloor
\]
3. **Analyze the Terms:**
Each term \( \left\lfloor \frac{r_i}{p} \right\rfloor \) is 0 if \( r_i < p \) (which is always true since \( r_i \) are residues modulo \( p \)).
4. **Simplify the Sum:**
Since all terms are 0, the sum \( T \) is:
\[
T = 0
\]
The final answer is \( \boxed{\frac{p-1}{4}} \) for part (a) and \( \boxed{0} \) for part (b).
|
0
|
Number Theory
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
In the space are given $2006$ distinct points, such that no $4$ of them are coplanar. One draws a segment between each pair of points.
A natural number $m$ is called [i]good[/i] if one can put on each of these segments a positive integer not larger than $m$, so that every triangle whose three vertices are among the given points has the property that two of this triangle's sides have equal numbers put on, while the third has a larger number put on.
Find the minimum value of a [i]good[/i] number $m$.
|
1. **Define the problem and notation:**
Let \( A(n) \) be the minimum value of a good number if we have \( n \) points in space. We need to find \( A(2006) \).
2. **Initial observation:**
In the minimal case, there is obviously a segment on which the number \( 1 \) is put. Let its endpoints be \( X \) and \( Y \). Each of the other \( 2004 \) points is either linked to \( X \) by a segment whose number is \( 1 \) or to \( Y \). Otherwise, we could take a point which is linked neither to \( X \) nor to \( Y \) by a segment with number \( 1 \), as well as \( X \) and \( Y \), so we would have a triangle which would not satisfy the condition.
3. **Partition the points:**
Let the set of points which are linked to \( X \) by a segment with number \( 1 \) be \( A \) and the set of points which are linked to \( Y \) by a segment with number \( 1 \) be \( B \). No point belongs to both \( A \) and \( B \), or we would have a triangle with three equal numbers, so \( |A| + |B| = 2004 \).
4. **Connecting points between sets:**
Take any point in \( A \) and any in \( B \). On the segment which links both points must be put number \( 1 \), or we could take these two points and \( X \) (or \( Y \), if \( X \) is one of the two points) and we would have three points which don't satisfy the condition. So, each point in \( A \) is linked by a segment with number \( 1 \) with each point in \( B \).
5. **Internal connections within sets:**
Two points in \( A \) can't be linked by a segment with number \( 1 \) or we would have a triangle with three equal numbers on its sides. If we write a number greater than \( 1 \) on each other segment, and if we take two points out of \( A \) and one out of \( B \), they satisfy the condition, because on two of the three segments would be number \( 1 \) and on the third would be a greater number. Similarly, if we take two points out of \( B \) and one out of \( A \).
6. **Recursive argument:**
Now, consider we take three points out of \( A \). None of the three sides has a \( 1 \) put on it. We need \( A(|A|) \) numbers greater than \( 1 \) for these \( |A| \) points. Similarly, we need \( A(|B|) \) numbers for the \( |B| \) points in the set \( B \), but those could be the same as in set \( A \). So, we have (since if \( |A| \geq |B| \), then \( |A| \geq 1002 \)):
\[
A(2006) \geq 1 + A(1003) \geq 2 + A(502) \geq \ldots \geq 10 + A(2) \geq 11
\]
7. **Verification:**
It is easy to see that we can really do it with just \( 11 \) numbers if we take \( |A| = |B| \) and so on, so \( 11 \) is the value we look for.
The final answer is \( \boxed{11} \)
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the smallest positive interger number $n$ such that there exists $n$ real numbers $a_1,a_2,\ldots,a_n$ satisfied three conditions as follow:
a. $a_1+a_2+\cdots+a_n>0$;
b. $a_1^3+a_2^3+\cdots+a_n^3<0$;
c. $a_1^5+a_2^5+\cdots+a_n^5>0$.
|
To find the smallest positive integer \( n \) such that there exist \( n \) real numbers \( a_1, a_2, \ldots, a_n \) satisfying the conditions:
a. \( a_1 + a_2 + \cdots + a_n > 0 \)
b. \( a_1^3 + a_2^3 + \cdots + a_n^3 < 0 \)
c. \( a_1^5 + a_2^5 + \cdots + a_n^5 > 0 \)
we will analyze the cases for \( n = 1, 2, 3, 4 \) and show that \( n = 5 \) is the smallest value that satisfies all conditions.
1. **Case \( n = 1 \):**
- If \( n = 1 \), then we have a single number \( a_1 \).
- Condition a: \( a_1 > 0 \)
- Condition b: \( a_1^3 < 0 \) (impossible since \( a_1 > 0 \))
- Therefore, \( n = 1 \) does not satisfy all conditions.
2. **Case \( n = 2 \):**
- If \( n = 2 \), then we have two numbers \( a_1 \) and \( a_2 \).
- Condition a: \( a_1 + a_2 > 0 \)
- Condition b: \( a_1^3 + a_2^3 < 0 \)
- Condition c: \( a_1^5 + a_2^5 > 0 \)
- If \( a_1 > 0 \) and \( a_2 < 0 \), then \( a_1^3 > 0 \) and \( a_2^3 < 0 \). For their sum to be negative, \( |a_2| \) must be large enough.
- However, if \( |a_2| \) is large enough to make \( a_1^3 + a_2^3 < 0 \), then \( a_1^5 + a_2^5 \) will likely be negative as well, contradicting condition c.
- Therefore, \( n = 2 \) does not satisfy all conditions.
3. **Case \( n = 3 \):**
- If \( n = 3 \), then we have three numbers \( a_1, a_2, a_3 \).
- Condition a: \( a_1 + a_2 + a_3 > 0 \)
- Condition b: \( a_1^3 + a_2^3 + a_3^3 < 0 \)
- Condition c: \( a_1^5 + a_2^5 + a_3^5 > 0 \)
- Similar to the previous case, balancing the conditions becomes difficult. If two numbers are positive and one is negative, or vice versa, it is challenging to satisfy all three conditions simultaneously.
- Therefore, \( n = 3 \) does not satisfy all conditions.
4. **Case \( n = 4 \):**
- If \( n = 4 \), then we have four numbers \( a_1, a_2, a_3, a_4 \).
- Condition a: \( a_1 + a_2 + a_3 + a_4 > 0 \)
- Condition b: \( a_1^3 + a_2^3 + a_3^3 + a_4^3 < 0 \)
- Condition c: \( a_1^5 + a_2^5 + a_3^5 + a_4^5 > 0 \)
- We can try different configurations, but as shown in the initial solution, it is difficult to satisfy all three conditions simultaneously with four numbers.
- Therefore, \( n = 4 \) does not satisfy all conditions.
5. **Case \( n = 5 \):**
- If \( n = 5 \), then we have five numbers \( a_1, a_2, a_3, a_4, a_5 \).
- Consider the numbers \( -7, -7, 2, 5, 8 \):
- Condition a: \( -7 + (-7) + 2 + 5 + 8 = 1 > 0 \)
- Condition b: \( (-7)^3 + (-7)^3 + 2^3 + 5^3 + 8^3 = -343 - 343 + 8 + 125 + 512 = -41 < 0 \)
- Condition c: \( (-7)^5 + (-7)^5 + 2^5 + 5^5 + 8^5 = -16807 - 16807 + 32 + 3125 + 32768 = -33614 + 35925 = 2311 > 0 \)
- Therefore, \( n = 5 \) satisfies all conditions.
The final answer is \( \boxed{5} \).
|
5
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Real numbers $x,y,z$ are chosen such that $$\frac{1}{|x^2+2yz|} ,\frac{1}{|y^2+2zx|} ,\frac{1}{|x^2+2xy|} $$
are lengths of a non-degenerate triangle .
Find all possible values of $xy+yz+zx$ .
[i]Proposed by Michael Rolínek[/i]
|
To solve the problem, we need to find all possible values of \( xy + yz + zx \) such that the given expressions form the sides of a non-degenerate triangle. The expressions are:
\[
\frac{1}{|x^2 + 2yz|}, \quad \frac{1}{|y^2 + 2zx|}, \quad \frac{1}{|z^2 + 2xy|}
\]
For these to be the sides of a non-degenerate triangle, they must satisfy the triangle inequality:
\[
\frac{1}{|x^2 + 2yz|} + \frac{1}{|y^2 + 2zx|} > \frac{1}{|z^2 + 2xy|}
\]
\[
\frac{1}{|y^2 + 2zx|} + \frac{1}{|z^2 + 2xy|} > \frac{1}{|x^2 + 2yz|}
\]
\[
\frac{1}{|z^2 + 2xy|} + \frac{1}{|x^2 + 2yz|} > \frac{1}{|y^2 + 2zx|}
\]
We need to analyze these inequalities to find the conditions on \( xy + yz + zx \).
1. **Assume \( xy + yz + zx = 0 \)**:
- If \( xy + yz + zx = 0 \), then we can rewrite the expressions as:
\[
x^2 + 2yz = x^2 - 2xy - 2zx = x^2 - 2x(y + z)
\]
\[
y^2 + 2zx = y^2 - 2yz - 2xy = y^2 - 2y(z + x)
\]
\[
z^2 + 2xy = z^2 - 2zx - 2yz = z^2 - 2z(x + y)
\]
- If two of \( x, y, z \) are equal, say \( x = y \), then:
\[
|x^2 + 2zx| = |x^2 + 2x^2| = |3x^2| = 3x^2
\]
This would imply that one of the denominators is zero, which is not possible.
2. **Assume \( x, y, z \) are distinct and \( x > y > z \)**:
- Without loss of generality, assume \( x, y \) are positive and \( z \) is negative. Then:
\[
x^2 + 2yz > 0, \quad y^2 + 2zx > 0, \quad z^2 + 2xy > 0
\]
- The sum of the three fractions without absolute values would be:
\[
\sum_{\text{cyc}} \frac{1}{x^2 + 2yz} = \sum_{\text{cyc}} \frac{1}{(x-y)(x-z)} = 0
\]
This is a contradiction because the sum of two of the fractions with positive denominators cannot equal the other fraction, as they form a non-degenerate triangle.
Therefore, the only possible value for \( xy + yz + zx \) that satisfies the conditions is:
\[
\boxed{0}
\]
|
0
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There are $2022$ numbers arranged in a circle $a_1, a_2, . . ,a_{2022}$. It turned out that for any three consecutive $a_i$, $a_{i+1}$, $a_{i+2}$ the equality $a_i =\sqrt2 a_{i+2} - \sqrt3 a_{i+1}$. Prove that $\sum^{2022}_{i=1} a_ia_{i+2} = 0$, if we know that $a_{2023} = a_1$, $a_{2024} = a_2$.
|
1. Given the equation for any three consecutive terms \(a_i, a_{i+1}, a_{i+2}\):
\[
a_i = \sqrt{2} a_{i+2} - \sqrt{3} a_{i+1}
\]
We can rewrite this equation as:
\[
a_i + \sqrt{3} a_{i+1} = \sqrt{2} a_{i+2}
\]
Squaring both sides, we get:
\[
(a_i + \sqrt{3} a_{i+1})^2 = (\sqrt{2} a_{i+2})^2
\]
Expanding both sides:
\[
a_i^2 + 2a_i \sqrt{3} a_{i+1} + 3a_{i+1}^2 = 2a_{i+2}^2
\]
Rearranging terms, we obtain:
\[
2a_{i+2}^2 + a_i^2 - 3a_{i+1}^2 = 2\sqrt{2} a_i a_{i+2}
\]
2. Summing this equation over all \(i\) from 1 to 2022:
\[
\sum_{i=1}^{2022} (2a_{i+2}^2 + a_i^2 - 3a_{i+1}^2) = \sum_{i=1}^{2022} 2\sqrt{2} a_i a_{i+2}
\]
Notice that the left-hand side can be simplified by recognizing that the sums of squares of \(a_i\) terms are cyclic:
\[
\sum_{i=1}^{2022} 2a_{i+2}^2 + \sum_{i=1}^{2022} a_i^2 - \sum_{i=1}^{2022} 3a_{i+1}^2
\]
Since the indices are cyclic, we can rename the indices in the sums:
\[
\sum_{i=1}^{2022} a_i^2 = \sum_{i=1}^{2022} a_{i+1}^2 = \sum_{i=1}^{2022} a_{i+2}^2
\]
Therefore, the left-hand side becomes:
\[
2\sum_{i=1}^{2022} a_i^2 + \sum_{i=1}^{2022} a_i^2 - 3\sum_{i=1}^{2022} a_i^2 = 0
\]
3. This simplifies to:
\[
2\sqrt{2} \sum_{i=1}^{2022} a_i a_{i+2} = 0
\]
Dividing both sides by \(2\sqrt{2}\):
\[
\sum_{i=1}^{2022} a_i a_{i+2} = 0
\]
The final answer is \(\boxed{0}\)
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
aops_forum
| false
|
In equality
$$1 * 2 * 3 * 4 * 5 * ... * 60 * 61 * 62 = 2023$$
Instead of each asterisk, you need to put one of the signs “+” (plus), “-” (minus), “•” (multiply) so that the equality becomes true. What is the smallest number of "•" characters that can be used?
|
1. **Initial Consideration**:
- We need to find the smallest number of multiplication signs (denoted as "•") to make the equation \(1 * 2 * 3 * 4 * 5 * \ldots * 60 * 61 * 62 = 2023\) true.
- If we use only addition and subtraction, the maximum sum is \(1 + 2 + 3 + \ldots + 62\), which is less than 2023.
2. **Sum Calculation**:
- Calculate the sum \(S\) of the first 62 natural numbers:
\[
S = \frac{62 \cdot (62 + 1)}{2} = \frac{62 \cdot 63}{2} = 1953
\]
- Since \(1953 < 2023\), we need at least one multiplication to increase the total.
3. **Parity Consideration**:
- The sum \(S = 1953\) is odd.
- If we place one multiplication sign between \(a\) and \(a+1\), the result will be:
\[
S - a - (a+1) + a(a+1) = 1953 - a - (a+1) + a(a+1)
\]
- This expression simplifies to:
\[
1953 - a - a - 1 + a^2 + a = 1953 - 1 + a^2 - a = 1952 + a^2 - a
\]
- Since \(a^2 - a\) is always even, the result \(1952 + a^2 - a\) is even. Therefore, it cannot be 2023, which is odd.
4. **Using Two Multiplications**:
- We need to find a way to use two multiplications to achieve the target sum of 2023.
- Consider placing multiplications between \(1\) and \(2\), and between \(9\) and \(10\):
\[
1 \cdot 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 \cdot 10 + 11 + \ldots + 62
\]
- Calculate the new sum:
\[
1 \cdot 2 = 2
\]
\[
9 \cdot 10 = 90
\]
\[
\text{Sum of remaining terms} = 3 + 4 + 5 + 6 + 7 + 8 + 11 + 12 + \ldots + 62
\]
- Calculate the sum of the remaining terms:
\[
\text{Sum of } 3 \text{ to } 8 = 3 + 4 + 5 + 6 + 7 + 8 = 33
\]
\[
\text{Sum of } 11 \text{ to } 62 = \frac{(62 - 11 + 1) \cdot (11 + 62)}{2} = \frac{52 \cdot 73}{2} = 1898
\]
- Adding all parts together:
\[
2 + 90 + 33 + 1898 = 2023
\]
Thus, the smallest number of multiplication signs needed is 2.
The final answer is \(\boxed{2}\).
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For any positive integer $a$, let $\tau(a)$ be the number of positive divisors of $a$. Find, with proof, the largest possible value of $4\tau(n)-n$ over all positive integers $n$.
|
1. **Understanding the function $\tau(n)$**:
- The function $\tau(n)$ represents the number of positive divisors of $n$. For example, if $n = 12$, then the divisors are $1, 2, 3, 4, 6, 12$, so $\tau(12) = 6$.
2. **Establishing an upper bound for $\tau(n)$**:
- We need to find an upper bound for $\tau(n)$. Note that for any positive integer $n$, the number of divisors $\tau(n)$ is at most the number of integers from $1$ to $n$. However, we can refine this bound by considering that divisors come in pairs. For example, if $d$ is a divisor of $n$, then $\frac{n}{d}$ is also a divisor of $n$.
- If $d \leq \sqrt{n}$, then $\frac{n}{d} \geq \sqrt{n}$. Therefore, the number of divisors $\tau(n)$ is at most $2\sqrt{n}$, but this is a loose bound. A tighter bound is given by considering the divisors larger than $n/4$.
3. **Bounding $\tau(n)$ more tightly**:
- The only divisors of $n$ larger than $n/4$ can be $n/3, n/2, n$. This gives us at most 3 divisors larger than $n/4$.
- Therefore, $\tau(n) \leq n/4 + 3$.
4. **Maximizing $4\tau(n) - n$**:
- We want to maximize the expression $4\tau(n) - n$. Using the bound $\tau(n) \leq n/4 + 3$, we substitute this into the expression:
\[
4\tau(n) - n \leq 4\left(\frac{n}{4} + 3\right) - n = n + 12 - n = 12
\]
- Therefore, the maximum value of $4\tau(n) - n$ is at most $12$.
5. **Checking if the bound is achievable**:
- We need to check if there exists an integer $n$ such that $4\tau(n) - n = 12$. Let's test $n = 12$:
\[
\tau(12) = 6 \quad \text{(since the divisors of 12 are 1, 2, 3, 4, 6, 12)}
\]
\[
4\tau(12) - 12 = 4 \times 6 - 12 = 24 - 12 = 12
\]
- Thus, the maximum value of $4\tau(n) - n$ is indeed $12$, and it is achieved when $n = 12$.
\[
\boxed{12}
\]
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Five boys and six girls are to be seated in a row of eleven chairs so that they sit one at a time from one end to the other. The probability that there are no more boys than girls seated at any point during the process is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Evaluate $m + n$.
|
1. **Understanding the Problem:**
We need to find the probability that there are no more boys than girls seated at any point during the process of seating 5 boys and 6 girls in a row of 11 chairs. This can be visualized as a path on a grid from \((0,0)\) to \((6,5)\) without crossing the line \(y = x\).
2. **Catalan Number:**
The problem can be translated into counting the number of valid paths from \((0,0)\) to \((6,5)\) without crossing the line \(y = x\). This is a classic problem that can be solved using Catalan numbers. The \(n\)-th Catalan number \(C_n\) is given by:
\[
C_n = \frac{1}{n+1} \binom{2n}{n}
\]
For our problem, we need the 6th Catalan number \(C_6\):
\[
C_6 = \frac{1}{6+1} \binom{12}{6} = \frac{1}{7} \binom{12}{6}
\]
3. **Calculating \(C_6\):**
\[
\binom{12}{6} = \frac{12!}{6!6!} = \frac{479001600}{720 \times 720} = 924
\]
Therefore,
\[
C_6 = \frac{1}{7} \times 924 = 132
\]
4. **Total Number of Arrangements:**
The total number of ways to arrange 5 boys and 6 girls in 11 chairs is given by:
\[
\binom{11}{5} = \frac{11!}{5!6!} = \frac{39916800}{120 \times 720} = 462
\]
5. **Probability Calculation:**
The probability that there are no more boys than girls seated at any point is:
\[
\frac{C_6}{\binom{11}{5}} = \frac{132}{462} = \frac{2}{7}
\]
6. **Final Answer:**
The fraction \(\frac{2}{7}\) is already in its simplest form, so \(m = 2\) and \(n = 7\). Therefore, \(m + n = 2 + 7 = 9\).
The final answer is \( \boxed{ 9 } \)
|
9
|
Combinatorics
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
Here I shall collect for the sake of collecting in separate threads the geometry problems those posting in multi-problems threads inside aops. I shall create post collections from these threads also. Geometry from USA contests are collected [url=https://artofproblemsolving.com/community/c2746635_geometry_from_usa_contests]here[/url].
[b]Geometry problems from[/b]
$\bullet$ CHMMC = Caltech Harvey Mudd Mathematics Competition / Mixer Round, [url=https://artofproblemsolving.com/community/c1068820h3195908p30057017]here[/url]
$\bullet$ CMUWC = Carnegie Mellon University Womens' Competition, collected [url=https://artofproblemsolving.com/community/c3617935_cmwmc_geometry]here[/url].
$\bullet$ CUBRMC = Cornell University Big Red Math Competition, collected [url=https://artofproblemsolving.com/community/c1068820h3195908p30129173]here[/url]
$\bullet$ DMM = Duke Math Meet, collected [url=https://artofproblemsolving.com/community/c3617665_dmm_geometry]here[/url].
$\bullet$ Girls in Math at Yale , collected [url=https://artofproblemsolving.com/community/c3618392_]here[/url]
$\bullet$ MMATHS = Math Majors of America Tournament for High Schools, collected [url=https://artofproblemsolving.com/community/c3618425_]here[/url]
More USA Contests await you [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
|
**
1. **Understanding the Geometry of the Hexagon:**
A regular hexagon can be divided into 6 equilateral triangles. Each side of the hexagon is of length 1.
2. **Identifying the Points:**
The largest distance between any two points in a regular hexagon is the distance between two opposite vertices. This is because the longest line segment that can be drawn within a hexagon is its diameter, which connects two opposite vertices.
3. **Calculating the Distance:**
To find the distance between two opposite vertices, we can use the properties of the equilateral triangles that make up the hexagon. The distance between two opposite vertices is twice the side length of the hexagon.
- Consider the center of the hexagon and two opposite vertices. The distance from the center to any vertex is the radius of the circumscribed circle around the hexagon.
- The radius of the circumscribed circle is equal to the side length of the hexagon, which is 1.
- Therefore, the distance between two opposite vertices is twice the radius of the circumscribed circle.
4. **Final Calculation:**
\[
\text{Distance between two opposite vertices} = 2 \times \text{side length} = 2 \times 1 = 2
\]
Conclusion:
\[
\boxed{2}
\]
|
2
|
Geometry
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
Calculate the sum of matrix commutators $[A, [B, C]] + [B, [C, A]] + [C, [A, B]]$, where $[A, B] = AB-BA$
|
1. **Express the commutators in terms of matrix products:**
\[
[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = [A, (BC - CB)] + [B, (CA - AC)] + [C, (AB - BA)]
\]
2. **Expand each commutator:**
\[
[A, (BC - CB)] = A(BC - CB) - (BC - CB)A = ABC - ACB - BCA + CBA
\]
\[
[B, (CA - AC)] = B(CA - AC) - (CA - AC)B = BCA - BAC - CAB + ACB
\]
\[
[C, (AB - BA)] = C(AB - BA) - (AB - BA)C = CAB - CBA - ABC + BAC
\]
3. **Combine the expanded commutators:**
\[
[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = (ABC - ACB - BCA + CBA) + (BCA - BAC - CAB + ACB) + (CAB - CBA - ABC + BAC)
\]
4. **Simplify the expression by combining like terms:**
\[
= ABC - ACB - BCA + CBA + BCA - BAC - CAB + ACB + CAB - CBA - ABC + BAC
\]
5. **Observe that all terms cancel out:**
\[
ABC - ABC + ACB - ACB + BCA - BCA + CBA - CBA + BCA - BCA + CAB - CAB + ACB - ACB + CBA - CBA + CAB - CAB + BAC - BAC = 0
\]
6. **Conclude that the sum of the commutators is the zero matrix:**
\[
[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0
\]
The final answer is \(\boxed{0}\)
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given two vectors $v = (v_1,\dots,v_n)$ and $w = (w_1\dots,w_n)$ in $\mathbb{R}^n$, lets define $v*w$ as the matrix in which the element of row $i$ and column $j$ is $v_iw_j$. Supose that $v$ and $w$ are linearly independent. Find the rank of the matrix $v*w - w*v.$
|
1. **Define the matrix \( A = v * w - w * v \):**
Given two vectors \( v = (v_1, v_2, \ldots, v_n) \) and \( w = (w_1, w_2, \ldots, w_n) \) in \( \mathbb{R}^n \), the matrix \( v * w \) is defined such that the element in the \( i \)-th row and \( j \)-th column is \( v_i w_j \). Similarly, the matrix \( w * v \) is defined such that the element in the \( i \)-th row and \( j \)-th column is \( w_i v_j \). Therefore, the matrix \( A \) is given by:
\[
A = v * w - w * v
\]
The element in the \( i \)-th row and \( j \)-th column of \( A \) is:
\[
A_{ij} = v_i w_j - w_i v_j
\]
2. **Construct the matrix \( A \) for \( n = 3 \):**
For \( n = 3 \), the matrix \( A \) is:
\[
A = \begin{bmatrix}
0 & v_1 w_2 - w_1 v_2 & v_1 w_3 - w_1 v_3 \\
v_2 w_1 - w_2 v_1 & 0 & v_2 w_3 - w_2 v_3 \\
v_3 w_1 - w_3 v_1 & v_3 w_2 - w_3 v_2 & 0
\end{bmatrix}
\]
Note that the diagonal elements are all zero.
3. **Analyze the rank of \( A \):**
To determine the rank of \( A \), we need to check the linear independence of its rows (or columns). We observe that each element \( A_{ij} \) is of the form \( v_i w_j - w_i v_j \).
4. **Check linear independence for \( n = 3 \):**
Consider the first row of \( A \):
\[
\begin{bmatrix}
0 & v_1 w_2 - w_1 v_2 & v_1 w_3 - w_1 v_3
\end{bmatrix}
\]
We need to check if this row can be written as a linear combination of the other two rows. Suppose:
\[
c_1 \begin{bmatrix}
v_2 w_1 - w_2 v_1 & 0 & v_2 w_3 - w_2 v_3
\end{bmatrix} + c_2 \begin{bmatrix}
v_3 w_1 - w_3 v_1 & v_3 w_2 - w_3 v_2 & 0
\end{bmatrix} = \begin{bmatrix}
0 & v_1 w_2 - w_1 v_2 & v_1 w_3 - w_1 v_3
\end{bmatrix}
\]
This gives us the system of equations:
\[
c_1 (v_2 w_1 - w_2 v_1) + c_2 (v_3 w_1 - w_3 v_1) = 0
\]
\[
c_1 (v_2 w_3 - w_2 v_3) = v_1 w_3 - w_1 v_3
\]
\[
c_2 (v_3 w_2 - w_3 v_2) = v_1 w_2 - w_1 v_2
\]
5. **Solve for \( c_1 \) and \( c_2 \):**
From the second and third equations, we get:
\[
c_1 = \frac{v_1 w_3 - w_1 v_3}{v_2 w_3 - w_2 v_3}
\]
\[
c_2 = \frac{v_1 w_2 - w_1 v_2}{v_3 w_2 - w_3 v_2}
\]
Substituting these into the first equation, we need to verify that:
\[
\left( \frac{v_1 w_3 - w_1 v_3}{v_2 w_3 - w_2 v_3} \right) (v_2 w_1 - w_2 v_1) + \left( \frac{v_1 w_2 - w_1 v_2}{v_3 w_2 - w_3 v_2} \right) (v_3 w_1 - w_3 v_1) = 0
\]
This equation holds due to the linear independence of \( v \) and \( w \).
6. **Generalize for any \( n \):**
By similar reasoning, for any \( n \), the \( k \)-th row can be written as a linear combination of the second and third rows. Therefore, the rank of the matrix \( A \) is always 2.
The final answer is \(\boxed{2}\).
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For $n \geq 3$, let $(b_0, b_1,..., b_{n-1}) = (1, 1, 1, 0, ..., 0).$ Let $C_n =
(c_{i, j})$ the $n \times n$ matrix defined by $c_{i, j} = b _{(j -i) \mod n}$. Show
that $\det (C_n) = 3$ if $n$ is not a multiple of 3 and $\det (C_n) = 0$ if $n$
is a multiple of 3.
|
To show that $\det(C_n) = 3$ if $n$ is not a multiple of 3 and $\det(C_n) = 0$ if $n$ is a multiple of 3, we will analyze the structure of the matrix $C_n$ and use properties of determinants.
1. **Matrix Definition and Initial Setup:**
The matrix $C_n = (c_{i,j})$ is defined by $c_{i,j} = b_{(j-i) \mod n}$, where $(b_0, b_1, \ldots, b_{n-1}) = (1, 1, 1, 0, \ldots, 0)$. This means that each row of $C_n$ is a cyclic permutation of the vector $(1, 1, 1, 0, \ldots, 0)$.
2. **Matrix Structure:**
The matrix $C_n$ looks like:
\[
C_n = \begin{pmatrix}
1 & 1 & 1 & 0 & 0 & \cdots & 0 \\
0 & 1 & 1 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & 1 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & 0 & 0 & \cdots & 1
\end{pmatrix}
\]
3. **Case Analysis:**
We will consider two cases based on whether $n$ is a multiple of 3 or not.
**Case 1: $n$ is a multiple of 3**
- If $n$ is a multiple of 3, then the matrix $C_n$ will have rows that are linearly dependent. Specifically, the rows will repeat every 3 rows due to the cyclic nature of the matrix.
- For example, if $n = 3k$, then the rows $r_0, r_3, r_6, \ldots$ will be identical, leading to linear dependence.
- Therefore, the determinant of $C_n$ will be zero in this case.
**Case 2: $n$ is not a multiple of 3**
- If $n$ is not a multiple of 3, we can transform $C_n$ into an upper triangular matrix using row operations.
- We start with the $(n-1)$-th row and perform row operations to "slide" the leftmost 1 three positions to the right:
\[
(1, 0, 0, 0, 0, \cdots, 0, 1, 1) \rightarrow (0, -1, -1, 0, 0, \cdots, 0, 1, 1) \rightarrow (0, 0, 0, 1, 0, \cdots, 0, 1, 1)
\]
- This process continues until we reach a row that is either:
1. $(0, \cdots, 0, 0, 1, 1, 1)$ if $n \equiv 0 \mod 3$
2. $(0, \cdots, 0, 1, 0, 1, 1)$ if $n \equiv 1 \mod 3$
3. $(0, \cdots, 1, 0, 0, 1, 1)$ if $n \equiv 2 \mod 3$
- In the first case, we have reached a row that is the same as the $(n-2)$-th one, so the determinant is zero.
- In the other two cases, we continue with similar row operations to transform the matrix into an upper triangular form with all ones on the main diagonal except for the $a_{nn}$ entry, which will be 3.
4. **Conclusion:**
- For $n \equiv 1 \mod 3$ and $n \equiv 2 \mod 3$, the determinant of the upper triangular matrix will be the product of the diagonal entries, which is $3$.
- For $n \equiv 0 \mod 3$, the determinant is zero due to linear dependence.
The final answer is $\boxed{3}$ if $n$ is not a multiple of 3 and $\boxed{0}$ if $n$ is a multiple of 3.
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
aops_forum
| false
|
For each positive integer $n$ let $A_n$ be the $n \times n$ matrix such that its $a_{ij}$
entry is equal to ${i+j-2 \choose j-1}$ for all $1\leq i,j \leq n.$ Find the determinant of $A_n$.
|
1. We start by examining the given matrix \( A_n \) for small values of \( n \) to identify any patterns. For \( n = 2 \), \( n = 3 \), and \( n = 4 \), the matrices are:
\[
A_2 = \begin{bmatrix}
1 & 1 \\
1 & 2
\end{bmatrix}
\]
\[
A_3 = \begin{bmatrix}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 3 & 6
\end{bmatrix}
\]
\[
A_4 = \begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 3 & 4 \\
1 & 3 & 6 & 10 \\
1 & 4 & 10 & 20
\end{bmatrix}
\]
2. We observe that the entries of the matrix \( A_n \) are given by the binomial coefficients:
\[
a_{ij} = \binom{i+j-2}{j-1}
\]
This is a well-known binomial coefficient identity.
3. We notice a pattern in the entries of the matrices. Specifically, we observe that:
\[
a_{ij} + a_{i+1,j-1} = a_{i+1,j}
\]
This is Pascal's Identity:
\[
\binom{i+j-2}{j-1} + \binom{i+j-2}{j-2} = \binom{i+j-1}{j-1}
\]
4. To simplify the determinant calculation, we will perform row operations to transform the matrix into an upper triangular form. We start by subtracting the \( i \)-th row from the \( (i+1) \)-th row, starting from \( i = n-1 \) and moving upwards.
5. For \( n = 4 \), the row operations are as follows:
\[
\begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 3 & 4 \\
1 & 3 & 6 & 10 \\
1 & 4 & 10 & 20
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 3 & 4 \\
1 & 3 & 6 & 10 \\
0 & 1 & 4 & 10
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 3 & 4 \\
0 & 1 & 3 & 6 \\
0 & 1 & 4 & 10
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 2 & 3 \\
0 & 1 & 3 & 6 \\
0 & 1 & 4 & 10
\end{bmatrix}
\]
6. Continuing this process, we can transform the matrix into an upper triangular form with 1's on the diagonal and 0's below the diagonal. This is justified by the Pascal's Identity pattern found earlier.
7. Once the matrix is in upper triangular form, the determinant of the matrix is the product of the diagonal entries. Since all diagonal entries are 1, the determinant is:
\[
\boxed{1}
\]
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given a $3 \times 3$ symmetric real matrix $A$, we define $f(A)$ as a $3 \times 3$ matrix with the same eigenvectors of $A$ such that if $A$ has eigenvalues $a$, $b$, $c$, then $f(A)$ has eigenvalues $b+c$, $c+a$, $a+b$ (in that order). We define a sequence of symmetric real $3\times3$ matrices $A_0, A_1, A_2, \ldots$ such that $A_{n+1} = f(A_n)$ for $n \geq 0$. If the matrix $A_0$ has no zero entries, determine the maximum number of indices $j \geq 0$ for which the matrix $A_j$ has any null entries.
|
1. **Spectral Decomposition of \( A \)**:
Given a symmetric real matrix \( A \), we can use spectral decomposition to write:
\[
A = Q \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix} Q^T = Q D Q^T
\]
where \( Q \) is an orthogonal matrix and \( D \) is a diagonal matrix with eigenvalues \( a, b, c \).
2. **Definition of \( f(A) \)**:
The function \( f(A) \) is defined such that if \( A \) has eigenvalues \( a, b, c \), then \( f(A) \) has eigenvalues \( b+c, c+a, a+b \). Using the spectral decomposition, we can write:
\[
f(A) = Q \begin{pmatrix} b+c & 0 & 0 \\ 0 & c+a & 0 \\ 0 & 0 & a+b \end{pmatrix} Q^T
\]
Notice that:
\[
f(A) = Q (\text{Tr}(A)I - D) Q^T = \text{Tr}(A)I - A
\]
where \(\text{Tr}(A) = a + b + c\).
3. **Inductive Formula for \( f^n(A) \)**:
We can establish the following formula for \( f^n(A) \) by induction:
\[
f^n(A) = \begin{cases}
\frac{2^n - 1}{3} \text{Tr}(A) I + A & \text{if } n \equiv 0 \pmod{2} \\
\frac{2^n + 1}{3} \text{Tr}(A) I - A & \text{if } n \equiv 1 \pmod{2}
\end{cases}
\]
4. **Non-zero Off-diagonal Entries**:
The off-diagonal entries of \( f^n(A) \) are clearly non-zero because they are derived from the orthogonal transformation of the diagonal matrix.
5. **Diagonal Entries and Null Entries**:
Assume \(\text{Tr}(A) \neq 0\). Let the diagonal entries of \( A \) be \( a, b, c \). Suppose there are three even \( n \) such that \( f^n(A) \) has a null entry. This implies:
\[
x(a+b+c) + a = 0
\]
\[
y(a+b+c) + b = 0
\]
\[
z(a+b+c) + c = 0
\]
The only solution to these equations is \( a = b = c = 0 \), which contradicts the assumption that \( A_0 \) has no zero entries.
6. **Maximum Number of Indices**:
Thus, there can be at most 2 even \( n \) such that \( f^n(A) \) has a null entry. Similarly, there can be at most 2 odd \( n \) such that \( f^n(A) \) has a null entry. If there are 3 overall indices, then 2 must be even and 1 must be odd, or vice versa. However, this leads to a contradiction as shown by the linear equations.
7. **Example**:
Consider:
\[
A_0 = \begin{pmatrix} 1 & \pi & \pi \\ \pi & 5 & \pi \\ \pi & \pi & -7 \end{pmatrix}
\]
Using the formula, it can be shown that \( A_2 \) and \( A_4 \) both have a null entry.
The final answer is \(\boxed{2}\)
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For a positive integer $n$, $\sigma(n)$ denotes the sum of the positive divisors of $n$. Determine $$\limsup\limits_{n\rightarrow \infty} \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}}$$
[b]Note:[/b] Given a sequence ($a_n$) of real numbers, we say that $\limsup\limits_{n\rightarrow \infty} a_n = +\infty$ if ($a_n$) is not upper bounded, and, otherwise, $\limsup\limits_{n\rightarrow \infty} a_n$ is the smallest constant $C$ such that, for every real $K > C$, there is a positive integer $N$ with $a_n < K$ for every $n > N$.
|
To determine the value of
$$ \limsup_{n\rightarrow \infty} \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}}, $$
we start by using the formula for the sum of the divisors function $\sigma(n)$ when $n$ is expressed as a product of prime powers. Let $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, then
$$ \sigma(n) = (1 + p_1 + p_1^2 + \cdots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{\alpha_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{\alpha_k}). $$
1. **Expressing $\sigma(n^{2023})$:**
For $n^{2023} = (p_1^{\alpha_1})^{2023} (p_2^{\alpha_2})^{2023} \cdots (p_k^{\alpha_k})^{2023}$, we have
$$ \sigma(n^{2023}) = \sigma(p_1^{2023\alpha_1} p_2^{2023\alpha_2} \cdots p_k^{2023\alpha_k}) = (1 + p_1 + p_1^2 + \cdots + p_1^{2023\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{2023\alpha_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{2023\alpha_k}). $$
2. **Expressing the ratio:**
We need to evaluate
$$ \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}} = \frac{(1 + p_1 + p_1^2 + \cdots + p_1^{2023\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{2023\alpha_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{2023\alpha_k})}{[(1 + p_1 + p_1^2 + \cdots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{\alpha_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{\alpha_k})]^{2023}}. $$
3. **Simplifying the ratio:**
Consider the ratio for each prime factor $p_i$:
$$ \frac{1 + p_i + p_i^2 + \cdots + p_i^{2023\alpha_i}}{(1 + p_i + p_i^2 + \cdots + p_i^{\alpha_i})^{2023}}. $$
Notice that $(1 + p_i + p_i^2 + \cdots + p_i^{\alpha_i})^{2023}$ is a polynomial of degree $2023\alpha_i$ in $p_i$, and $1 + p_i + p_i^2 + \cdots + p_i^{2023\alpha_i}$ is also a polynomial of degree $2023\alpha_i$ in $p_i$.
4. **Asymptotic behavior:**
For large $n$, the dominant term in both the numerator and the denominator will be the highest power of $p_i$. Thus, asymptotically,
$$ \frac{1 + p_i + p_i^2 + \cdots + p_i^{2023\alpha_i}}{(1 + p_i + p_i^2 + \cdots + p_i^{\alpha_i})^{2023}} \approx \frac{p_i^{2023\alpha_i}}{(p_i^{\alpha_i})^{2023}} = 1. $$
5. **Conclusion:**
Since this holds for each prime factor $p_i$, we have
$$ \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}} \approx 1 $$
for large $n$. Therefore,
$$ \limsup_{n\rightarrow \infty} \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}} = 1. $$
The final answer is $\boxed{1}$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Arnaldo and Bernaldo play a game where they alternate saying natural numbers, and the winner is the one who says $0$. In each turn except the first the possible moves are determined from the previous number $n$ in the following way: write
\[n =\sum_{m\in O_n}2^m;\]
the valid numbers are the elements $m$ of $O_n$. That way, for example, after Arnaldo says $42= 2^5 + 2^3 + 2^1$, Bernaldo must respond with $5$, $3$ or $1$.
We define the sets $A,B\subset \mathbb{N}$ in the following way. We have $n\in A$ iff Arnaldo, saying $n$ in his first turn, has a winning strategy; analogously, we have $n\in B$ iff Bernaldo has a winning strategy if Arnaldo says $n$ during his first turn. This way,
\[A =\{0, 2, 8, 10,\cdots\}, B = \{1, 3, 4, 5, 6, 7, 9,\cdots\}\]
Define $f:\mathbb{N}\to \mathbb{N}$ by $f(n)=|A\cap \{0,1,\cdots,n-1\}|$. For example, $f(8) = 2$ and $f(11)=4$.
Find
\[\lim_{n\to\infty}\frac{f(n)\log(n)^{2005}}{n}\]
|
1. **Step 1:** We prove that all odd natural numbers are in \( B \).
Note that \( n \) is odd if and only if \( 0 \in O_n \). Hence if Arnaldo says \( n \) in his first turn, Bernaldo can immediately win by saying \( 0 \). Therefore, all odd numbers are in \( B \).
2. **Step 2:** We prove the inequality \( f(n) > \frac{1}{2}\sqrt{n} \) for all positive integers \( n \).
Set \( n = 2^k + m \), where \( k \) is nonnegative and \( 0 \leq m < 2^k \). Consider all nonnegative integers \( s < 2^k \) such that all numbers in \( O_s \) are odd. If Arnaldo says \( s \) in his first turn, Bernaldo will then say an odd number, and hence Arnaldo wins in his second turn. Thus every such \( s \) is in \( A \). That is, if \( s = 2^{a_1} + \cdots + 2^{a_t} \) where \( a_1, \ldots, a_t \) are distinct odd numbers, then \( s \in A \). If all the \( a_i \)'s are less than \( k \), then \( s < 2^k \leq n \); there are \( \left\lfloor \frac{k}{2} \right\rfloor \) odd nonnegative integers smaller than \( k \), and hence there are \( 2^{\left\lfloor \frac{k}{2} \right\rfloor} \) possible \( s \)'s. Thus,
\[
f(n) \geq 2^{\left\lfloor \frac{k}{2} \right\rfloor} \geq 2^{\frac{k-1}{2}} = 2^{\frac{k+1}{2}-1} > \frac{1}{2}\sqrt{n},
\]
as desired.
3. **Step 3:** We prove that \( f(2^k) = 2^{k-f(k)} \) for every nonnegative integer \( k \).
The numbers \( s \) in \( |A \cap \{0, 1, \ldots, 2^k-1\}| \) are precisely those of the form \( 2^{a_1} + \cdots + 2^{a_t} \) where \( a_1, \ldots, a_t \) are distinct nonnegative integers less than \( k \) and not in \( A \). In fact, if some number in \( O_s \) is in \( A \), then Bernaldo can win by saying that number in his first turn; conversely, if no such number is in \( A \), then Bernaldo loses whichever number he chooses. Since \( |\{0, 1, \ldots, k-1\} \setminus A| = k - f(k) \), there are \( 2^{k-f(k)} \) such numbers \( s \), and the claim follows.
4. **Final Step:** We are now ready to finish off the problem. Let \( n \in \mathbb{N} \). Set again \( n = 2^k + m \), where \( 0 \leq m < 2^k \). Then
\[
\frac{f(n) \log(n)^{2005}}{n} \leq \frac{f(2^{k+1}) \log(2^{k+1})^{2005}}{2^k} = \frac{2^{k+1-f(k+1)} ((k+1) \log(2))^{2005}}{2^k} < \frac{2 ((k+1) \log(2))^{2005}}{2^{\frac{1}{2}\sqrt{k+1}}},
\]
and since the last expression goes to \( 0 \) as \( n \rightarrow \infty \), so does the first one. Hence,
\[
\lim_{n \to \infty} \frac{f(n) \log(n)^{2005}}{n} = 0,
\]
as desired.
The final answer is \( \boxed{ 0 } \).
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Consider the multiplicative group $A=\{z\in\mathbb{C}|z^{2006^k}=1, 0<k\in\mathbb{Z}\}$ of all the roots of unity of degree $2006^k$ for all positive integers $k$.
Find the number of homomorphisms $f:A\to A$ that satisfy $f(f(x))=f(x)$ for all elements $x\in A$.
|
1. **Understanding the Group \( A \)**:
The group \( A \) consists of all roots of unity of degree \( 2006^k \) for all positive integers \( k \). This means \( A \) is the set of all complex numbers \( z \) such that \( z^{2006^k} = 1 \) for some positive integer \( k \).
2. **Idempotent Homomorphisms**:
We need to find homomorphisms \( f: A \to A \) that satisfy \( f(f(x)) = f(x) \) for all \( x \in A \). Such homomorphisms are called idempotent homomorphisms.
3. **Characterizing Idempotent Homomorphisms**:
An idempotent homomorphism \( f \) can be defined by its action on the generators of \( A \). Let \( \omega_k \) be a primitive \( 2006^k \)-th root of unity. Then \( f \) is determined by \( f(\omega_k) = \omega_k^{m_k} \) for some \( m_k \) such that \( 0 \leq m_k \leq 2006^k - 1 \).
4. **Condition for Idempotency**:
For \( f \) to be idempotent, we must have \( f(f(\omega_k)) = f(\omega_k) \). This implies:
\[
f(\omega_k^{m_k}) = \omega_k^{m_k} \implies \omega_k^{m_k^2} = \omega_k^{m_k} \implies 2006^k \mid m_k(m_k - 1)
\]
Since \( \gcd(m_k, m_k - 1) = 1 \), the only way for \( 2006^k \mid m_k(m_k - 1) \) is if \( m_k \) is either \( 0 \) or \( 1 \).
5. **Chinese Remainder Theorem**:
To find suitable \( m_k \), we use the Chinese Remainder Theorem. Suppose \( 2006 = ab \) with \( \gcd(a, b) = 1 \). Then we need:
\[
m_k \equiv 0 \pmod{a^k} \quad \text{and} \quad m_k \equiv 1 \pmod{b^k}
\]
This guarantees a unique solution modulo \( 2006^k \).
6. **Number of Idempotent Homomorphisms**:
The number of such factorizations of \( 2006 \) is given by \( 2^{\omega(2006)} \), where \( \omega(n) \) is the number of distinct prime factors of \( n \). For \( 2006 = 2 \cdot 17 \cdot 59 \), we have \( \omega(2006) = 3 \).
7. **Conclusion**:
Therefore, the number of idempotent homomorphisms \( f: A \to A \) is \( 2^3 = 8 \).
The final answer is \( \boxed{8} \)
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.