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1,601
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_26
| 2
|
If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$
|
adding $\log_{10} n$ to both sides: \[\log_{10} m + \log_{10} n=b\] using the logarithm property: $\log_a {b} + \log_a {c}=\log_a{bc}$ \[\log_{10} {mn}=b\] rewriting in exponential notation: \[10^b=mn\] \[m=\boxed{10}\] ~Vndom
| 10
|
1,602
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_27
| 1
|
A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:
$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37\text{ mph}$
|
The car takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{30 \text{ miles }}=4 \text{ hr}$ to get from $A$ to $B$ . Also, it takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{40 \text{ miles }}=3 \text{ hr}$ to get from $B$ to $A$ . Therefore, the average speed is $\dfrac{240\text{ miles }}{7 \text{ hr}}=34\dfrac{2}{7}\text{ mph}$ , which is closest to $\boxed{34}$
| 34
|
1,603
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_28
| 1
|
Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$ $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph} \qquad \textbf{(C)}\ 12\text{ mph} \qquad \textbf{(D)}\ 16\text{ mph} \qquad \textbf{(E)}\ 20\text{ mph}$
|
Let the speed of boy $A$ be $a$ , and the speed of boy $B$ be $b$ . Notice that $A$ travels $4$ miles per hour slower than boy $B$ , so we can replace $b$ with $a+4$
Now let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to make an equation, where we set the time to be equal: \[\frac{48}{a}=\frac{72}{a+4}\] Cross-multiplying gives $48a+192=72a$ . Isolating the variable $a$ , we get the equation $24a=192$ , so $a=\boxed{8}$
| 8
|
1,604
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_30
| 1
|
From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{None of these}$
|
Let us represent the number of boys $b$ , and the number of girls $g$
From the first sentence, we get that $2(g-15)=b$
From the second sentence, we get $5(b-45)=g-15$
Expanding both equations and simplifying, we get $2g-30 = b$ and $5b = g+210$
Substituting $b$ for $2g-30$ , we get $5(2g-30)=g+210$ . Solving for $g$ , we get $g = \boxed{40}$
| 40
|
1,605
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_32
| 1
|
$25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$
|
By the Pythagorean triple $(7,24,25)$ , the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \boxed{8}$
| 8
|
1,606
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_32
| 2
|
$25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$
|
We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.
\[x^2 + 7^2 = 25^2\] \[x^2 = 625 - 49\] \[x^2 = 576\] \[x = 24\]
Since the top of the ladder slipped by 4 ft the new height is $24 - 4 = 20 ft$ . The base of the ladder has moved so the new base is say $(7+y)$ . The hypotenuse remains the same at 25ft. So,
\[20^2 + (7+y)^2 = 25^2\] \[400 + 49 + y^2 + 14y = 625\] \[y^2 + 14y - 176 = 0\] \[y^2 + 22y - 8y - 176\] \[x(y+22) - 8(y+22)\] \[(y-8)(y+22)\]
Disregarding the negative solution to equation the solution to the problem is $\boxed{8}$
| 8
|
1,607
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_33
| 1
|
The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:
$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$
|
It must be assumed that the pipes have an equal height.
We can represent the amount of water carried per unit time by cross sectional area.
Cross sectional of Pipe with diameter $6 in$ \[\pi r^2 = \pi \cdot 3^2 = 9\pi\]
Cross sectional area of pipe with diameter $1 in$
\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}\]
So number of 1 in pipes required is the number obtained by dividing their cross sectional areas
\[\frac{9\pi}{\frac{\pi}{4}} = 36\]
So the answer is $\boxed{36}$
| 36
|
1,608
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35
| 1
|
In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$
|
The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$ . Therefore the inradius is $\boxed{4}$
| 4
|
1,609
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35
| 2
|
In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$
|
Since this is a right triangle, we have \[\frac{a+b-c}{2}=\boxed{4}\]
| 4
|
1,610
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_36
| 1
|
A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?
$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(C)}\ 120\% \qquad \textbf{(D)}\ 80\% \qquad \textbf{(E)}\ 75\%$
|
Without loss of generality, we can set the list price equal to $100$ . The merchant buys the goods for $100*.75=75$ . Let $x$ be the marked price.
We then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\boxed{125}$
| 125
|
1,611
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_38
| 1
|
If the expression $\begin{pmatrix}a & c\\ d & b\end{pmatrix}$ has the value $ab-cd$ for all values of $a, b, c$ and $d$ , then the equation $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$
$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \textbf{(C)}\ \text{Is satisified for no values of }x\qquad\\ \textbf{(D)}\ \text{Is satisfied for an infinite number of values of }x\qquad\\ \textbf{(E)}\ \text{None of these.}$
|
By $\begin{pmatrix}a & c\\ d & b\end{pmatrix}=ab-cd$ , we have $2x^2-x=3$ . Subtracting $3$ from both sides, giving $2x^2-x-3=0$ . This factors to $(2x-3)(x+1)=0$ . Thus, $x=\dfrac{3}{2},-1$ , so the equation is $\boxed{2}$
| 2
|
1,612
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_40
| 1
|
The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$
|
Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$ , using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$ .
Thus, the answer is $\boxed{2}$
| 2
|
1,613
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_40
| 2
|
The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$
|
The numerator of $\frac {x^2-1}{x-1}$ can be factored as $(x+1)(x-1)$ . The $x-1$ terms in the numerator and denominator cancel, so the expression is equal to $x+1$ so long as $x$ does not equal $1$ . Looking at the function's behavior near 1, we see that as $x$ approaches one, the expression approaches $\boxed{2}$
| 2
|
1,614
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45
| 1
|
The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$
|
Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be $\binom{100}{2}=4950$ . However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\boxed{4850}$
| 850
|
1,615
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45
| 2
|
The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$
|
The formula for the number of diagonals of a polygon with $n$ sides is $n(n-3)/2$ . Taking $n=100$ , we see that the number of diagonals that may be drawn in this polygon is $100(97)/2$ or $\boxed{4850}$
| 850
|
1,616
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_46
| 1
|
In triangle $ABC$ $AB=12$ $AC=7$ , and $BC=10$ . If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:
$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The median is unchanged} \qquad\\ \textbf{(E)}\ \text{The area of the triangle is 0}$
|
If you double sides $AB$ and $AC$ , they become $24$ and $14$ respectively. If $BC$ remains $10$ , then this triangle has area $0$ because ${14} + {10} = {24}$ , so two sides overlap the third side. Therefore the answer is $\boxed{0}$
| 0
|
1,617
|
https://artofproblemsolving.com/wiki/index.php/2020_USOMO_Problems/Problem_4
| 1
|
Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs.
|
Let's start off with just $(a_1, b_1), (a_2, b_2)$ and suppose that it satisfies the given condition. We could use $(1, 1), (1, 2)$ for example. We should maximize the number of conditions that the third pair satisfies. We find out that the third pair should equal $(a_1+a_2, b_1+b_2)$
We know this must be true: \[|a_1b_2-a_2b_1| = 1\]
So \[a_1b_2-a_2b_1 = 1\]
We require the maximum conditions for $(a_3, b_3)$ \[|a_3b_2-a_2b_3| = 1\] \[|a_3b_1-a_1b_3| = 1\]
Then one case can be: \[a_3b_2-a_2b_3 = 1\] \[a_3b_1-a_1b_3 = -1\]
We try to do some stuff such as solving for $a_3$ with manipulations: \[a_3b_2a_1-a_2b_3a_1 = a_1\] \[a_3b_1a_2-a_1b_3a_2 = -a_2\] \[a_3(a_1b_2-a_2b_1) = a_1+a_2\] \[a_3 = a_1+a_2\] \[a_3b_2b_1-a_2b_3b_1 = b_1\] \[a_3b_1b_2-a_1b_3b_2 = -b_2\] \[b_3(a_1b_2-a_2b_1) = b_1+b_2\] \[b_3 = b_1+b_2\]
We showed that 3 pairs are a complete graph; however, 4 pairs are not a complete graph. We will now show that: \[a_4 = a_1+2a_2\] \[b_4 = b_1+2b_2\] \[|a_1b_1+2a_2b_1-a_1b_1-2a_1b_2| = 1\] \[2|a_2b_1-a_1b_2| = 1\]
This is clearly impossible because $1$ is not even and also $|a_2b_1-a_1b_2| = 1$ .
The answer is as follows: \[0+1+2+\ldots+2\] $a_1$ has $0$ subtractions that follow condition while $a_2$ has $1$ and then the rest has $2$ .
There are $n$ terms, so our answer be $2n-3$ and in case of $n=100$ that means \[\boxed{197}.\] ~Lopkiloinm
| 197
|
1,618
|
https://artofproblemsolving.com/wiki/index.php/2020_USOMO_Problems/Problem_4
| 2
|
Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs.
|
We claim the answer is $197$
Study the points $(0, 0), (a_i, b_i), (a_j, b_j)$ . If we let these be the vertices of a triangle, applying shoelace theorem gives us an area of $\frac{1}{2}|0\times{b_i}+{a_i}\times{b_j}+{b_i}\times{0}-0\times{a_i}-{b_i}\times{a_j}-{b_j}\times{0} = \frac{1}{2}|a_ib_j - a_j b_i| = \frac{1}{2}$ . Therefore, the triangle formed by the points $(0, 0), (a_i, b_i), (a_j, b_j)$ must have an area of $\frac{1}{2}$
Two cases follow.
Case 1: Both $(a_i, b_i), (a_j, b_j)$ have exactly one coordinate equal to $0$ .
Here, one point must be on the $x$ axis and the other on the $y$ axis in order for the triangle to have a positive area. For the area of the triangle to be $\frac{1}{2}$ , it follows that the points must be $(1, 0), (0, 1)$ in some order.
Case 2: At least one of $(a_i, b_i), (a_j, b_j)$ does not have exactly one coordinate equal to $0$ . Define $S[l]$ to be a list of lines such that each line in the list has some two lattice points that, with $(0, 0)$ , form a triangle with area $\frac{1}{2}$ . Note that for any such line that passes through such two lattice points, we may trivially generate infinite lattice points on the line that have nonnegative coordinates.
Note that lines $y=1$ and $x=1$ are included in $S[l]$ , because the points $(1, 1), (2, 1)$ serve as examples for $y=1$ and $(1, 1), (1, 2)$ serve as examples for $x=1$ . For the optimal construction, include the points $(1, 0)$ and all the points $(0, 1), (0, 2), (0, 3), ... , (0, 99)$ , in that order. In this case, every adjacent pair of points would count ( $98$ ), as well as picking $(0, 1)$ and a nonadjacent point ( $99$ ), so this would be $98+99=197$
To prove that this is the maximum, consider the case where some $n$ number of points were neither on $x=1$ nor on $y=1$ . In this case, we would be removing $n$ adjacent pairs and $n$ options to choose from after choosing $(0, 0)$ , resulting in a net loss of $2n$ . By having $n$ points on some other combination of lines in $S[l]$ , we would trivially have a maximum gain of $n-1$ pairs of points on the lines such that there are no lattice points between those pairs. Because these points are not on $x=1$ or $y=1$ , the altitude from a given point to the line formed by $(0, 0)$ and $(0, 1)$ and $(1, 0)$ is not $1$ , and so the area of the triangle cannot be $\frac{1}{2}$ . Thus, by not having all points on lines $x=1$ and $y=1$ , we cannot exceed the maximum of $197$ . Thus, $\boxed{197}$ is our answer.
| 197
|
1,619
|
https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_4
| 1
|
problem_id
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object
|
We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$
| 6
|
1,620
|
https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1
| 1
|
problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object
|
Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to show the existence of acute triangles in $\{b\}.$ Note that $b_1 = 1$ and for all $i$ $b_i \le n.$
Now three arbitrary sidelengths $x$ $y$ , and $z$ , with $x \le y \le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$ 's for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$ ).
We now make another substitution: $c_i \equiv b_i ^2.$ So $c_1 = 1$ and for all $i$ $c_i \le n^2.$ Now we examine the smallest possible sets $\{c\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$ , then the smallest possible set, call it $\{s_3\},$ is trivially $\{1,1,2\}$ , since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\{s_n\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\{s_n\} = \{F_0, F_1, ... F_n\}$ , then $\{s_{n+1}\} = \{F_0, F_1, ... F_n, c_{n+1}\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\{s_n\}$ which are $F_{n-1}$ and $F_n$ . But these sum to $F_{n+1}$ so $\{s_{n+1}\} = \{F_0, F_1, ... F_{n+1}\}$ and our induction is complete.
Now since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\{c\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\{c\}$ is bounded between $1$ and $n^2$ , then the conditions of the problem are met if and only if $F_{n-1} > n^2$ . The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\boxed{13}$
| 13
|
1,621
|
https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1
| 2
|
problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object
|
Outline:
1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$
2. If the chosen $n$ is such that $F_n \le n^2$ , then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$ . It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \sqrt{F_a}, y = \sqrt{F_b}, z = \sqrt{F_c}$ with $a<b<c$ $x^2 + y^2 = F_a + F_b \le F_{b-1} + F_b = F_{b+1} \le F_c = z^2$ , x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \le n^2$ do not work.
3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \ge F_1a_1^2, a_3^2 \ge a_2^2 + a_1^2 \ge F_2a^2$ , and by induction $a_n^2 > F_na_1^2$ , a contradiction to the condition's inequality.)
4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$ . It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$ . Thus, $\boxed{13}$ is the desired solution set.
| 13
|
1,622
|
https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_4
| 1
|
Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.
|
We claim that $n = 1999$ is the smallest such number. For $n \le 1998$ , we can simply color any of the $1998$ squares forming the top row and the left column, but excluding the top left corner square.
We now show that no configuration with no colored right triangles exists for $n = 1999$ . We call a row or column filled if all $1000$ of its squares are colored. Then any of the remaining $999$ colored squares must share a column or row, respectively, with one of the colored squares in a filled row or column. These two squares, and any other square in the filled row or column, form a colored right triangle, giving us a contradiction. Hence, no filled row or column may exist.
Let $m$ be the number of columns with $1$ colored square. Then there are $1999-m$ colored squares in the remaining columns, and in each of these $< 1999-m$ columns that have a colored square must have at least two colored squares in them. These two colored squares will form a triangle with any other colored square in either of the rows containing the colored squares. Hence, each of the $1999-m$ colored squares must be placed in different rows, but as there are only $1000$ rows, the inequality $1999 - m \le 1000 \Longrightarrow m \ge 999$ holds. If $m = 1000$ , then each column only has $1$ colored square, leaving no place for the remaining $999$ , contradiction. If $m = 999$ , then each of the $1000$ rows has $1$ black square, leaving no place for the other $999$ , contradiction. Hence $n = \boxed{1999}$ is the minimal value.
| 999
|
1,623
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https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1
| 1
|
In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse , and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter
|
[asy] import olympiad; pair A, B, C, D, extensionAC; real angleABC; path braceBC; A = (0, 0); B = (2, 0); D = (1, .5); angleABC = atan(.5); //y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here: C = (6/11, 8/11); braceBC = brace(C, B, .1); label("$\mathsf{A}$", A, W); label("$\mathsf{B}$", B, E); label("$\mathsf{C}$", C, N); label("$\mathsf{D}$", D, S); label("$\mathsf{a}$", braceBC, NE); label("$\mathsf{b}$", A--C, NW); label("$\mathsf{c}$", A--B, S); label("$\mathsf{x}$", A--D, N); draw(A--B--C--cycle); draw(A--D); draw(anglemark(C, B, A)); draw(anglemark(B, A, D)); draw(anglemark(D, A, C)); draw(braceBC); [/asy] (diagram by integralarefun)
After drawing the triangle, also draw the angle bisector of $\angle A$ , and let it intersect $\overline{BC}$ at $D$ . Notice that $\triangle ADC\sim \triangle BAC$ , and let $AD=x$ . Now from similarity, \[x=\frac{bc}{a}\] However, from the angle bisector theorem, we have \[BD=\frac{ac}{b+c}\] but $\triangle ABD$ is isosceles, so \[x=BD\Longrightarrow \frac{bc}{a}=\frac{ac}{b+c}\Longrightarrow a^2=b(b+c)\] so all sets of side lengths which satisfy the conditions also meet the boxed condition.
Notice that $\text{gcd}(a, b, c)=1$ or else we can form a triangle by dividing $a, b, c$ by their greatest common divisor to get smaller integer side lengths, contradicting the perimeter minimality. Since $a$ is squared, $b$ must also be a square because if it isn't, then $b$ must share a common factor with $b+c$ , meaning it also shares a common factor with $c$ , which means $a, b, c$ share a common factor—a contradiction. Thus we let $b = x^2, b+c = y^2$ , so $a = xy$ , and we want the minimal pair $(x,y)$
By the Law of Cosines \[b^2 = a^2 + c^2 - 2ac\cos B\]
Substituting $a^2 = b^2 + bc$ yields $\cos B = \frac{b+c}{2a} = \frac{y}{2x}$ . Since $\angle C > 90^{\circ}$ $0^{\circ} < \angle B < 30^{\circ} \Longrightarrow \sqrt{3} < \frac{y}{x} < 2$ . For $x \le 3$ there are no integer solutions. For $x = 4$ , we have $y = 7$ that works, so the side lengths are $(a, b, c)=(28, 16, 33)$ and the minimal perimeter is $\boxed{77}$
| 77
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1,624
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https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1
| 2
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In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse , and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter
|
In $\triangle ABC$ let $\angle B = \beta, \angle A = 2\beta, \angle C = 180^{\circ} - 3\beta$ . From the law of sines, we have \[\frac{a}{\sin 2\beta} = \frac{b}{\sin \beta} = \frac{c} {\sin (180^{\circ} - 3\beta)} = \frac{c}{\sin 3\beta}\] Thus the ratio \[b : a : c = \sin\beta : \sin 2\beta : \sin 3\beta\] We can simplify \[\frac{\sin 2\beta}{\sin\beta} = \frac{2\sin\beta\cos\beta}{\sin\beta} = 2\cos\beta\] Likewise, \[\frac{\sin 3\beta}{\sin\beta} = \frac{\sin 2\beta\cos\beta + \sin\beta\cos 2\beta}{\sin\beta} = \frac{2\sin\beta\cos^2\beta + \sin\beta(\cos^2\beta - \sin^2\beta)}{\sin\beta}\] \[= {2 \cos^2 \beta + \cos^2 \beta - \sin^2 \beta} = 4\cos^2 \beta - 1\] Letting $\gamma = \cos\beta$ , rewrite \[b : a : c = 1 : 2\gamma : 4\gamma^2 - 1\]
We find that to satisfy the conditions for an obtuse triangle, $\beta \in (0^\circ, 30^\circ)$ and therefore $\gamma \in \left(\frac{\sqrt{3}}{2}, 1\right)$
The rational number with minimum denominator (in order to minimize scaling to obtain integer solutions) above $\frac{\sqrt{3}}{2}$ is $\frac{7}{8}$ , which also has a denominator divisible by 2 (to take advantage of the coefficients of 2 and 4 in the ratio and further minimize scaling).
Inserting $\gamma = \frac{7}{8}$ into the ratio, we find $b : a : c = 1 : \frac{7}{4} : \frac{33}{16}$ . When scaled minimally to obtain integer side lengths, we find \[b, a, c = 16, 28, 33\] and that the perimeter is $\boxed{77}$
| 77
|
1,625
|
https://artofproblemsolving.com/wiki/index.php/1986_USAMO_Problems/Problem_3
| 1
|
What is the smallest integer $n$ , greater than one, for which the root-mean-square of the first $n$ positive integers is an integer?
$\mathbf{Note.}$ The root-mean-square of $n$ numbers $a_1, a_2, \cdots, a_n$ is defined to be \[\left[\frac{a_1^2 + a_2^2 + \cdots + a_n^2}n\right]^{1/2}\]
|
Let's first obtain an algebraic expression for the root mean square of the first $n$ integers, which we denote $I_n$ . By repeatedly using the identity $(x+1)^3 = x^3 + 3x^2 + 3x + 1$ , we can write \[1^3 + 3\cdot 1^2 + 3 \cdot 1 + 1 = 2^3,\] \[1^3 + 3 \cdot(1^2 + 2^2) + 3 \cdot (1 + 2) + 1 + 1 = 3^3,\] and \[1^3 + 3\cdot(1^2 + 2^2 + 3^2) + 3 \cdot (1 + 2 + 3) + 1 + 1 + 1 = 4^3.\] We can continue this pattern indefinitely, and thus for any
positive integer $n$ \[1 + 3\sum_{j=1}^n j^2 + 3 \sum_{j=1}^n j^1 + \sum_{j=1}^n j^0 = (n+1)^3.\] Since $\sum_{j=1}^n j = n(n+1)/2$ , we obtain \[\sum_{j=1}^n j^2 = \frac{2n^3 + 3n^2 + n}{6}.\] Therefore, \[I_n = \left(\frac{1}{n} \sum_{j=1}^n j^2\right)^{1/2} = \left(\frac{2n^2 + 3n + 1}{6}\right)^{1/2}.\] Requiring that $I_n$ be an integer, we find that \[(2n+1 ) (n+1) = 6k^2,\] where $k$ is an integer. Using the Euclidean algorithm, we see that $\gcd(2n+1, n+1) = \gcd(n+1,n) = 1$ , and so $2n+1$ and $n+1$ share no
factors greater than 1. The equation above thus implies that $2n+1$ and $n+1$ is each proportional to a perfect square. Since $2n+1$ is
odd, there are only two possible cases:
Case 1: $2n+1 = 3 a^2$ and $n+1 = 2b^2$ , where $a$ and $b$ are integers.
Case 2: $2n+1 = a^2$ and $n+1 = 6b^2$
In Case 1, $2n+1 = 4b^2 -1 = 3a^2$ . This means that $(4b^2 -1)/3 = a^2$ for some integers $a$ and $b$ . We proceed by checking whether $(4b^2-1)/3$ is a perfect square for $b=2, 3, 4, \dots$ . (The solution $b=1$ leads to $n=1$ , and we are asked to find a value of $n$ greater than 1.) The smallest positive integer $b$ greater than 1 for
which $(4b^2-1)/3$ is a perfect square is $b=13$ , which results in $n=337$
In Case 2, $2n+1 = 12b^2 - 1 = a^2$ . Note that $a^2 = 2n+1$ is an odd square, and hence is congruent to $1 \pmod 4$ . But $12b^2 -1 \equiv 3 \pmod 4$ for any $b$ , so Case 2 has no solutions.
Alternatively, one can proceed by checking whether $12b^2 -1$ is a perfect square for $b=1, 2 ,3 ,\dots$ . We find that $12b^2 -1$ is not a perfect square for $b = 1,2, 3, ..., 7, 8$ , and $n= 383$ when $b=8$ . Thus the smallest positive integers $a$ and $b$ for which $12b^2- 1 = a^2$ result in a value of $n$ exceeding the value found in Case 1, which was 337.
In summary, the smallest value of $n$ greater than 1 for which $I_n$ is an integer is $\boxed{337}$
| 337
|
1,626
|
https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_5
| 1
|
Let $a_1,a_2,a_3,\cdots$ be a non-decreasing sequence of positive integers. For $m\ge1$ , define $b_m=\min\{n: a_n \ge m\}$ , that is, $b_m$ is the minimum value of $n$ such that $a_n\ge m$ . If $a_{19}=85$ , determine the maximum value of $a_1+a_2+\cdots+a_{19}+b_1+b_2+\cdots+b_{85}$
|
We create an array of dots like so: the array shall go out infinitely to the right and downwards, and at the top of the $i$ th column we fill the first $a_i$ cells with one dot each. Then the $19$ th row shall have 85 dots. Now consider the first 19 columns of this array, and consider the first 85 rows. In row $j$ , we see that the number of blank cells is equal to $b_j-1$ . Therefore the number of filled cells in the first 19 columns of row $j$ is equal to $20-b_j$
We now count the number of cells in the first 19 columns of our array, but we do it in two different ways. First, we can sum the number of dots in each column: this is simply $a_1+\cdots+a_{19}$ . Alternatively, we can sum the number of dots in each row: this is $(20-b_1)+\cdots +(20-b_{85})$ . Since we have counted the same number in two different ways, these two sums must be equal. Therefore \[a_1+\cdots +a_{19}+b_1+\cdots +b_{85}=20\cdot 85=\boxed{1700}.\] Note that this shows that the value of the desired sum is constant.
| 700
|
1,627
|
https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1
| 1
|
In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$
|
Using Vieta's formulas, we have:
\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*}
From the last of these equations, we see that $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$ . Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$ , and so $ac+ad+bc+bd=k-30$ . The key insight is now to factor the left-hand side as a product of two binomials: $(a+b)(c+d)=k-30$ , so that we now only need to determine $a+b$ and $c+d$ rather than all four of $a,b,c,d$
Let $p=a+b$ and $q=c+d$ . Plugging our known values for $ab$ and $cd$ into the third Vieta equation, $-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)$ , we have $-200 = -32(c+d) + 62(a+b) = 62p-32q$ . Moreover, the first Vieta equation, $a+b+c+d=18$ , gives $p+q=18$ . Thus we have two linear equations in $p$ and $q$ , which we solve to obtain $p=4$ and $q=14$
Therefore, we have $(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30$ , yielding $k=4\cdot 14+30 = \boxed{86}$
| 86
|
1,628
|
https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1
| 2
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In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$
|
We start as before: $ab=-32$ and $cd=62$ . We now observe that a and b must be the roots of a quadratic, $x^2+rx-32$ , where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic $x^2+sx+62$
Now
\begin{align*}x^4-18x^3+kx^2+200x-1984 =& (x^2+rx-32)(x^2+sx+62)\\ =& x^4+(r+s)x^3+(62-32+rs)x^2\\ &+(62s-32r)x-1984.\end{align*}
Equating the coefficients of $x^3$ and $x$ with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of $x^2$ and get $k=\boxed{86}.$
| 86
|
1,629
|
https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1
| 3
|
In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$
|
Let the roots of the equation be $a,b,c,$ and $d$ . By Vieta's, \begin{align*} a+b+c+d &= 18\\ ab+ac+ad+bc+bd+cd &= k\\ abc+abd+acd+bcd &=-200\\ abcd &=-1984.\\ \end{align*} Since $abcd=-1984$ and $ab=-32$ , then, $cd=62$ . Notice that \[abc + abd + acd + bcd = -200\] can be factored into \[ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).\] From the first equation, $c+d=18-a-b$ . Substituting it back into the equation, \[-32(18-a-b)+62(a+b)=-200\] Expanding, \[-576+32a+32b+62a+62b=-200 \implies 94a+94b=376\] So, $a+b=4$ and $c+d=14$ . Notice that \[ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)\] Plugging all our values in, \[-32+62+4(14)=\boxed{86}.\]
| 86
|
1,630
|
https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1
| 4
|
In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$
|
Since two of the roots have product $-32,$ the equation can be factored in the form \[x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + c).\] Expanding, we get \[x^4 - 18x^3 + kx^2 + 200x - 1984 = x^4 + (a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.\] Matching coefficients, we get
\begin{align*}
a + b &= -18, \\
ab + c - 32 &= k, \\
ac - 32b &= 200, \\
-32c &= -1984.
\end{align*}Then $c = \frac{-1984}{-32} = 62,$ so $62a - 32b = 200.$ With $a + b = -18,$ we can solve to find $a = -4$ and $b = -14.$ Then \[k = ab + c - 32 = \boxed{86}.\]
| 86
|
1,631
|
https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_1
| 1
|
In a party with $1982$ people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?
|
We induct on $n$ to prove that in a party with $n$ people, there must be at least $(n-3)$ people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people $A, B, C$ , who do not know each other.)
Base case: $n = 4$ is obvious.
Inductive step: Suppose in a party with $k$ people (with $k \ge 4$ ), at least $(k-3)$ people know everyone else. Consider a party with $(k+1)$ people. Take $k$ of the people (leaving another person, $A$ , out) and apply the inductive step to conclude that at least $(k-3)$ people know everyone else in the $k$ -person group, $G$
Now suppose that everyone in the group $G$ knows each other. Then take $3$ of these people and $A$ to deduce that $A$ knows a person $B \in G$ , which means $B$ knows everyone else. Then apply the inductive step on the remaining $k$ people (excluding $B$ ) to find $(k-3)$ people out of them that know everyone else (including $B$ , of course). Then these $(k-3)$ people and $B$ , which enumerate $(k-2)$ people, know everyone else.
Suppose that there exist two people $B, C \in G$ who do not know each other. Because $k-3 \ge 1$ , there exist at least one person in $G$ , person $D$ , who knows everyone else in $G$ . Now, take $A, B, C, D$ and observe that because $B, C$ do not know each other, either $A$ or $D$ knows everyone else of $A, B, C, D$ (by the problem condition), so in particular $A$ and $D$ know each other. Then apply the inductive step on the remaining $k$ people (excluding $D$ ) to find $(k-3)$ people out of them that know everyone else (including $D$ , of course). Then these $(k-3)$ people and $D$ , which enumerate $(k-2)$ people, know everyone else.
This completes the inductive step and thus the proof of this stronger result, which easily implies that at least $1982 - 3 = \boxed{1979}$ people know everyone else.
| 979
|
1,632
|
https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_2
| 1
|
Let $S_r=x^r+y^r+z^r$ with $x,y,z$ real. It is known that if $S_1=0$
$(*)$ $\frac{S_{m+n}}{m+n}=\frac{S_m}{m}\frac{S_n}{n}$
for $(m,n)=(2,3),(3,2),(2,5)$ , or $(5,2)$ . Determine all other pairs of integers $(m,n)$ if any, so that $(*)$ holds for all real numbers $x,y,z$ such that $x+y+z=0$
|
Claim Both $m,n$ can not be even.
Proof $x+y+z=0$ $\implies x=-(y+z)$
Since $\frac{S_{m+n}}{m+n} = \frac{S_m S_n}{mn}$
by equating cofficient of $y^{m+n}$ on LHS and RHS ,get
$\frac{2}{m+n}=\frac{4}{mn}$
$\implies \frac{m}{2} + \frac {n}{2} = \frac{m\cdot n}{2\cdot2}$
So we have, $\frac{m}{2} \biggm{|} \frac{n}{2}$ and $\frac{n}{2} \biggm{|} \frac{m}{2}$
$\implies m=n=4$
So we have $S_8=2(S_4)^2$
Now since it will true for all real $x,y,z,x+y+z=0$ .
So choose $x=1,y=-1,z=0$
$S_8=2$ and $S_4=2$ so $S_8 \neq 2 S_4^2$
This is contradiction. So, at least one of $m,n$ must be odd. WLOG assume $n$ is odd and m is even. The coefficient of $y^{m+n-1}$ in $\frac{S_{m+n}}{m+n}$ is $\frac{\binom{m+n}{1} }{m+n} =1$
The coefficient of $y^{m+n-1}$ in $\frac{S_m\cdot S_n}{m\cdot n}$ is $\frac{2}{m}$
Therefore, $\boxed{2}$
| 2
|
1,633
|
https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_4
| 1
|
Prove that there exists a positive integer $k$ such that $k\cdot2^n+1$ is composite for every integer $n$
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Indeed, $\boxed{2935363331541925531}$ has the requisite property.
| 531
|
1,634
|
https://artofproblemsolving.com/wiki/index.php/2021_USAJMO_Problems/Problem_1
| 1
|
Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]
|
The answer is $\boxed{1}$ , which works.
To show it is necessary, we first get $f(1)=f(1)^2$ , so $f(1)=1$ .
Then, we get $f(2)=f(1^2 + 1^2)=f(1)^2 =1$
| 1
|
1,635
|
https://artofproblemsolving.com/wiki/index.php/2020_USOJMO_Problems/Problem_5
| 1
|
Suppose that $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_{100},b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i,j)$ satisfying $1\leq i<j\leq 100$ and $|a_ib_j-a_jb_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs.
|
Call the pair $(i, j)$ good if $1\leq i < j \leq 100$ and $|a_ib_j-a_jb_i|=1$ . Note that we can reorder the pairs $(a_1, b_1), (a_2, b_2), \ldots, (a_{100}, b_{100})$ without changing the number of good pairs. Thus, we can reorder them so that $a_1\leq a_2\leq\ldots\leq a_{100}$ . Furthermore, reorder them so that if $a_i=a_j$ for some $i<j$ , then $b_i<b_j$
Now I claim the maximum value of $N$ is $\boxed{197}$ . First, we will show $N\leq 197$
| 197
|
1,636
|
https://artofproblemsolving.com/wiki/index.php/2016_USAJMO_Problems/Problem_4
| 1
|
Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,...,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$
|
Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$ . Then the smallest possible sum of $2016$ of the remaining elements is \[2017+2018+\cdots + 4032 = 1008 \cdot 6049 = 6097392\] so clearly $N\ge 6097392$ . We will show that $N=6097392$ works.
$\vspace{0.2 in}$
$\{1,2\cdots 6097392\}$ contain the integers from $1$ to $6048$ , so pair these numbers as follows:
\[1, 6048\]
\[2, 6047\]
\[3, 6046\]
\[\cdots\]
\[3024, 3025\]
When we remove any $2016$ integers from the set $\{1,2,\cdots N\}$ , clearly we can remove numbers from at most $2016$ of the $3024$ pairs above, leaving at least $1008$ complete pairs. To get a sum of $N$ , simply take these $1008$ pairs, all of which sum to $6049$ . The sum of these $2016$ elements is $1008 \cdot 6049 = 6097392$ , as desired.
$\vspace{0.2 in}$
We have shown that $N$ must be at least $6097392$ , and that this value is attainable. Therefore our answer is $\boxed{6097392}$
| 392
|
1,637
|
https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_4
| 2
|
problem_id
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object
|
We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$
| 6
|
1,638
|
https://artofproblemsolving.com/wiki/index.php/2013_USAJMO_Problems/Problem_4
| 1
|
Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd.
|
First of all, note that $f(n)$ $\sum_{i=0}^{k} f(n-2^{i})$ where $k$ is the largest integer such that $2^k \le n$ . We let $f(0) = 1$ for convenience.
From here, we proceed by induction, with our claim being that the only $n$ such that $f(n)$ is odd are $n$ representable of the form $2^{a} - 1, a \in \mathbb{Z}$
We induct on $a$ . It is trivially true for $a = 0$ and $a = 1$ . From here, we show that, if the only numbers $n \le 2^{a-1} - 1$ where $f(n)$ is odd are of the form described above, then the only numbers $n \le 2^{a} -1$ that are odd are of that form. We first consider all numbers $b$ , such that $2^{a-1} \le b \le 2^{a} - 2$ , going from the lower bound to the upper bound (a mini induction, you might say). We know that $f(b) = \sum_{i=0}^{a-1} f(b-2^{i})$ . For a number in this summation to be odd, $b - 2^i = 2^m -1 \rightarrow b = 2^i + 2^m - 1$ . However, we know that $b > 2^{a-1}$ , so $m$ must be equal to $a-1$ , or else $b$ cannot be in that interval. Now, from this, we know that $i < a-1$ , as $b<2^{a} - 1$ . Therefore, $i$ and $m$ are distinct, and thus $f(b - 2^i)$ and $f(b- 2^{a-1})$ are odd; since there are just two odd numbers, the ending sum for any $b$ is even. Finally, considering $2^{a} - 1$ , the only odd number is $f(2^{a} - 1 - 2^{a-1})$ , so the ending sum is odd. $\Box$
The smallest $n$ greater than $2013$ expressible as $2^d - 1, d \in \mathbb{N}$ is $2^{11} -1 = \boxed{2047}$
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1,639
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https://artofproblemsolving.com/wiki/index.php/2013_USAJMO_Problems/Problem_4
| 2
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Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd.
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Of course, as with any number theory problem, use actual numbers to start, not variables! By plotting out the first few sums (do it!) and looking for patterns, we observe that $f(n)=\sum_{\textrm{power}=0}^{\textrm{pow}_{\textrm{larg}}} f(n-2^{\textrm{power}})$ , where $\textrm{pow}_{\textrm{larg}}$ represents the largest power of $2$ that is smaller than $n$ . I will call this sum the Divine Sign, or DS.
But wait a minute... we are trying to determine odd/even of $f(n)$ . Why not call all the evens 0 and odds 1, basically using mod 2? Sounds so simple. Draw a small table for the values: as $n$ goes up from $0$ , you get: $1,1,0,1,0,0,0,1,0...$ . We have to set $f(0)=1$ for this to work. Already it looks like $f(n)$ is only odd if $n=2^{\textrm{power}}-1$
The only tool here is induction. The base case is clearly established. Then let's assume we successfully made our claim up to $2^n-1$ . We need to visit numbers from $2^n$ to $2^{n+1}-1$ . Realize that $2^n$ has $0$ for $f$ because there will be two numbers in DS that give a $f$ of one: $2^{n-1}$ and $1$
But to look at whether a value of $f(\textrm{number})$ is 1 or 0, we need to revisit our first equation. We can answer this rather natural question: When will a number to be inducted upon, say $2^n+k$ , ever have a 1 as $f(\textrm{number})$ in the DS equation? Well- because by our assumption of the claim up to $2^n-1$ , we know that the only way for that to happen is if $2^n+k-2^{\textrm{power}}$ in the DS is equal to $2^{\textrm{Some power}} - 1$ . Clearly $1 \leq k \leq 2^n - 1$
Finally, we can simplify. Using our last equation, $2^n+k-2^{\textrm{power}}=2^{\textrm{Some power}}-1$ , regrouping gives $2^n+k=2^{\textrm{power}}+2^{\textrm{Some power}}-1$
Most importantly, realize that $\textrm{power}$ can be from $0$ to $n$ , because of the restraints on $k$ mentioned earlier. Same with $\textrm{Some power}$ . Immediately at least one of $\textrm{power}$ and $\textrm{Some power}$ has to be $n$ . If both were smaller, LHS is greater, contradiction. If both were greater, RHS is greater, contradiction.
Therefore, by setting one of $\textrm{power}$ or $\textrm{Some power}$ to $n$ , we realize $k=2^{\textrm{A certain power}}-1$
The conclusion is clear, right? Each $k$ from $1$ to $2^{n-1}-1$ yields two distinct cases: one of $\textrm{power}$ and $\textrm{Some power}$ is equal to $n$ , while the other is LESS THAN $n$ . But for $k=2^n-1$ , there is ONE CASE: BOTH values have to equal $n$ . Therefore, the only $k$ that has $f(2^n+k)$ as odd must only be $2^n-1$ , because the other ones yield a $f$ of 1+1=0 in our mod. That proves our induction for a new power of 2, namely $n+1$ , meaning that $f(\textrm{number})$ is only odd if $\textrm{number} = 2^{\textrm{Power of two}} - 1$ , and we are almost done...
Thus, the answer is $2^{11}-1=\boxed{2047}$
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1,640
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https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1
| 3
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problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object
|
Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to show the existence of acute triangles in $\{b\}.$ Note that $b_1 = 1$ and for all $i$ $b_i \le n.$
Now three arbitrary sidelengths $x$ $y$ , and $z$ , with $x \le y \le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$ 's for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$ ).
We now make another substitution: $c_i \equiv b_i ^2.$ So $c_1 = 1$ and for all $i$ $c_i \le n^2.$ Now we examine the smallest possible sets $\{c\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$ , then the smallest possible set, call it $\{s_3\},$ is trivially $\{1,1,2\}$ , since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\{s_n\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\{s_n\} = \{F_0, F_1, ... F_n\}$ , then $\{s_{n+1}\} = \{F_0, F_1, ... F_n, c_{n+1}\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\{s_n\}$ which are $F_{n-1}$ and $F_n$ . But these sum to $F_{n+1}$ so $\{s_{n+1}\} = \{F_0, F_1, ... F_{n+1}\}$ and our induction is complete.
Now since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\{c\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\{c\}$ is bounded between $1$ and $n^2$ , then the conditions of the problem are met if and only if $F_{n-1} > n^2$ . The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\boxed{13}$
| 13
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1,641
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https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1
| 4
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problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object
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Outline:
1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$
2. If the chosen $n$ is such that $F_n \le n^2$ , then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$ . It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \sqrt{F_a}, y = \sqrt{F_b}, z = \sqrt{F_c}$ with $a<b<c$ $x^2 + y^2 = F_a + F_b \le F_{b-1} + F_b = F_{b+1} \le F_c = z^2$ , x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \le n^2$ do not work.
3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \ge F_1a_1^2, a_3^2 \ge a_2^2 + a_1^2 \ge F_2a^2$ , and by induction $a_n^2 > F_na_1^2$ , a contradiction to the condition's inequality.)
4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$ . It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$ . Thus, $\boxed{13}$ is the desired solution set.
| 13
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1,642
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https://artofproblemsolving.com/wiki/index.php/2012_USAJMO_Problems/Problem_5
| 1
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For distinct positive integers $a$ $b < 2012$ , define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$ , where $a$ and $b$ range over all pairs of distinct positive integers less than 2012. Determine $S$
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Let $ak \equiv r_{a} \pmod{2012}$ and $bk \equiv r_{b} \pmod{2012}$ . Notice that this means $a(2012 - k) \equiv 2012 - r_{a} \pmod{2012}$ and $b(2012 - k) \equiv 2012 - r_{b} \pmod{2012}$ . Thus, for every value of $k$ where $r_{a} > r_{b}$ , there is a value of $k$ where $r_{b} > r_{a}$ . Therefore, we merely have to calculate $\frac{1}{2}$ times the number of values of $k$ for which $r_{a} \neq r_{b}$ and $r_{a} \neq 0$
However, the answer is NOT $\frac{1}{2}(2012) = 1006$ ! This is because we must count the cases where the value of $k$ makes $r_{a} = r_{b}$ or where $r_{a} = 0$
So, let's start counting.
If $k$ is even, we have either $a \equiv 0 \pmod{1006}$ or $a - b \equiv 0 \pmod{1006}$ . So, $a = 1006$ or $a = b + 1006$ . We have $1005$ even values of $k$ (which is all the possible even values of $k$ , since the two above requirements don't put any bounds on $k$ at all).
If $k$ is odd, if $k = 503$ or $k = 503 \cdot 3$ , then $a \equiv 0 \pmod{4}$ or $a \equiv b \pmod{4}$ . Otherwise, $ak \equiv 0 \pmod{2012}$ or $ak \equiv bk \pmod{2012}$ , which is impossible to satisfy, given the domain $a, b < 2012$ . So, we have $2$ values of $k$
In total, we have $2 + 1005 = 1007$ values of $k$ which makes $r_{a} = r_{b}$ or $r_{a} = 0$ , so there are $2011 - 1007 = 1004$ values of $k$ for which $r_{a} \neq r_{b}$ and $r_{a} \neq 0$ . Thus, by our reasoning above, our solution is $\frac{1}{2} \cdot 1004 = \boxed{502}$
| 502
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1,643
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https://artofproblemsolving.com/wiki/index.php/2011_USAJMO_Problems/Problem_1
| 1
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Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.
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We will first take the expression modulo $3$ . We get $2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3$
Lemma 1: All perfect squares are equal to $0$ or $1$ modulo $3$ .
We can prove this by testing the residues modulo $3$ . We have $0^2 \equiv 0 \pmod 3$ $1^2 \equiv 1 \pmod 3$ , and $2^2 \equiv 1 \pmod 3$ , so the lemma is true.
We know that if $n$ is odd, $-1^n+1^n \equiv 0 \pmod 3$ , which satisfies the lemma's conditions. However, if $n$ is even, we get $2 \pmod 3$ , which does not satisfy the lemma's conditions. So, we can conclude that $n$ is odd.
Now, we take the original expression modulo $4$ . For right now, we will assume that $n>1$ , and test $n=1$ later. For $n>1$ $2^n \equiv 0 \pmod 4$ , so $2^n+12^n+2011^n=-1^n \pmod 4$
Lemma 2: All perfect squares are equal to $0$ or $1$ modulo $4$ .
We can prove this by testing the residues modulo $4$ . We have $0^2 \equiv 0 \pmod 4$ $1^2 \equiv 1 \pmod 4$ $2^2 \equiv 0 \pmod 4$ , and $3^2 \equiv 1 \pmod 4$ , so the lemma is true.
We know that if $n$ is even, $-1^n \equiv 0 \pmod 4$ , which satisfies the lemma's conditions. However, if $n$ is odd, $-1^n \equiv -1 \equiv 3 \pmod 4$ , which does not satisfy the lemma's conditions. Therefore, $n$ must be even.
However, a number cannot be even and odd at the same time, so this is impossible. Now, we only have to test $n=1$ . We know that $2^1+12^1+2011^1=45^2$ , so the only integer is $\boxed{1}$
| 1
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1,644
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https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1
| 1
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A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1\leq k\leq n$ . Find with proof the smallest $n$ such that $P(n)$ is a multiple of $2010$
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We claim that the smallest $n$ is $67^2 = \boxed{4489}$
| 489
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1,645
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https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1
| 2
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A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1\leq k\leq n$ . Find with proof the smallest $n$ such that $P(n)$ is a multiple of $2010$
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This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as "equivalence relation":
It is possible to write all positive integers $n$ in the form $p\cdot m^2$ , where $m^2$ is the largest perfect square dividing $n$ , so $p$ is not divisible by the square of any prime. Obviously, one working permutation of $[n]$ is simply $(1, 2, \ldots, n)$ ; this is acceptable, as $ka_k$ is always $k^2$ in this sequence.
Lemma 1. We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities $p$
Proof. Let $p_k$ and $m_k$ be the values of $p$ and $m$ , respectively, for a given $k$ as defined above, such that $p$ is not divisible by the square of any prime. We can obviously permute two numbers which have the same $p$ , since if $p_j = p_w$ where $j$ and $w$ are 2 values of $k$ , then $j\cdot w = p_j^2\cdot m_j^2\cdot m_w^2$ , which is a perfect square. This proves that we can permute any numbers with the same value of $p$
End Lemma
Lemma 2. We will prove the converse of Lemma 1: Let one number have a $p$ value of $\phi$ and another, $\gamma$ $\phi\cdot f$ and $\gamma\cdot g$ are both perfect squares.
Proof. $\phi\cdot f$ and $\gamma\cdot g$ are both perfect squares, so for $\phi\cdot \gamma$ to be a perfect square, if $g$ is greater than or equal to $f$ $g/f$ must be a perfect square, too. Thus $g$ is $f$ times a square, but $g$ cannot divide any squares besides $1$ , so $g = 1f$ $g = f$ . Similarly, if $f\geq g$ , then $f = g$ for our rules to keep working.
End Lemma
We can permute $l$ numbers with the same $p$ in $l!$ ways. We must have at least 67 numbers with a certain $p$ so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as $h$ , in general, we need numbers all the way up to $h\cdot 67^2$ , so obviously, $67^2$ is the smallest such number such that we can get a $67!$ term; here 67 $p$ terms are 1. Thus we need the integers $1, 2, \ldots, 67^2$ , so $67^2$ , or $\boxed{4489}$ , is the answer.
| 489
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1,646
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https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1
| 3
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A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1\leq k\leq n$ . Find with proof the smallest $n$ such that $P(n)$ is a multiple of $2010$
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It's well known that there exists $f(n)$ and $g(n)$ such that $n = f(n) \cdot g(n)$ , no square divides $f(n)$ other than 1, and $g(n)$ is a perfect square.
We prove first: If $f(k) = f(a_k)$ $k \cdot a_k$ is a perfect square.
$k \cdot a_k = f(k) \cdot g(k) \cdot f(a_k) \cdot g(a_k) = f(k)^2 \cdot g(k) \cdot g(a_k)$ , which is a perfect square.
We will now prove: If $k \cdot a_k$ is a perfect square, $f(k) = f(a_k)$
We do proof by contrapositive: If $f(k) \neq f(a_k)$ $k \cdot a_k$ is not a perfect square.
$v_p(k)$ is the p-adic valuation of k. (Basically how many factors of p you can take out of k)
Note that if $f(k) \neq f(a_k)$ , By the Fundamental Theorem of Arithmetic, $f(k)$ and $f(a_k)$ 's prime factorization are different, and thus there exists a prime p, such that $v_p(f(k)) \neq v_p(f(a_k))$ . Also, since $f(k)$ and $f(a_k)$ is squarefree, $v_p(k), v_p(a_k) \leq 1$ . Thus, $v_p(k \cdot a_k) = 1$ , making $k \cdot a_k$ not a square.
Thus, we can only match k with $a_k$ if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the $a_k$ with f value 1, then 2, ... Thus, our answer is:
$P(n) = \prod_{1 \leq i \leq n, g(i) = 1} \left\lfloor \sqrt{\frac{n}{i}} \right \rfloor !$
For all $n < 67^2$ $P(n)$ doesn't have a factor of 67. However, if $n = 67^2$ , the first term will be a multiple of 2010, and thus the answer is $67^2 = \boxed{4489}$
| 489
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1,647
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1
| 1
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Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
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This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$ . The $18x$ is Alicia so $18\times1.5=\boxed{27}$
| 27
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1,648
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1
| 2
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Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
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The relative speed of the two is $18+12=30$ , so $\frac{3}{2}$ hours would be required to travel $45$ miles. $d=st$ , so $x=18\cdot\frac{3}{2}=\boxed{27}$
| 27
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1,649
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1
| 3
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Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
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Since $18$ mph is $\frac{3}{2}$ times $12$ mph, Alicia will travel $\frac{3}{2}$ times as far as Beth. If $x$ is the distance Beth travels, \[\frac{3}{2}x+x=45\] \[\frac{5}{2}x=45\] \[x=18\] Since this is the amount Beth traveled, the amount that Alicia traveled was \[45-18=\boxed{27}\]
| 27
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1,650
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1
| 4
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Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
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Alice and Barbara close in on each other at 30mph. Since they are 45 miles apart, they will meet in t = d/s = 45miles / 30mph = 3/2 hours. We can either calculate the distance Alice travels at 18mph or the distance Barbara travels at 12mph; since we want the distance from Alice, we go with the former. Alice (and Barbara) will meet in 1 1/2 hours at 18mph x 3/2 hours = 27 miles from A. $\boxed{27}$
| 27
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1,651
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1
| 5
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Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
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We know that Beth starts 45 miles away from City A, let’s create two equations:
Alice-> $18t=d$ Beth-> $-12t+45=d$ [-12 is the slope; 45 is the y-intercept]
Solve the system:
$18t=-12t+45 30t=45 t=1.5$
So, $18(1.5)=$ $\boxed{27}$
| 27
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1,652
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1
| 6
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Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
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Since Alicia and Beth's speeds are constant, they are directly proportional to their distances covered, so the ratio of their speeds is equal to the ratio of their covered distances. Since Alicia travels $\frac{18}{30} = \frac{3}{5}$ of their combined speed, she travels $\frac{3}{5}\cdot 45 = \boxed{27}$ of the total distance.
| 27
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1,653
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3
| 1
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How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$
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Note that $40^2=1600$ but $45^{2}=2025$ (which is over our limit of $2023$ ). Therefore, the list is $5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2$ . There are $8$ elements, so the answer is $\boxed{8}$
| 8
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1,654
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3
| 4
|
How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$
|
We know the highest value would be at least $40$ but less than $50$ so we check $45$ , prime factorizing 45. We get $3^2 \cdot 5$ . We square this and get $81 \cdot 25$ . We know that $80 \cdot 25 = 2000$ , then we add 25 and get $2025$ , which does not satisfy our requirement of having the square less than $2023$ . The largest multiple of $5$ that satisfies this is $40$ and the smallest multiple of $5$ that works is $5$ so all multiples of $5$ from $5$ to $40$ satisfy the requirements. Now we divide each element of the set by $5$ and get $1-8$ so there are $\boxed{8}$ solutions.
| 8
|
1,655
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4
| 1
|
A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$
|
Let's use the triangle inequality. We know that for a triangle, the sum of the 2 shorter sides must always be longer than the longest side. This is because if the longest side were to be as long as the sum of the other sides, or longer, we would only have a line.
Similarly, for a convex quadrilateral, the sum of the shortest 3 sides must always be longer than the longest side. Thus, the answer is $\frac{26}{2}-1=13-1=\boxed{12}$
| 12
|
1,656
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4
| 2
|
A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$
|
Say the chosen side is $a$ and the other sides are $b,c,d$
By the Generalised Polygon Inequality, $a<b+c+d$ . We also have $a+b+c+d=26\Rightarrow b+c+d=26-a$
Combining these two, we get $a<26-a\Rightarrow a<13$
The largest length that satisfies this is $a=\boxed{12}$
| 12
|
1,657
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4
| 3
|
A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$
|
The quadrilateral can by cyclic only when it is an isosceles triangle. Without Loss of Generality, lets assume that this quadrilateral is a trapezoid. We can assume this as if we inscribe a trapezoid in a triangle, the base can be the diameter of the circle which is the longest chord in the circle, therefore maximizing the side length. By Brahmagupta's Formula, the area of the quadrilateral is defined by $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semi-perimeter. If the perimeter of the quadrilateral is $26$ , then the semi-perimeter will be $13$ . The area of the quadrilateral must be positive so the difference between the semi-perimeter and a side length must be greater than $0$ as otherwise, the area will be $0$ or negative. Therefore, the longest a side can be in this quadrilateral is $\boxed{12}$
| 12
|
1,658
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4
| 4
|
A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$
|
This is an AMC 10 problem 4, so there is no need for any complex formulas. The largest singular side length from a quadrilateral comes from a trapezoid. So we can set the $2$ sides of the trapezoid equal to $4$ . Next we can split the trapezoid into $5$ triangles, where each base length of the triangle equals $4$ . So the top side equals $8$ , and the bottom side length equals $4+4+4$ $=$ $\boxed{12}$ ~ kabbybear
| 12
|
1,659
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_5
| 1
|
How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$
|
Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$
$10^{15}$ gives us $15$ digits and $243$ gives us $3$ digits. $15+3=\text{\boxed{18}$
| 18
|
1,660
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6
| 1
|
An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$
|
Each of the vertices is counted $3$ times because each vertex is shared by three different edges.
Each of the edges is counted $2$ times because each edge is shared by two different faces.
Since the sum of the integers assigned to all vertices is $21$ , the final answer is $21\times3\times2=\boxed{126}$
| 126
|
1,661
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6
| 2
|
An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$
|
Note that each vertex is counted $2\times 3=6$ times. Thus, the answer is $21\times6=\boxed{126}$
| 126
|
1,662
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6
| 3
|
An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$
|
Just set one vertice equal to $21$ , it is trivial to see that there are $3$ faces with value $42$ , and $42 \cdot 3=\boxed{126}$
| 126
|
1,663
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6
| 4
|
An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$
|
Since there are 8 vertices in a cube, there are $\dfrac{21}4$ vertices for two edges. There are $4$ edges per face, and $6$ faces in a cube, so the value of the cube is $\dfrac{21}4 \cdot 24 = \boxed{126}$
| 126
|
1,664
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6
| 5
|
An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$
|
Set each vertex to value 1, so the sum of the vertices is 8. We find that the value of the cube, if all vertices are 1, is 48. We conclude that the value of the cube is 6 times the value of the sum of the vertices. Therefore, we choose $21\times6=\boxed{126}$
| 126
|
1,665
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6
| 6
|
An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$
|
The wording of the problem implies that the answer should hold for any valid combination of integers. Thus, we choose the numbers $21, 0, 0, 0, 0, 0, 0, 0$ , which are indeed $8$ integers that add to $21$ . Doing this, we find three edges that have a value of $21$ , and from there, we get three faces with a value of $42$ (while the other three faces have a value of $0$ ). Adding the three faces together, we get $42+42+42 = \boxed{126}$
| 126
|
1,666
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9
| 2
|
A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$
|
There is one $3$ , so we need one more (three more means that either the month or units digit of the day is $3$ ). For the same reason, we need one more $0$
If $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two digits must match ( $11, 22$ ). For the second (tens digit of the day), we must have the other two be $1$ , as a month can't start with $2$ or $0$ . There are $3$ successes this way.
If $3$ is the tens digit of the day, then $0$ can be either the tens digit of the month or the units digit of the day. For the first case, $1$ must go in the other slots. For the second, the other two slots must be $1$ as well. There are $2$ successes here.
If $3$ is the units digit of the day, then $0$ could go in any of the $3$ remaining slots again. If it's the tens digit of the day, then the other digits must be $1$ . If $0$ is the units digit of the day, then the other two slots must both be $1$ . If $0$ is the tens digit of the month, then the other two slots can be either both $1$ or both $2$ . In total, there are $4$ successes here.
Summing through all cases, there are $3 + 2 + 4 = \boxed{9}$ dates.
| 9
|
1,667
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9
| 3
|
A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$
|
We start with $2023----$ we need an extra $0$ and an extra $3$ . So we have at least one of those extras in the days, except we can have the month $03$ . We now have $6$ possible months $01,02,03,10,11,12$ . For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$ . For month $2$ we need an extra $2$ so we can have the day $23$ note that we can't use $32$ because it is to large. Now for month $3$ we can have any number and multiply it by $11$ so we have the solution $11,22$ . For October we need a $1$ and a $3$ so we have $13,31$ as our choices. For November we have two choices which are $03,30$ .Now for December we have $0$ options. Summing $2+1+2+2+2$ we get $\boxed{9}$ solutions.
| 9
|
1,668
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10
| 1
|
Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$
|
Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously.
We can write the following equations:
\[\frac{ax+11}{a+1}=x+1\qquad (1)\] \[\frac{ax+33}{a+3}=x+2\qquad (2)\]
Multiplying $(x+1)$ by $(a+1)$ and solving, we get: \[ax+11=ax+a+x+1\] \[11=a+x+1\] \[a+x=10\qquad (3)\]
Multiplying $(2)$ by $(a+3)$ and solving, we get: \[ax+33=ax+2a+3x+6\] \[33=2a+3x+6\] \[2a+3x=27\qquad (4)\]
Solving the system of equations for $(3)$ and $(4)$ , we find that $a=3$ and $x=\boxed{7}$
| 7
|
1,669
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10
| 2
|
Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$
|
Suppose Maureen took $n$ tests with an average of $m$
If she takes another test, her new average is $\frac{(nm+11)}{(n+1)}=m+1$
Cross-multiplying: $nm+11=nm+n+m+1$ , so $n+m=10$
If she takes $3$ more tests, her new average is $\frac{(nm+33)}{(n+3)}=m+2$
Cross-multiplying: $nm+33=nm+2n+3m+6$ , so $2n+3m=27$
But $2n+3m$ can also be written as $2(n+m)+m=20+m$ . Therefore $m=27-20=\boxed{7}$
| 7
|
1,670
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10
| 3
|
Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$
|
Let $s$ represent the sum of Maureen's test scores previously and $t$ be the number of scores taken previously.
So, $\frac{s+11}{t+1} = \frac{s}{t}+1$ and $\frac{s+33}{t+3} = \frac{s}{t}+2$
We can use the first equation to write $s$ in terms of $t$
We then substitute this into the second equation: $\frac{-t^2+10t+33}{t+3} = \frac{-t^2+10}{t}+2$
From here, we solve for t, getting $t=3$
We substitute this to get $s=21$
Therefore, the solution to the problem is $\frac{21}{3}=$ $\boxed{7}$
| 7
|
1,671
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_12
| 1
|
How many three-digit positive integers $N$ satisfy the following properties?
$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$
|
Multiples of $5$ will always end in $0$ or $5$ , and since the numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with $5$ . Since the numbers must be divisible by 7, all possibilities have to be in the range from $7 \cdot 72$ to $7 \cdot 85$ inclusive.
$85 - 72 + 1 = 14$ $\boxed{14}$
| 14
|
1,672
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_12
| 4
|
How many three-digit positive integers $N$ satisfy the following properties?
$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$
|
The key point is that when reversed, the number must start with a $0$ or a $5$ based on the second restriction. But numbers can't start with a $0$
So the problem is simply counting the number of multiples of $7$ in the $500$ s.
$7 \times 72 = 504$ , so the first multiple is $7 \times 72$
$7 \times 85 = 595$ , so the last multiple is $7 \times 85$
Now, we just have to count $7\times 72, 7\times 73, 7\times 74,\cdots, 7\times 85$
We have a set that numbers $85-71=\boxed{14}$
| 14
|
1,673
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15
| 1
|
An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$
[asy] filldraw(circle((0,0),8),gray); filldraw(circle((-1,0),7),white); filldraw(circle((-2,0),6),gray); filldraw(circle((-3,0),5),white); filldraw(circle((-4,0),4),gray); filldraw(circle((-5,0),3),white); filldraw(circle((-6,0),2),gray); filldraw(circle((-7,0),1),white); [/asy]
$\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64$
|
Notice that the area of the shaded region is $(2^2\pi-1^2\pi)+(4^2\pi-3^2\pi)+(6^2\pi-5^2\pi)+ \cdots + (n^2\pi-(n-1)^2 \pi)$ for any even number $n$
Using the difference of squares, this simplifies to $(1+2+3+4+\cdots+n) \pi$ . So, we are basically finding the smallest $n$ such that $\frac{n(n+1)}{2}>2023 \Rightarrow n(n+1) > 4046$ . Since $60(61) > 60^2=3600$ , the only option higher than $60$ is $\boxed{64}$
| 64
|
1,674
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15
| 2
|
An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$
[asy] filldraw(circle((0,0),8),gray); filldraw(circle((-1,0),7),white); filldraw(circle((-2,0),6),gray); filldraw(circle((-3,0),5),white); filldraw(circle((-4,0),4),gray); filldraw(circle((-5,0),3),white); filldraw(circle((-6,0),2),gray); filldraw(circle((-7,0),1),white); [/asy]
$\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64$
|
After first observing the problem, we can work out a few of the areas.
1st area = $4\pi-\pi = 3\pi$
2nd area = $16\pi-9\pi = 7\pi$
3rd area = $36\pi-25\pi = 11\pi$
4th area = $64\pi-49\pi = 15\pi$
We can see that the pattern is an arithmetic sequence with first term $3$ and common difference $4$ . From here, we can start from the bottom of the answer choices and work our way up. As the question asks for the least number of circles needed total, we have to divide the number of total circles by 2.
We can find the sum of the first $32$ terms of the arithmetic sequence by using the formula.
The last term is: $3 + 4\cdot(32-1) = 127$
Then, we can find the sum: $(3+127)/2 \cdot 32 = 65\cdot32 = 2080$ . It is clear that $64$ works.
The next answer choice is $60$ , which we have to divide by 2 to get $30$
The last term is: $3 + 4\cdot(30-1) = 119$
The sum is: $(3+119)/2 \cdot 30 = 61\cdot30 = 1830$ . This does not work.
As answer choice $D$ does not work and $E$ does, we can conclude that the answer is $\boxed{64}$
| 64
|
1,675
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15
| 3
|
An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$
[asy] filldraw(circle((0,0),8),gray); filldraw(circle((-1,0),7),white); filldraw(circle((-2,0),6),gray); filldraw(circle((-3,0),5),white); filldraw(circle((-4,0),4),gray); filldraw(circle((-5,0),3),white); filldraw(circle((-6,0),2),gray); filldraw(circle((-7,0),1),white); [/asy]
$\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64$
|
We can easily see that all of the answer choices are even, which helps us solve this problem a little.
Lets just not consider the $\pi$ , since it is not that important, so let's just cancel that out.
When we plug in 64, we get $64^2-63^2+62^2-61^2+\cdots +4^2-3^2+2^2-1^2$ . By difference of squares, we get $1+2+3+\cdots+62+63+64$ , which by the sum of an arithmetic sequence, is $\frac{64(64+1)}{2}$ , which is $2080$
Similarly, we can use this for answer choice $D$ , and we have $\frac{60(60+1)}{2}$ which is $1830$
So, we see that answer choice $D$ is too small to satisfy the requirements, so we conclude the answer is $\boxed{64}$
| 64
|
1,676
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15
| 4
|
An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$
[asy] filldraw(circle((0,0),8),gray); filldraw(circle((-1,0),7),white); filldraw(circle((-2,0),6),gray); filldraw(circle((-3,0),5),white); filldraw(circle((-4,0),4),gray); filldraw(circle((-5,0),3),white); filldraw(circle((-6,0),2),gray); filldraw(circle((-7,0),1),white); [/asy]
$\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64$
|
We can consider making a table.
If there is 1 circle, the area of the shaded region is 0π. (We can write this as 0π.)
If there are 2 circles, the area of the shaded region is 3π. (We can write this as (1+2)π).
If there are 3 circles, the area of the shaded region is 3π. (We can write this as (1+2)π).
If there are 4 circles, the area of the shaded region is 10π. (We can write this as (1+2+3+4)π).
If there are 5 circles, the area of the shaded region is 10π. (We can write this as (1+2+3+4)π).
If there are 6 circles, the area of the shaded region is 21π. (We can write this as (1+2+3+4+5+6)π).
Now the pattern emerges. When $n$ is even, the area of the shaded region is $(1+2+3+\cdots+n)\pi$ , or $\left( \frac{n(n+1)}{2} \right) \pi$ . But remember that the problem stated that there are an even number of circles. So now we are solving the equation \[\left( \frac{n(n+1)}{2} \right) \pi\ge2023\pi.\] Dividing by $\pi$ and multiplying by 2 on both sides, we get $n(n+1)\ge4046$ . Now we can plug in the answer choices, and we start off with $60$ because it is the only answer choice that is a multiple of $10$ . Plugging in we get $60\cdot61=3660$ , and this is not quite yet more than $4046$ . But only option $(\text{E})$ is bigger, so we know that the solution can only be $\boxed{64}$
| 64
|
1,677
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15
| 5
|
An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$
[asy] filldraw(circle((0,0),8),gray); filldraw(circle((-1,0),7),white); filldraw(circle((-2,0),6),gray); filldraw(circle((-3,0),5),white); filldraw(circle((-4,0),4),gray); filldraw(circle((-5,0),3),white); filldraw(circle((-6,0),2),gray); filldraw(circle((-7,0),1),white); [/asy]
$\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64$
|
Denote $S$ the area of shaded region and $W$ the area of white region.
$S > W$ and $S \approx W$ if $R$ is big.
Therefore \[\pi R^2 = S + W \approx 2S \ge 4046 \pi \implies R \approx 64.\] So we conclude the answer is $\boxed{64}$
| 64
|
1,678
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_16
| 1
|
In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$
|
We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$ , and since $l = 1.4r$ $g = 2.4r$ . Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is $n(n-1)/2$ , the sum of the first $n-1$ triangular numbers. Now, setting 36 and 48 equal to the equation will show that two consecutive numbers must have a product of 72 or 96. Clearly, $72=8*9$ , so the answer is $\boxed{36}$
| 36
|
1,679
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_16
| 2
|
In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$
|
First, we know that every player played every other player, so there's a total of $\dbinom{n}{2}$ games since each pair of players forms a bijection to a game. Therefore, that rules out D. Also, if we assume the right-handed players won a total of $x$ games, the left-handed players must have won a total of $\dfrac{7}{5}x$ games, meaning that the total number of games played was $\dfrac{12}{5}x$ . Thus, the total number of games must be divisible by $12$ . Therefore leaving only answer choices B and D. Since answer choice D doesn't satisfy the first condition, the only answer that satisfies both conditions is $\boxed{36}$
| 36
|
1,680
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_16
| 3
|
In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$
|
Let $r$ be the amount of games the right-handed won. Since the left-handed won $1.4r$ games, the total number of games played can be expressed as $(2.4)r$ , or $12/5r$ , meaning that the answer is divisible by 12. This brings us down to two answer choices, $B$ and $D$ .
We note that the answer is some number $x$ choose $2$ . This means the answer is in the form $x(x-1)/2$ . Since answer choice D gives $48 = x(x-1)/2$ , and $96 = x(x-1)$ has no integer solutions, we know that $\boxed{36}$ is the only possible choice.
| 36
|
1,681
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_16
| 4
|
In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$
|
If there are $n$ players, the total number of games played must be $\binom{n}{2}$ , so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is $7:5$ , so the number of games played must also be divisible by $12$ . Finally, we notice that only $\boxed{36}$ satisfies both of these conditions.
| 36
|
1,682
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18
| 2
|
A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
|
Let $x$ be the number of vertices with three edges, and $y$ be the number of vertices with four edges. Since there are $\frac{4*12}{2}=24$ edges on the polyhedron, we can see that $\frac{3x+4y}{2}=24$ . Then, $3x+4y=48$ . Notice that by testing the answer choices, (D) is the only one that yields an integer solution for $y$ . Thus, the answer is $\boxed{8}$
| 8
|
1,683
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18
| 3
|
A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
|
With $12$ rhombi, there are $4\cdot12=48$ total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have $\dfrac{48}2=24$ total edges.
Let $A$ be the number of vertices with $3$ edges (this is what the problem asks for) and $B$ be the number of vertices with $4$ edges. We have $3A + 4B = 48$
Euler's formula states that, for all convex polyhedra, $V-E+F=2$ . In our case, $V-24+12=2\implies V=14.$ We know that $A+B$ is the total number of vertices as we are given that all vertices are connected to either $3$ or $4$ edges. Therefore, $A+B=14.$
We now have a system of two equations. There are many ways to solve for $A$ ; choosing one yields $A=\boxed{8}$
| 8
|
1,684
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18
| 4
|
A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
|
Note that Euler's formula is $V+F-E=2$ . We know $F=12$ from the question. We also know $E = \frac{12 \cdot 4}{2} = 24$ because every face has $4$ edges and every edge is shared by $2$ faces. We can solve for the vertices based on this information.
Using the formula we can find: \[V + 12 - 24 = 2\] \[V = 14\] Let $t$ be the number of vertices with $3$ edges and $f$ be the number of vertices with $4$ edges. We know $t+f = 14$ from the question and $3t + 4f = 48$ . The second equation is because the total number of points is $48$ because there are 12 rhombuses of $4$ vertices.
Now, we just have to solve a system of equations. \[3t + 4f = 48\] \[3t + 3f = 42\] \[f = 6\] \[t = 8\] Our answer is simply just $t$ , which is $\boxed{8}$ ~musicalpenguin
| 8
|
1,685
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18
| 6
|
A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
|
Note that a rhombic dodecahedron is formed when a cube is turned inside out (as seen here ), thus there are 6 4-vertices (corresponding to each face of the cube) and 8 3-vertices (corresponding to each corner of the cube). Thus the answer is $\boxed{8}$
| 8
|
1,686
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18
| 7
|
A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
|
Let $m$ be the number of $4$ -edge vertices, and $n$ be the number of $3$ -edge vertices. The total number of vertices is $m+n$ . Now, we know that there are $4 \cdot 12 = 48$ vertices, but we have overcounted. We have overcounted $m$ vertices $3$ times and overcounted $n$ vertices $2$ times. Therefore, we subtract $3m$ and $2n$ from $48$ and set it equal to our original number of vertices. \[48 - 3m - 2n = m+n\] \[4m + 3n = 48\] From here, we reduce both sides modulo $4$ . The $4m$ disappears, and the left hand side becomes $3n$ . The right hand side is $0$ , meaning that $3n$ must be divisible by $4$ . Looking at the answer choices, this is only possible for $n = \boxed{8}$
| 8
|
1,687
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_19
| 1
|
The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$
|
Due to rotations preserving an equal distance, we can bash the answer with the distance formula. $D(A, P) = D(A', P)$ , and $D(B, P) = D(B',P)$ .
Thus we will square our equations to yield: $(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2$ , and $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$ .
Canceling $(3-s)^2$ from the second equation makes it clear that $r$ equals $3.5$
Substituting will yield
\begin{align*}(2.5)^2+(2-s)^2 &= (-0.5)^2+(1-s)^2 \\ 6.25+4-4s+s^2 &= 0.25+1-2s+s^2 \\ 2s &= 9 \\ s &=4.5 \\ \end{align*}
Now $|r-s| = |3.5-4.5| = \boxed{1}$
| 1
|
1,688
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_19
| 2
|
The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$
|
Due to rotations preserving distance, we have that $BP = B^\prime P$ , as well as $AP = A^\prime P$ . From here, we can see that P must be on the perpendicular bisector of $\overline{BB^\prime}$ due to the property of perpendicular bisectors keeping the distance to two points constant.
From here, we proceed to find the perpendicular bisector of $\overline{BB^\prime}$ . We can see that this is just a horizontal line segment with midpoint at $(3.5, 3)$ . This means that the equation of the perpendicular bisector is $x = 3.5$
Similarly, we find the perpendicular bisector of $\overline{AA^\prime}$ . We find the slope to be $\frac{1-2}{3-1} = -\frac12$ , so our new slope will be $2$ . The midpoint of $A$ and $A^\prime$ is $(2, \frac32)$ , which we can use with our slope to get another equation of $y = 2x - \frac52$
Now, point P has to lie on both of these perpendicular bisectors, meaning that it has to satisfy both equations. Plugging in the value of $x$ we found earlier, we find that $y=4.5$ . This means that $|r - s| = |3.5 - 4.5| = \boxed{1}$
| 1
|
1,689
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_19
| 3
|
The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$
|
To find the center of rotation, we find the intersection point of the perpendicular bisectors of $\overline{AA^\prime}$ and $\overline{BB^\prime}$
We can find that the equation of the line $\overline{AA^\prime}$ is $y = -\frac{1}{2}x + \frac{5}{2}$ , and that the equation of the line $\overline{BB^\prime}$ is $y = 3$
When we solve for the perpendicular bisector of $y = -\frac{1}{2}x + \frac{5}{2}$ , we determine that it has a slope of 2, and it runs through $(2, 1.5)$ . Plugging in $1.5 = 2(2)-n$ , we get than $n = \frac{5}{2}$ . Therefore our perpendicular bisector is $y=2x-\frac{5}{2}$ . Next, we solve for the perpendicular of $y = 3$ . We know that it has an undefined slope, and it runs through $(3.5, 3)$ . We can determine that our second perpendicular bisector is $x = 3.5$
Setting the equations equal to each other, we get $2x-\frac{5}{2} = 3.5$ . Solving for x, we get that $x = \frac{9}{2}$ . Therefore, $|r - s| = |3.5 - 4.5| = \boxed{1}$
| 1
|
1,690
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_19
| 4
|
The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$
|
We use the complex numbers approach to solve this problem.
Denote by $\theta$ the angle that $AB$ rotates about $P$ in the counterclockwise direction.
Thus, $A' - P = e^{i \theta} \left( A - P \right)$ and $B' - P = e^{i \theta} \left( B - P \right)$
Taking ratio of these two equations, we get \[ \frac{A' - P}{A - P} = \frac{B' - P}{B - P} . \]
By solving this equation, we get $P = \frac{7}{2} + i \frac{9}{2}$ .
Therefore, $|s-t| = \left| \frac{7}{2} - \frac{9}{2} \right| = \boxed{1}$
| 1
|
1,691
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20
| 1
|
Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("B", (19.5,7.5)); label("R", (22.5,7.5)); label("B", (25.5,7.5)); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$
|
Let a "tile" denote a $1\times1$ square and "square" refer to $2\times2$
We first have $4!=24$ possible ways to fill out the top left square. We then fill out the bottom right tile. In the bottom right square, we already have one corner filled out (from our initial coloring), and we now have $3$ options left to pick from.
We then look at the right middle tile. It is part of two squares: the top right and top left. Among these squares, $3$ colors have already been used, so we only have one more option for it. Similarly, every other square only has one more option, so we have a total of $3\cdot4!=\boxed{72}$ ways.
| 72
|
1,692
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20
| 2
|
Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("B", (19.5,7.5)); label("R", (22.5,7.5)); label("B", (25.5,7.5)); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$
|
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("R", (1.5,1.5)); label("B", (4.5,1.5)); label("R", (7.5,1.5)); label("G", (1.5,4.5)); label("W", (4.5,4.5)); label("G", (7.5,4.5)); label("B", (1.5,7.5)); label("R", (4.5,7.5)); label("B", (7.5,7.5)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("R", (19.5,7.5)); label("B", (22.5,7.5)); label("R", (25.5,7.5)); [/asy] We can split this problem into $2$ cases as shown above. We can swap a set of equal colors for another set of equal colors to create a new square.
Square 1:
The first square can be rotated to create another square so we have to multiply the number of arrangements by $2$ . We have $4! = 24$ arrangements without rotating and $24\cdot 2 = 48$ arrangements in total for the first square.
Square 2:
There are $4! = 24$ ways to arrange the colors.
In total, we have $48 + 24 = \boxed{72}$ arrangements.
| 72
|
1,693
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20
| 3
|
Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("B", (19.5,7.5)); label("R", (22.5,7.5)); label("B", (25.5,7.5)); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$
|
Let’s call the top-right corner color A, the top-middle color B, the top-right color C, and so on, with color D being the middle row, and right corner square, and color G being the bottom-left square’s color. WLOG A=Red, B=White, D=Blue, and E=Green. We will now consider squares C and F’s colors.
Case 1 : C=Red and F=Blue
In this case, we get that G and H have to be Red and White in some order, and the same for H and I. We can color this in 2 ways.
Case 2 : C=Blue and F=Red
In this case, one of G and H needs to be White and Red, and H and I needs to be White and Blue. There is 1 way to color this.
In total, we get 24*(2+1)=72 ways to color the grid. $\boxed{72}$
| 72
|
1,694
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20
| 4
|
Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("B", (19.5,7.5)); label("R", (22.5,7.5)); label("B", (25.5,7.5)); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$
|
We will choose colors step-by-step:
1. There are $4$ ways to choose a color in the center.
2. Then we select any corner and there would be $3$ ways to choose a color as we can't use the same color as the one in the center.
3. Consider the $2\times 2$ square that contains the center and the corner we have selected. For the other $2$ squares, there are $2$ ways to choose colors.
4. Now, consider how many configurations it makes sense to construct the $2\times 2$ square opposite to the corner we have selected using the $2$ other $2\times 2$ squares, and we get $3$ configurations.
Finally, the answer is $4 \cdot 3 \cdot 2 \cdot 3 = \boxed{72}$
| 72
|
1,695
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20
| 5
|
Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("B", (19.5,7.5)); label("R", (22.5,7.5)); label("B", (25.5,7.5)); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$
|
Note that there can be no overlap between colors in each square.
Then, we can choose $1$ color to be in the center. ${4 \choose 1}$ = 4
Now, we have some casework:
Case 1: 1 color is placed in 4 corners and then others are placed on opposite edges. $232$ $414$ $232$ There's $3!=6$ ways to do this.
Case 2: 2 colors are placed with 2 in adjacent corners and 1 edge opposite them. The final color is placed in the remaining 2 edges. $232$ $414$ $323$ The orientation of the 2 colors has 2 possibilities, and there are $3!$ color permutations. $2*3!=12$
There can't be any more ways to do this, as we have combined all cases such that each color is used once and only once per $2*2$ square.
We multiply the start with the sum of the 2 cases: $4(6+12)=\boxed{72}$
| 72
|
1,696
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20
| 6
|
Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("B", (19.5,7.5)); label("R", (22.5,7.5)); label("B", (25.5,7.5)); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$
|
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("2", (1.5,1.5)); label("1", (4.5,1.5)); label("1", (7.5,1.5)); label("2", (1.5,4.5)); label("3", (4.5,4.5)); label("1", (7.5,4.5)); label("3", (1.5,7.5)); label("1", (4.5,7.5)); label("2", (7.5,7.5)); [/asy]
Note that we could fill the 3 by 3 square with numbers of choices, rather than letters or color names. The top-left corner receives a 3 because there are 3 total choices to choose from: R, G and B. The squares bordering them has values of 2 and 1, regardless of order. 2 indicates that the small square can have any color excluding the one existing color, 1 indicates the remaining color of the 2 by 2 square. Finally, the middle square receives 3, since the first 2 by 2 square has RGB already, and it does not matter what color it has. Moving onto the next 2 by 2 square, we see that there are already 2 decided colors, so the other squares must be 2 and 1, again, regardless of the order. The same applys to the third 2 by 2 square, and finally the last square has the value of one, as it is the only square left. Multiplying the numbers, $2\times2\times2\times3\time3\times3$ $\boxed{72}$
| 72
|
1,697
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_21
| 1
|
Let $P(x)$ be the unique polynomial of minimal degree with the following properties:
The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. What is $m+n$
$\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(C) }45\qquad\textbf{(D) }47\qquad\textbf{(E) }49$
|
From the problem statement, we know $P(2-2)=0$ $P(9)=0$ and $4P(4)=0$ . Therefore, we know that $0$ $9$ , and $4$ are roots. So, we can factor $P(x)$ as $x(x - 9)(x - 4)(x - a)$ , where $a$ is the unknown root. Since $P(x) - 1 = 0$ , we plug in $x = 1$ which gives $1(-8)(-3)(1 - a) = 1$ , therefore $24(1 - a) = 1 \implies 1 - a = 1/24 \implies a = 23/24$ . Therefore, our answer is $23 + 24 =\boxed{47}$
| 47
|
1,698
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_21
| 2
|
Let $P(x)$ be the unique polynomial of minimal degree with the following properties:
The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. What is $m+n$
$\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(C) }45\qquad\textbf{(D) }47\qquad\textbf{(E) }49$
|
We proceed similarly to solution one. We get that $x(x-9)(x-4)(x-a)=1$ . Expanding, we get that $x(x-9)(x-4)(x-a)=x^4-(a+13)x^3+(13a+36)x^2-36ax$ . We know that $P(1)=1$ , so the sum of the coefficients of the cubic expression is equal to one. Thus $1+(a+13)+(13a+36)-36a=1$ . Solving for a, we get that a=23/24. Therefore, our answer is $23 + 24 =\boxed{47}$
| 47
|
1,699
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_23
| 2
|
If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ by $23$ . What is the sum of the digits of $N$
$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
|
We have 4 integers in our problem. Let's call the smallest of them $a$ $a(a+23) =$ either $(a+1)(a+21)$ or $(a+2)(a+22)$ . So, we have the following:
$a^2 + 23a = a^2 + 22a +21$ or
$a^2+23a = a^2 + 24a +44$
The second equation has negative solutions, so we discard it. The first equation has $a = 21$ , and so $a + 23 = 44$ . If we check $(a+1)(a+21)$ we get $22 \cdot 42 = 21 \cdot 44$ $44$ is $2$ times $22$ , and $42$ is $2$ times $21$ , so our solution checks out. Multiplying $21$ by $44$ , we get $924$ => $9 + 2 + 4 = \boxed{15}$
| 15
|
1,700
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_23
| 3
|
If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ by $23$ . What is the sum of the digits of $N$
$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
|
From the problems, it follows that
\begin{align*} x(x+20)&=y(y+23) = N\\ x^2+20x&=y^2+23y\\ 4x^2+4\cdot20x &= 4y^2+4\cdot23y\\ 4x^2+4\cdot20x+20^2-20^2 &= 4y^2+4\cdot23y+23^2-23^2\\ (2x+20)^2-20^2 &= (2y+23)^2-23^2\\ 23^2-20^2 &= (2y+23)^2-(2x+20)^2\\ (23+20)(23-20) &= (2y+23+2x+20)(2y+23-2x-20)\\ 43\cdot 3 &= (2y+2x+43)(2y-2x+3)\\ 129\cdot 1 &= (2y+2x+43)(2y-2x+3)\\ \end{align*} Since both $(2y+2x+43)$ and $(2y-2x+3)$ must be integer, we get two equations. \begin{align} 129 or 43 &= (2y+2x+43)\\ 1 or 3 &= 2y-2x+3\\ \end{align} 43 & 1 yields (0,0) which is not what we want.
129 & 1 yields (22,21) which is more interesting.
Simplifying the equations, we get: \begin{align*} x+y &= 43\\ x-y &= 1\\ x=22&, y=21\\ N &= (22)(22+20) = 924. \end{align*}
So, the answer is $\boxed{15}$
| 15
|
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