problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|
Find the number of pairs of integers $x$ and $y$ such that $x^2 + xy + y^2 = 28$. | To find the number of pairs of integers \( (x, y) \) such that \( x^2 + xy + y^2 = 28 \), we start by analyzing the given equation.
1. **Rewrite the equation**:
\[
x^2 + xy + y^2 = 28
\]
2. **Express \( x \) in terms of \( y \)**:
\[
x^2 + xy + y^2 - 28 = 0
\]
This is a quadratic equation in \( x... | 4 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Suppose you are given that for some positive integer $n$, $1! + 2! + \ldots + n!$ is a perfect square. Find the sum of all possible values of $n$. | 1. We start by examining the sum of factorials for small values of \( n \) to determine if the sum is a perfect square. We will check \( n = 1, 2, 3 \) and then analyze the behavior for \( n \geq 4 \).
2. For \( n = 1 \):
\[
1! = 1
\]
Since \( 1 \) is a perfect square (\( 1^2 = 1 \)), \( n = 1 \) works.
3... | 4 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Three people, John, Macky, and Rik, play a game of passing a basketball from one to another. Find the number of ways of passing the ball starting with Macky and reaching Macky again at the end of the seventh pass. | 1. **Define the problem and initial conditions:**
Let \( a_n \) denote the total number of ways the ball can get back to Macky in \( n \) turns. We need to find \( a_7 \), the number of ways to pass the ball starting with Macky and reaching Macky again at the end of the seventh pass.
2. **Establish the recurrence r... | 42 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
It is known that a certain mechanical balance can measure any object of integer mass anywhere between 1 and 2009 (both included). This balance has $k$ weights of integral values. What is the minimum $k$ for which there exist weights that satisfy this condition? | To solve this problem, we need to determine the minimum number of weights, \( k \), such that any integer mass from 1 to 2009 can be measured using these weights on a mechanical balance. The key insight is that the weights can be placed on either side of the balance, allowing us to use the concept of ternary (base-3) r... | 8 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
How many strings of ones and zeroes of length 10 are there such that there is an even number of ones, and no zero follows another zero? | 1. **Understanding the problem**: We need to find the number of binary strings of length 10 that have an even number of ones and no two consecutive zeros.
2. **Counting the number of ones**: Since the number of ones must be even, the possible counts of ones are 0, 2, 4, 6, 8, and 10.
3. **No two consecutive zeros**: ... | 72 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
We divide up the plane into disjoint regions using a circle, a rectangle and a triangle. What is the greatest number of regions that we can get? | 1. **Understanding the Problem:**
We need to find the maximum number of regions created by a circle, a rectangle, and a triangle on a plane. Each shape can intersect with the others, and we aim to maximize these intersections.
2. **Intersections Between Shapes:**
- **Triangle and Rectangle:**
Each side of t... | 21 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
You are given that \[17! = 355687ab8096000\] for some digits $a$ and $b$. Find the two-digit number $\overline{ab}$ that is missing above. | 1. **Given Information:**
We are given that \(17! = 355687ab8096000\), where \(a\) and \(b\) are unknown digits. We need to find the two-digit number \(\overline{ab}\).
2. **Divisibility by 11:**
According to the divisibility rule for 11, a number is divisible by 11 if the alternating sum of its digits is divisi... | 75 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Find the sum of all prime numbers $p$ which satisfy \[p = a^4 + b^4 + c^4 - 3\] for some primes (not necessarily distinct) $a$, $b$ and $c$. | 1. We start by analyzing the given equation \( p = a^4 + b^4 + c^4 - 3 \) where \( p \) is a prime number and \( a, b, c \) are primes.
2. First, we note that \( p \geq 2^4 + 2^4 + 2^4 - 3 = 45 \). Since \( p \) is a prime number greater than 45, \( p \) must be odd. This implies that the sum \( a^4 + b^4 + c^4 - 3 \... | 719 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of ordered pairs $(a, b)$ of positive integers that are solutions of the following equation: \[a^2 + b^2 = ab(a+b).\] | To find the number of ordered pairs \((a, b)\) of positive integers that satisfy the equation \(a^2 + b^2 = ab(a + b)\), we will follow these steps:
1. **Rearrange the given equation:**
\[
a^2 + b^2 = ab(a + b)
\]
We can rewrite this equation as:
\[
a^2 + b^2 = a^2b + ab^2
\]
2. **Move all terms ... | 1 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
The real numbers $x$, $y$, $z$, and $t$ satisfy the following equation:
\[2x^2 + 4xy + 3y^2 - 2xz -2 yz + z^2 + 1 = t + \sqrt{y + z - t} \]
Find 100 times the maximum possible value for $t$. | 1. We start with the given equation:
\[
2x^2 + 4xy + 3y^2 - 2xz - 2yz + z^2 + 1 = t + \sqrt{y + z - t}
\]
2. To simplify this equation, we aim to complete the square. Let's rewrite the left-hand side in a form that allows us to do this:
\[
2x^2 + 4xy + 3y^2 - 2xz - 2yz + z^2 + 1
\]
3. We can group a... | 125 | Inequalities | math-word-problem | Yes | Yes | aops_forum | false |
Find the largest positive integer $k$ such that $\phi ( \sigma ( 2^k)) = 2^k$. ($\phi(n)$ denotes the number of positive integers that are smaller than $n$ and relatively prime to $n$, and $\sigma(n)$ denotes the sum of divisors of $n$). As a hint, you are given that $641|2^{32}+1$. | 1. We need to find the largest positive integer \( k \) such that \( \phi(\sigma(2^k)) = 2^k \). Here, \(\phi(n)\) denotes Euler's totient function, and \(\sigma(n)\) denotes the sum of divisors of \( n \).
2. First, we calculate \(\sigma(2^k)\):
\[
\sigma(2^k) = 1 + 2 + 2^2 + \cdots + 2^k = \sum_{i=0}^k 2^i = 2... | 31 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of functions $f:\mathbb{Z}\mapsto\mathbb{Z}$ for which $f(h+k)+f(hk)=f(h)f(k)+1$, for all integers $h$ and $k$. | 1. **Substitute \( k = 1 \) into the given functional equation:**
\[
f(h+1) + f(h) = f(h)f(1) + 1 \quad \text{(Equation 1)}
\]
2. **Substitute \( h = k = 0 \) into the given functional equation:**
\[
f(0+0) + f(0) = f(0)f(0) + 1 \implies 2f(0) = f(0)^2 + 1
\]
Solving the quadratic equation \( f(0)... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
There are $n$ players in a round-robin ping-pong tournament (i.e. every two persons will play exactly one game). After some matches have been played, it is known that the total number of matches that have been played among any $n-2$ people is equal to $3^k$ (where $k$ is a fixed integer). Find the sum of all possible v... | 1. We start with the given information that the total number of matches played among any \( n-2 \) people is equal to \( 3^k \), where \( k \) is a fixed integer. The number of matches played among \( n-2 \) people can be represented by the binomial coefficient \( \binom{n-2}{2} \).
2. The binomial coefficient \( \bin... | 5 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
We have a $6 \times 6$ square, partitioned into 36 unit squares. We select some of these unit squares and draw some of their diagonals, subject to the condition that no two diagonals we draw have any common points. What is the maximal number of diagonals that we can draw? | To solve the problem of finding the maximal number of diagonals that can be drawn in a $6 \times 6$ grid of unit squares such that no two diagonals intersect, we can use a combinatorial approach and a coloring argument.
1. **Understanding the Grid and Diagonals**:
- A $6 \times 6$ grid consists of 36 unit squares.
... | 18 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
We randomly choose 5 distinct positive integers less than or equal to 90. What is the floor of 10 times the expected value of the fourth largest number? | To solve the problem, we need to find the expected value of the fourth largest number when choosing 5 distinct positive integers from the set \(\{1, 2, \ldots, 90\}\). We will then multiply this expected value by 10 and take the floor of the result.
1. **Understanding the Problem:**
We are choosing 5 distinct integ... | 606 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Consider the solid with 4 triangles and 4 regular hexagons as faces, where each triangle borders 3 hexagons, and all the sides are of length 1. Compute the [i]square[/i] of the volume of the solid. Express your result in reduced fraction and concatenate the numerator with the denominator (e.g., if you think that the sq... | 1. **Identify the solid**: The solid described has 4 triangles and 4 regular hexagons as faces, where each triangle borders 3 hexagons. This solid is known as a truncated tetrahedron.
2. **Volume of a regular tetrahedron**: The volume \( V \) of a regular tetrahedron with side length \( a \) is given by:
\[
V = ... | 52972 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Suppose that for some positive integer $n$, the first two digits of $5^n$ and $2^n$ are identical. Suppose the first two digits are $a$ and $b$ in this order. Find the two-digit number $\overline{ab}$. | 1. Let \( 5^n = \overline{aba_k a_{k-1} \cdots a_0} \) and \( 2^n = \overline{abb_k b_{k-1} \cdots b_0} \), where \( \overline{ab} \) represents the first two digits of \( 5^n \) and \( 2^n \).
2. Note that \( 5^n \cdot 2^n = 10^n \). Let \( \overline{ab} = n \).
3. For some positive integers \( x \) and \( y \), we ha... | 31 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $S=\{p/q| q\leq 2009, p/q <1257/2009, p,q \in \mathbb{N} \}$. If the maximum element of $S$ is $p_0/q_0$ in reduced form, find $p_0+q_0$. | 1. **Express the given fraction as a continued fraction:**
\[
\frac{1257}{2009} = 0 + \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{22+\cfrac{1}{2+\cfrac{1}{5}}}}}}}
\]
2. **Analyze the continued fraction:**
- The continued fraction representation helps us find the best rational approximations ... | 595 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $(x_n)$ be a sequence of positive integers defined as follows: $x_1$ is a fixed six-digit number and for any $n \geq 1$, $x_{n+1}$ is a prime divisor of $x_n + 1$. Find $x_{19} + x_{20}$. | 1. **Initial Setup and Sequence Definition:**
Let $(x_n)$ be a sequence of positive integers where $x_1$ is a fixed six-digit number. For any $n \geq 1$, $x_{n+1}$ is a prime divisor of $x_n + 1$.
2. **Bounding the Sequence:**
Since $x_1$ is a six-digit number, we have $x_1 \leq 999999$. Therefore, $x_1 + 1 \leq... | 5 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $p_1 = 2, p_2 = 3, p_3 = 5 ...$ be the sequence of prime numbers. Find the least positive even integer $n$ so that $p_1 + p_2 + p_3 + ... + p_n$ is not prime. | 1. We need to find the least positive even integer \( n \) such that the sum of the first \( n \) prime numbers is not a prime number.
2. Let's start by calculating the sums for even values of \( n \) until we find a composite number.
- For \( n = 2 \):
\[
p_1 + p_2 = 2 + 3 = 5 \quad (\text{prime})
\... | 8 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
A train car held $6000$ pounds of mud which was $88$ percent water. Then the train car sat in the sun, and some of the water evaporated so that now the mud is only $82$ percent water. How many pounds does the mud weigh now? | 1. Initially, the train car held $6000$ pounds of mud, which was $88\%$ water. This means that $12\%$ of the mud was not water.
\[
\text{Weight of non-water mud} = 12\% \times 6000 = 0.12 \times 6000 = 720 \text{ pounds}
\]
2. After some water evaporated, the mud is now $82\%$ water. This means that $18\%$ of... | 4000 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
In five years, Tom will be twice as old as Cindy. Thirteen years ago, Tom was three times as old as Cindy. How many years ago was Tom four times as old as Cindy? | 1. Let \( t \) be Tom's current age and \( c \) be Cindy's current age.
2. According to the problem, in five years, Tom will be twice as old as Cindy:
\[
t + 5 = 2(c + 5)
\]
3. Simplify the equation:
\[
t + 5 = 2c + 10 \implies t = 2c + 5
\]
4. The problem also states that thirteen years ago, Tom was ... | 19 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Find the least positive integer $n$ such that for every prime number $p, p^2 + n$ is never prime. | 1. **Understanding the problem**: We need to find the smallest positive integer \( n \) such that for every prime number \( p \), the expression \( p^2 + n \) is never a prime number.
2. **Initial consideration**: Let's consider the nature of \( p^2 \). For any prime \( p \), \( p^2 \) is always an odd number except w... | 5 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
There are three bags of marbles. Bag two has twice as many marbles as bag one. Bag three has three times as many marbles as bag one. Half the marbles in bag one, one third the marbles in bag two, and one fourth the marbles in bag three are green. If all three bags of marbles are dumped into a single pile, $\frac{m}{n}$... | 1. Let Bag 1 have \( x \) marbles. Then Bag 2 has \( 2x \) marbles and Bag 3 has \( 3x \) marbles.
2. Calculate the number of green marbles in each bag:
- Bag 1: \(\frac{1}{2}x\) green marbles.
- Bag 2: \(\frac{1}{3}(2x) = \frac{2x}{3}\) green marbles.
- Bag 3: \(\frac{1}{4}(3x) = \frac{3x}{4}\) green marbles.... | 95 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Bill bought 13 notebooks, 26 pens, and 19 markers for 25 dollars. Paula bought 27 notebooks, 18 pens, and 31 markers for 31 dollars. How many dollars would it cost Greg to buy 24 notebooks, 120 pens, and 52 markers? | 1. We start with the given equations:
\[
13N + 26P + 19M = 25
\]
\[
27N + 18P + 31M = 31
\]
2. We need to find the cost for Greg to buy 24 notebooks, 120 pens, and 52 markers. This translates to finding the value of:
\[
24N + 120P + 52M
\]
3. To solve this, we will use linear combinations o... | 88 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
The four points $A(-1,2), B(3,-4), C(5,-6),$ and $D(-2,8)$ lie in the coordinate plane. Compute the minimum possible value of $PA + PB + PC + PD$ over all points P . | 1. **Identify the coordinates of points \(A\), \(B\), \(C\), and \(D\):**
\[
A(-1, 2), \quad B(3, -4), \quad C(5, -6), \quad D(-2, 8)
\]
2. **Find the equations of the lines \(AC\) and \(BD\):**
- For line \(AC\):
\[
\text{slope of } AC = \frac{-6 - 2}{5 - (-1)} = \frac{-8}{6} = -\frac{4}{3}
... | 23 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
What is the least possible sum of two positive integers $a$ and $b$ where $a \cdot b = 10! ?$ | 1. **Factorize \(10!\):**
\[
10! = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1
\]
Breaking it down into prime factors:
\[
10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7
\]
2. **Apply the AM-GM Inequality:**
The Arithmetic Mean-Geometric Mean (AM-GM) Inequality states that ... | 3960 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Greta is completing an art project. She has twelve sheets of paper: four red, four white, and four blue. She also has twelve paper stars: four red, four white, and four blue. She randomly places one star on each sheet of paper. The probability that no star will be placed on a sheet of paper that is the same color as th... | 1. **Define the problem and constraints**: We need to find the probability that no star will be placed on a sheet of paper that is the same color as the star. We have 12 sheets of paper and 12 stars, each divided equally into three colors: red, white, and blue.
2. **Calculate the total number of arrangements**: The to... | -4250 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Let the complex number $z = \cos\tfrac{1}{1000} + i \sin\tfrac{1}{1000}.$ Find the smallest positive integer $n$ so that $z^n$ has an imaginary part which exceeds $\tfrac{1}{2}.$ | 1. Given the complex number \( z = \cos\left(\frac{1}{1000}\right) + i \sin\left(\frac{1}{1000}\right) \), we can write it in polar form as \( z = \text{cis}\left(\frac{1}{1000}\right) \), where \(\text{cis} \theta = \cos \theta + i \sin \theta \).
2. By De Moivre's Theorem, for any integer \( n \), we have:
\[
... | 524 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
How many ordered triples $(a, b, c)$ of odd positive integers satisfy $a + b + c = 25?$ | 1. Since \(a, b, c\) are odd positive integers, we can express them in the form \(a = 2x + 1\), \(b = 2y + 1\), and \(c = 2z + 1\) where \(x, y, z\) are non-negative integers. This is because any odd integer can be written as \(2k + 1\) for some integer \(k\).
2. Substitute these expressions into the equation \(a + b ... | 78 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
If $a$ and $b$ are complex numbers such that $a^2 + b^2 = 5$ and $a^3 + b^3 = 7$, then their sum, $a + b$, is real. The greatest possible value for the sum $a + b$ is $\tfrac{m+\sqrt{n}}{2}$ where $m$ and $n$ are integers. Find $n.$ | 1. Let \( a + b = x \). We are given two equations involving \( a \) and \( b \):
\[
a^2 + b^2 = 5
\]
\[
a^3 + b^3 = 7
\]
2. We can use the identity for the sum of cubes:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
Substituting \( a + b = x \) and \( a^2 + b^2 = 5 \), we get:
\[
a^3 +... | 57 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
How many distinct four letter arrangements can be formed by rearranging the letters found in the word [b]FLUFFY[/b]? For example, FLYF and ULFY are two possible arrangements. | To determine the number of distinct four-letter arrangements that can be formed by rearranging the letters in the word **FLUFFY**, we need to consider the different cases based on the number of F's used in the arrangement.
The letters available are: F, F, F, L, U, Y.
### Case 1: Using 1 F
If we use 1 F, we need to ch... | 72 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of non-congruent scalene triangles whose sides all have integral length, and the longest side has length $11$. | To find the number of non-congruent scalene triangles with integral side lengths where the longest side is 11, we need to consider the triangle inequality theorem. The triangle inequality theorem states that for any triangle with sides \(a\), \(b\), and \(c\) (where \(a \leq b \leq c\)), the following must hold:
1. \(a... | 20 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Aisha went shopping. At the first store she spent $40$ percent of her money plus four dollars. At the second store she spent $50$ percent of her remaining money plus $5$ dollars. At the third store she spent $60$ percent of her remaining money plus six dollars. When Aisha was done shopping at the three stores, she had ... | 1. **Determine the amount of money before entering the third store:**
- Let \( x \) be the amount of money Aisha had before entering the third store.
- At the third store, she spent \( 60\% \) of her remaining money plus 6 dollars.
- After shopping at the third store, she had 2 dollars left.
- Therefore, we... | 90 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
One plant is now $44$ centimeters tall and will grow at a rate of $3$ centimeters every $2$ years. A second plant is now $80$ centimeters tall and will grow at a rate of $5$ centimeters every $6$ years. In how many years will the plants be the same height? | 1. Define the height functions for both plants:
- The height of the first plant as a function of time \( t \) (in years) is given by:
\[
h_1(t) = 44 + \frac{3}{2}t
\]
- The height of the second plant as a function of time \( t \) (in years) is given by:
\[
h_2(t) = 80 + \frac{5}{6}t
... | 54 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Let $ d_n$ be the determinant of the $ n\times n$ matrix whose entries, from left to right and then from top to bottom, are $ \cos 1,\cos 2,\dots,\cos n^2.$ (For example, $ d_3 \equal{} \begin{vmatrix}\cos 1 & \cos2 & \cos3 \\
\cos4 & \cos5 & \cos 6 \\
\cos7 & \cos8 & \cos 9\end{vmatrix}.$ The argument of $ \cos$ i... | 1. Consider the \( n \times n \) matrix \( A_n \) whose entries are \( \cos 1, \cos 2, \ldots, \cos n^2 \). We need to evaluate the limit of the determinant of this matrix as \( n \) approaches infinity, i.e., \( \lim_{n \to \infty} d_n \).
2. To understand the behavior of \( d_n \), we need to analyze the structure o... | 0 | Calculus | other | Yes | Yes | aops_forum | false |
Say that a polynomial with real coefficients in two variable, $ x,y,$ is [i]balanced[/i] if the average value of the polynomial on each circle centered at the origin is $ 0.$ The balanced polynomials of degree at most $ 2009$ form a vector space $ V$ over $ \mathbb{R}.$ Find the dimension of $ V.$ | 1. **Understanding the Problem:**
We need to find the dimension of the vector space \( V \) of balanced polynomials of degree at most 2009. A polynomial is balanced if its average value on each circle centered at the origin is 0.
2. **Homogeneous Polynomials:**
Split the polynomial into its homogeneous parts. Fo... | 2020050 | Other | math-word-problem | Yes | Yes | aops_forum | false |
How many integer solutions $ (x_0$, $ x_1$, $ x_2$, $ x_3$, $ x_4$, $ x_5$, $ x_6)$ does the equation
\[ 2x_0^2\plus{}x_1^2\plus{}x_2^2\plus{}x_3^2\plus{}x_4^2\plus{}x_5^2\plus{}x_6^2\equal{}9\]
have? | To find the number of integer solutions to the equation
\[ 2x_0^2 + x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 = 9, \]
we will consider different cases based on the values of \(x_0, x_1, x_2, x_3, x_4, x_5,\) and \(x_6\).
1. **Case 1: \(x_0 = \pm 2\) and one of \(x_i = \pm 1\) for \(1 \leq i \leq 6\), and all othe... | 996 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$ be nine points in space such that $ABCDE$, $ABFGH$, and $GFCDI$ are each regular pentagons with side length $1$. Determine the lengths of the sides of triangle $EHI$. | 1. **Understanding the Problem:**
We are given three regular pentagons \(ABCDE\), \(ABFGH\), and \(GFCDI\) with side length 1. We need to determine the lengths of the sides of triangle \(EHI\).
2. **Analyzing the Geometry:**
- Since \(ABCDE\) is a regular pentagon, all its sides are equal to 1.
- Similarly, \... | 1 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let be a real number $ a\in \left[ 2+\sqrt 2,4 \right] . $ Find $ \inf_{\stackrel{z\in\mathbb{C}}{|z|\le 1}} \left| z^2-az+a \right| . $ | 1. **Define the function and the domain:**
We need to find the infimum of the function \( f(z) = |z^2 - az + a| \) where \( z \in \mathbb{C} \) and \( |z| \leq 1 \).
2. **Consider the polynomial \( P(z) = z^2 - az + a \):**
We need to analyze the behavior of \( P(z) \) on the boundary of the unit disk \( |z| = ... | 1 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Find the natural numbers $ n\ge 2 $ which have the property that the ring of integers modulo $ n $ has exactly an element that is not a sum of two squares. | 1. **Initial Setup and Definitions:**
We need to find natural numbers \( n \ge 2 \) such that the ring of integers modulo \( n \), denoted \( \mathbb{Z}/n\mathbb{Z} \), has exactly one element that is not a sum of two squares.
2. **Multiplicative Closure of Sums of Two Squares:**
The set of elements which, modul... | 4 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Find the largest natural number $ n$ for which there exist different sets $ S_1,S_2,\ldots,S_n$ such that:
$ 1^\circ$ $ |S_i\cup S_j|\leq 2004$ for each two $ 1\leq i,j\le n$ and
$ 2^\circ$ $ S_i\cup S_j\cup S_k\equal{}\{1,2,\ldots,2008\}$ for each three integers $ 1\le i<j<k\le n$. | 1. **Define the universal set and complementary sets:**
Let \( U = \{1, 2, \ldots, 2008\} \) and define \( A_k = U \setminus S_k \) for each \( k \). This means \( S_k = U \setminus A_k \).
2. **Reformulate the conditions:**
- The first condition \( |S_i \cup S_j| \leq 2004 \) translates to \( |U \setminus (A_i ... | 32 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Determine the smallest integer $n > 1$ with the property that $n^2(n - 1)$ is divisible by 2009. | To determine the smallest integer \( n > 1 \) such that \( n^2(n - 1) \) is divisible by 2009, we first need to factorize 2009.
1. **Factorize 2009:**
\[
2009 = 7 \times 7 \times 41 = 7^2 \times 41
\]
Therefore, \( n^2(n - 1) \) must be divisible by \( 7^2 \times 41 \).
2. **Divisibility by \( 7^2 \):**
... | 42 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
A game is played on a board with an infinite row of holes labelled $0, 1, 2, \dots$. Initially, $2009$ pebbles are put into hole $1$; the other holes are left empty. Now steps are performed according to the following scheme:
(i) At each step, two pebbles are removed from one of the holes (if possible), and one pebble ... | To show that the game always terminates and that the number of pebbles in hole \(0\) at the end of the game is independent of the specific sequence of steps, we will proceed as follows:
1. **Invariant and Termination**:
- Define the total number of pebbles on the board as \(N\).
- Initially, \(N = 2009\).
- E... | 1953 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
In an ${8}$×${8}$ squares chart , we dig out $n$ squares , then we cannot cut a "T"shaped-5-squares out of the surplus chart .
Then find the mininum value of $n$ . | To solve this problem, we need to ensure that after removing \( n \) squares from an \( 8 \times 8 \) grid, it is impossible to form a "T"-shaped figure using any 5 of the remaining squares. A "T"-shaped figure consists of a central square with four adjacent squares forming the arms of the "T".
1. **Understanding the ... | 32 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Let $z_1$ and $z_2$ be the zeros of the polynomial $f(x) = x^2 + 6x + 11$. Compute $(1 + z^2_1z_2)(1 + z_1z_2^2)$. | 1. Given the polynomial \( f(x) = x^2 + 6x + 11 \), we know that the roots \( z_1 \) and \( z_2 \) satisfy Vieta's formulas:
\[
z_1 + z_2 = -\text{coefficient of } x = -6
\]
\[
z_1 z_2 = \text{constant term} = 11
\]
2. We need to compute \( (1 + z_1^2 z_2)(1 + z_1 z_2^2) \). First, let's expand this ... | 1266 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
No math tournament exam is complete without a self referencing question. What is the product of
the smallest prime factor of the number of words in this problem times the largest prime factor of the
number of words in this problem | 1. First, we need to count the number of words in the problem statement. The problem statement is:
"No math tournament exam is complete without a self referencing question. What is the product of the smallest prime factor of the number of words in this problem times the largest prime factor of the number of words in... | 1681 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
The dollar is now worth $\frac{1}{980}$ ounce of gold. After the $n^{th}$ 7001 billion dollars bailout package passed by congress, the dollar gains $\frac{1}{2{}^2{}^{n-1}}$ of its $(n-1)^{th}$ value in gold. After four bank bailouts, the dollar is worth $\frac{1}{b}(1-\frac{1}{2^c})$ in gold, where $b, c$ are positive... | 1. Let \( a_n \) be the value of the dollar in gold after the \( n \)-th bailout. The sequence is defined recursively by the equation:
\[
a_n = a_{n-1} \left( \frac{2^{2^{n-1}} + 1}{2^{2^{n-1}}} \right)
\]
We are given that \( a_0 = \frac{1}{980} \).
2. We need to find \( a_4 \). Using the recursive formul... | 506 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
All of the roots of $x^3+ax^2+bx+c$ are positive integers greater than $2$, and the coefficients satisfy $a+b+c+1=-2009$. Find $a$ | 1. Given the polynomial \( P(x) = x^3 + ax^2 + bx + c \) with roots \( r_1, r_2, r_3 \), we know from Vieta's formulas that:
\[
r_1 + r_2 + r_3 = -a
\]
\[
r_1r_2 + r_2r_3 + r_3r_1 = b
\]
\[
r_1r_2r_3 = -c
\]
2. We are also given that \( a + b + c + 1 = -2009 \). This can be rewritten as:
... | -58 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
The sum of all of the interior angles of seven polygons is $180\times17$. Find the total number of sides of the polygons. | 1. Let \( n \) be the total number of sides of the seven polygons.
2. The sum of the interior angles of a polygon with \( s \) sides is given by the formula:
\[
180(s-2) \text{ degrees}
\]
3. For seven polygons, the sum of all the interior angles is:
\[
180(s_1 - 2) + 180(s_2 - 2) + \cdots + 180(s_7 - 2)... | 31 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
$ABCD$ forms a rhombus. $E$ is the intersection of $AC$ and $BD$. $F$ lie on $AD$ such that $EF$ is perpendicular to $FD$. Given
$EF=2$ and $FD=1$. Find the area of the rhombus $ABCD$ | 1. Given that $ABCD$ forms a rhombus, we know that the diagonals $AC$ and $BD$ intersect at right angles at point $E$ and bisect each other. Therefore, $\triangle AED$ is a right triangle with $E$ as the right angle.
2. Since $EF$ is perpendicular to $FD$, $\triangle EFD$ is a right triangle with $EF$ as one leg and $F... | 20 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
In the future, each country in the world produces its Olympic athletes via cloning and strict training
programs. Therefore, in the fi nals of the 200 m free, there are two indistinguishable athletes from each
of the four countries. How many ways are there to arrange them into eight lanes? | 1. We need to arrange 8 athletes into 8 lanes, where there are 2 indistinguishable athletes from each of the 4 countries.
2. First, we choose 2 lanes out of 8 for the first country's athletes. The number of ways to do this is given by the binomial coefficient:
\[
\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \time... | 2520 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
If $x$ and $y$ are positive integers, and $x^4+y^4=4721$, find all possible values of $x+y$ | 1. We start with the equation \( x^4 + y^4 = 4721 \) and note that \( x \) and \( y \) are positive integers.
2. To find possible values of \( x \) and \( y \), we observe that one of \( x^4 \) or \( y^4 \) must be greater than half of 4721, which is 2360.5. This is because if both \( x^4 \) and \( y^4 \) were less tha... | 13 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
A number $N$ has 2009 positive factors. What is the maximum number of positive factors that $N^2$ could have? | 1. **Understanding the problem**: We need to find the maximum number of positive factors of \( N^2 \) given that \( N \) has exactly 2009 positive factors.
2. **Prime factorization and number of factors**: If \( N \) has a prime factorization of the form \( N = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \), then the number ... | 7857 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $a$ and $b$ be integer solutions to $17a+6b=13$. What is the smallest possible positive value for $a-b$? | 1. We start with the given equation:
\[
17a + 6b = 13
\]
We need to find integer solutions for \(a\) and \(b\).
2. Express \(a\) in terms of \(b\):
\[
a = \frac{13 - 6b}{17}
\]
For \(a\) to be an integer, the numerator \(13 - 6b\) must be divisible by 17. Therefore, we need:
\[
13 - 6b \e... | 17 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
The functions $ f(x) ,\ g(x)$ satify that $ f(x) \equal{} \frac {x^3}{2} \plus{} 1 \minus{} x\int_0^x g(t)\ dt,\ g(x) \equal{} x \minus{} \int_0^1 f(t)\ dt$.
Let $ l_1,\ l_2$ be the tangent lines of the curve $ y \equal{} f(x)$, which pass through the point $ (a,\ g(a))$ on the curve $ y \equal{} g(x)$.
Find the mini... | 1. **Determine the form of \( g(x) \):**
Given \( g(x) = x - \int_0^1 f(t) \, dt \), we assume \( g(x) \) is linear and of the form:
\[
g(x) = x + \alpha
\]
2. **Substitute \( g(x) \) into the equation for \( f(x) \):**
\[
f(x) = \frac{x^3}{2} + 1 - x \int_0^x g(t) \, dt
\]
Substituting \( g(t)... | 0 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Let $ f(x)\equal{}x^2\plus{}3$ and $ y\equal{}g(x)$ be the equation of the line with the slope $ a$, which pass through the point $ (0,\ f(0))$ .
Find the maximum and minimum values of $ I(a)\equal{}3\int_{\minus{}1}^1 |f(x)\minus{}g(x)|\ dx$. | Given the function \( f(x) = x^2 + 3 \) and the line \( y = g(x) \) with slope \( a \) passing through the point \( (0, f(0)) \), we need to find the maximum and minimum values of \( I(a) = 3 \int_{-1}^1 |f(x) - g(x)| \, dx \).
First, note that \( f(0) = 3 \), so the line \( g(x) \) passing through \( (0, 3) \) with s... | 0 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Consider the polynomial $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$, with degree less than or equal to 2.
When $ f$ varies with subject to the constrain $ f(0) \equal{} 0,\ f(2) \equal{} 2$, find the minimum value of $ S\equal{}\int_0^2 |f'(x)|\ dx$. | 1. Given the polynomial \( f(x) = ax^2 + bx + c \) with the constraints \( f(0) = 0 \) and \( f(2) = 2 \), we need to find the minimum value of \( S = \int_0^2 |f'(x)| \, dx \).
2. From the constraint \( f(0) = 0 \), we have:
\[
f(0) = a(0)^2 + b(0) + c = 0 \implies c = 0
\]
Thus, the polynomial simplifies... | 2 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Find the length of the curve expressed by the polar equation: $ r\equal{}1\plus{}\cos \theta \ (0\leq \theta \leq \pi)$. | To find the length of the curve given by the polar equation \( r = 1 + \cos \theta \) for \( 0 \leq \theta \leq \pi \), we use the formula for the length of a curve in polar coordinates:
\[
L = \int_{\alpha}^{\beta} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta
\]
1. **Calculate \(\frac{dr}{d\theta}\)... | 4 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Let $ I_n\equal{}\int_0^{\sqrt{3}} \frac{1}{1\plus{}x^{n}}\ dx\ (n\equal{}1,\ 2,\ \cdots)$.
(1) Find $ I_1,\ I_2$.
(2) Find $ \lim_{n\to\infty} I_n$. | 1. **Finding \( I_1 \) and \( I_2 \):**
For \( I_1 \):
\[
I_1 = \int_0^{\sqrt{3}} \frac{1}{1 + x} \, dx
\]
This is a standard integral:
\[
I_1 = \left[ \ln(1 + x) \right]_0^{\sqrt{3}} = \ln(1 + \sqrt{3}) - \ln(1 + 0) = \ln(1 + \sqrt{3}) - \ln(1) = \ln(1 + \sqrt{3})
\]
For \( I_2 \):
\[
... | 1 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
For $ n \equal{} 1,\ 2,\ 3,\ \cdots$, let $ (p_n,\ q_n)\ (p_n > 0,\ q_n > 0)$ be the point of intersection of $ y \equal{} \ln (nx)$ and $ \left(x \minus{} \frac {1}{n}\right)^2 \plus{} y^2 \equal{} 1$.
(1) Show that $ 1 \minus{} q_n^2\leq \frac {(e \minus{} 1)^2}{n^2}$ to find $ \lim_{n\to\infty} q_n$.
(2) Find ... | 1. To find the point of intersection \((p_n, q_n)\) of the curves \(y = \ln(nx)\) and \((x - \frac{1}{n})^2 + y^2 = 1\), we start by setting \(y = q_n\) and \(x = p_n\). Thus, we have:
\[
q_n = \ln(np_n) \implies p_n = \frac{1}{n} e^{q_n}
\]
Substituting \(p_n\) into the circle equation:
\[
\left(\fra... | 1 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Let $C: y=\ln x$. For each positive integer $n$, denote by $A_n$ the area of the part enclosed by the line passing through two points $(n,\ \ln n),\ (n+1,\ \ln (n+1))$ and denote by $B_n$ that of the part enclosed by the tangent line at the point $(n,\ \ln n)$, $C$ and the line $x=n+1$. Let $g(x)=\ln (x+1)-\ln x$.
(1)... | 1. To express \(A_n\) and \(B_n\) in terms of \(n\) and \(g(n)\), we need to analyze the areas described in the problem.
**Step 1: Express \(A_n\) in terms of \(n\) and \(g(n)\)**
The area \(A_n\) is the area enclosed by the line passing through the points \((n, \ln n)\) and \((n+1, \ln (n+1))\), and the curve ... | 0 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
In the $ xyz$ space with the origin $ O$, given a cuboid $ K: |x|\leq \sqrt {3},\ |y|\leq \sqrt {3},\ 0\leq z\leq 2$ and the plane $ \alpha : z \equal{} 2$. Draw the perpendicular $ PH$ from $ P$ to the plane. Find the volume of the solid formed by all points of $ P$ which are included in $ K$ such that $ \overline{OP}... | 1. **Identify the region \( K \) and the plane \( \alpha \):**
- The cuboid \( K \) is defined by the inequalities \( |x| \leq \sqrt{3} \), \( |y| \leq \sqrt{3} \), and \( 0 \leq z \leq 2 \).
- The plane \( \alpha \) is given by \( z = 2 \).
2. **Determine the condition \( \overline{OP} \leq \overline{PH} \):**
... | 6 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let $ f(x)\equal{}\frac{1}{\sin x\sqrt{1\minus{}\cos x}}\ (0<x<\pi)$.
(1) Find the local minimum value of $ f(x)$.
(2) Evaluate $ \int_{\frac{\pi}{2}}^{\frac{2\pi}{3}} f(x)\ dx$. | ### Part 1: Finding the Local Minimum Value of \( f(x) \)
1. **Define the function:**
\[
f(x) = \frac{1}{\sin x \sqrt{1 - \cos x}}
\]
2. **Simplify the expression:**
Using the identity \(1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)\), we get:
\[
f(x) = \frac{1}{\sin x \sqrt{2 \sin^2 \left(\frac{x}... | 1 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Each of $10$ identical jars contains some milk, up to $10$ percent of its capacity. At any time, we can tell the precise amount of milk in each jar. In a move, we may pour out an exact amount of milk from one jar into each of the other $9$ jars, the same amount in each case. Prove that we can have the same amount of mi... | 1. **Initial Setup**: Let the amount of milk in the jars be denoted by \(a_1, a_2, \ldots, a_{10}\), where \(0 \leq a_i \leq 10\) for all \(i\). Assume without loss of generality that the jars are sorted in ascending order of milk volume, i.e., \(a_1 \leq a_2 \leq \cdots \leq a_{10}\).
2. **Move Description**: In each... | 9 | Logic and Puzzles | proof | Yes | Yes | aops_forum | false |
We only know that the password of a safe consists of $7$ different digits. The safe will open if we enter $7$ different digits, and one of them matches the corresponding digit of the password. Can we open this safe in less than $7$ attempts?
[i](5 points for Juniors and 4 points for Seniors)[/i] | To determine if we can open the safe in less than 7 attempts, we need to ensure that each attempt has at least one digit in the correct position. We will use a strategy to cover all possible digits in each position over multiple attempts.
1. **Understanding the Problem:**
- The password consists of 7 different digi... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
The integer $n$ has exactly six positive divisors, and they are: $1<a<b<c<d<n$. Let $k=a-1$. If the $k$-th divisor (according to above ordering) of $n$ is equal to $(1+a+b)b$, find the highest possible value of $n$. | 1. **Identify the structure of \( n \)**:
- Given that \( n \) has exactly six positive divisors, we can infer that \( n \) must be of the form \( p^5 \) or \( p^2 q \) where \( p \) and \( q \) are distinct primes. This is because the number of divisors of \( n \) is given by the product of one more than each of th... | 2009 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
If $1<k_1<k_2<...<k_n$ and $a_1,a_2,...,a_n$ are integers such that for every integer $N,$ $k_i \mid N-a_i$ for some $1 \leq i \leq n,$ find the smallest possible value of $n.$ | To solve the problem, we need to find the smallest possible value of \( n \) such that for every integer \( N \), there exists an integer \( k_i \) from the set \( \{k_1, k_2, \ldots, k_n\} \) that divides \( N - a_i \) for some \( 1 \leq i \leq n \).
1. **Initial Assumptions and Setup:**
- Given \( 1 < k_1 < k_2 ... | 5 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
A necklace consists of 100 blue and several red beads. It is known that every segment of the necklace containing 8 blue beads contain also at least 5 red beads. What minimum number of red beads can be in the necklace?
[i]Proposed by A. Golovanov[/i] | 1. **Assigning Values to Beads:**
- Let each blue bead have a value of \(1\).
- Let each red bead have a value of \(-1\).
2. **Defining the Function \( f(s) \):**
- For any segment \( s \) of the necklace, \( f(s) \) is the sum of the values of the beads in that segment.
3. **Condition Given in the Problem:*... | 65 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
On the side $ AB$ of a cyclic quadrilateral $ ABCD$ there is a point $ X$ such that diagonal $ BD$ bisects $ CX$ and diagonal $ AC$ bisects $ DX$. What is the minimum possible value of $ AB\over CD$?
[i]Proposed by S. Berlov[/i] | 1. **Define the problem and setup the geometry:**
- We are given a cyclic quadrilateral \(ABCD\) with a point \(X\) on side \(AB\).
- Diagonal \(BD\) bisects \(CX\) and diagonal \(AC\) bisects \(DX\).
- We need to find the minimum possible value of \(\frac{AB}{CD}\).
2. **Introduce parallel lines and intersec... | 2 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
$ P(x)$ is a quadratic trinomial. What maximum number of terms equal to the sum of the two preceding terms can occur in the sequence $ P(1)$, $ P(2)$, $ P(3)$, $ \dots?$
[i]Proposed by A. Golovanov[/i] | 1. Let \( P(x) = ax^2 + bx + c \) be a quadratic polynomial. We need to find the maximum number of integers \( n \ge 2 \) such that \( P(n+1) = P(n) + P(n-1) \).
2. First, express \( P(n+1) \), \( P(n) \), and \( P(n-1) \):
\[
P(n+1) = a(n+1)^2 + b(n+1) + c = a(n^2 + 2n + 1) + b(n + 1) + c = an^2 + (2a + b)n + (... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
Archimedes planned to count all of the prime numbers between $2$ and $1000$ using the Sieve of Eratosthenes as follows:
(a) List the integers from $2$ to $1000$.
(b) Circle the smallest number in the list and call this $p$.
(c) Cross out all multiples of $p$ in the list except for $p$ itself.
(d) Let $p$ be the smalles... | 1. **List the integers from $2$ to $1000$.**
- This step is straightforward and involves writing down all integers from $2$ to $1000$.
2. **Circle the smallest number in the list and call this $p$.**
- The smallest number is $2$, so we circle $2$.
3. **Cross out all multiples of $p$ in the list except for $p$ i... | 331 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $ABCD$ be a convex quadrilateral with $AC \perp BD$, and let $P$ be the intersection of $AC$ and $BD$. Suppose that the distance from $P$ to $AB$ is $99$, the distance from $P$ to $BC$ is $63$, and the distance from $P$ to $CD$ is $77$. What is the distance from $P$ to $AD$? | 1. Let \( a = PA \), \( b = PB \), \( c = PC \), and \( d = PD \). We are given the perpendicular distances from \( P \) to the sides \( AB \), \( BC \), and \( CD \) as 99, 63, and 77 respectively. We need to find the distance from \( P \) to \( AD \).
2. Since \( AC \perp BD \), \( P \) is the orthocenter of the qua... | 231 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let $a$ and $b$ be positive integers such that all but $2009$ positive integers are expressible in the form $ma + nb$, where $m$ and $n$ are nonnegative integers. If $1776 $is one of the numbers that is not expressible, find $a + b$. | 1. **Assume \(a \geq b\)**: Without loss of generality, we assume \(a \geq b\).
2. **Lemma 1: \((a, b) = 1\)**:
- **Proof**: If \((a, b) = d\) where \(d > 1\), then \(am + bn = d(xm + yn)\) for some integers \(x\) and \(y\). This implies that only multiples of \(d\) can be expressed in the form \(ma + nb\), which ... | 133 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Alice has three daughters, each of whom has two daughters, each of Alice's six grand-daughters has one daughter. How many sets of women from the family of $16$ can be chosen such that no woman and her daughter are both in the set? (Include the empty set as a possible set.) | To solve this problem, we need to count the number of sets of women from Alice's family such that no woman and her daughter are both in the set. We will use casework and the principle of multiplication to achieve this.
1. **Case 1: Alice is in the set.**
- If Alice is in the set, none of her daughters (denoted as \... | 793 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
The sequence $ \{x_n\}$ is defined by \[ \left\{ \begin{array}{l}x_1 \equal{} \frac{1}{2} \\x_n \equal{} \frac{{\sqrt {x_{n \minus{} 1} ^2 \plus{} 4x_{n \minus{} 1} } \plus{} x_{n \minus{} 1} }}{2} \\\end{array} \right.\]
Prove that the sequence $ \{y_n\}$, where $ y_n\equal{}\sum_{i\equal{}1}^{n}\frac{1}{{{x}_{i}... | 1. **Define the sequence and initial conditions:**
The sequence $\{x_n\}$ is defined by:
\[
\begin{cases}
x_1 = \frac{1}{2} \\
x_n = \frac{\sqrt{x_{n-1}^2 + 4x_{n-1}} + x_{n-1}}{2} \quad \text{for } n \geq 2
\end{cases}
\]
2. **Simplify the recursive formula:**
Let's simplify the expression for... | 6 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Let $ M \ge 3$ be an integer and let $ S \equal{} \{3,4,5,\ldots,m\}$. Find the smallest value of $ m$ such that for every partition of $ S$ into two subsets, at least one of the subsets contains integers $ a$, $ b$, and $ c$ (not necessarily distinct) such that $ ab \equal{} c$.
[b]Note[/b]: a partition of $ S$ is a ... | To solve this problem, we need to find the smallest value of \( m \) such that for every partition of \( S = \{3, 4, 5, \ldots, m\} \) into two subsets, at least one of the subsets contains integers \( a \), \( b \), and \( c \) (not necessarily distinct) such that \( ab = c \).
1. **Initial Consideration**:
- We s... | 243 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Let $ \mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $ |8 \minus{} x| \plus{} y \le 10$ and $ 3y \minus{} x \ge 15$. When $ \mathcal{R}$ is revolved around the line whose equation is $ 3y \minus{} x \equal{} 15$, the volume of the resulting solid is $ \frac {m\pi}{n... | 1. **Identify the region $\mathcal{R}$:**
- The inequality $|8 - x| + y \le 10$ can be split into two cases:
- Case 1: $x \leq 8 \implies 8 - x + y \le 10 \implies y \le x + 2$
- Case 2: $x > 8 \implies x - 8 + y \le 10 \implies y \le -x + 18$
- The inequality $3y - x \ge 15$ can be rewritten as $y \ge ... | 365 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Define an ordered triple $ (A, B, C)$ of sets to be minimally intersecting if $ |A \cap B| \equal{} |B \cap C| \equal{} |C \cap A| \equal{} 1$ and $ A \cap B \cap C \equal{} \emptyset$. For example, $ (\{1,2\},\{2,3\},\{1,3,4\})$ is a minimally intersecting triple. Let $ N$ be the number of minimally intersecting order... | 1. **Choosing the elements for intersections:**
- We need to choose three distinct elements from the set \(\{1, 2, 3, 4, 5, 6, 7\}\) to be the elements of \(A \cap B\), \(B \cap C\), and \(C \cap A\).
- The number of ways to choose 3 distinct elements from 7 is given by the combination formula \(\binom{7}{3}\):
... | 760 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Positive integers $ a$, $ b$, $ c$, and $ d$ satisfy $ a > b > c > d$, $ a \plus{} b \plus{} c \plus{} d \equal{} 2010$, and $ a^2 \minus{} b^2 \plus{} c^2 \minus{} d^2 \equal{} 2010$. Find the number of possible values of $ a$. | 1. Given the equations:
\[
a + b + c + d = 2010
\]
\[
a^2 - b^2 + c^2 - d^2 = 2010
\]
2. We can rewrite the second equation using the difference of squares:
\[
a^2 - b^2 + c^2 - d^2 = (a+b)(a-b) + (c+d)(c-d)
\]
3. Since \(a > b > c > d\) and all are positive integers, we can assume \(a - b ... | 501 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Jackie and Phil have two fair coins and a third coin that comes up heads with probability $ \frac47$. Jackie flips the three coins, and then Phil flips the three coins. Let $ \frac{m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $ m$ and $ n$ are relatively prime positive integers. Fi... | 1. **Determine the probability distribution for the number of heads in one flip of the three coins:**
- Let \( X \) be the number of heads in one flip of the three coins.
- The two fair coins each have a probability of \( \frac{1}{2} \) for heads, and the biased coin has a probability of \( \frac{4}{7} \) for hea... | 515 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Find the remainder when \[9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}\] is divided by $ 1000$. | 1. We need to find the remainder when the product \(9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}\) is divided by 1000.
2. Notice that each number in the product can be written in the form \(10^k - 1\), where \(k\) is the number of digits in the number. For example:
- \(9 = 10... | 109 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
In right triangle $ ABC$ with right angle at $ C$, $ \angle BAC < 45$ degrees and $ AB \equal{} 4$. Point $ P$ on $ AB$ is chosen such that $ \angle APC \equal{} 2\angle ACP$ and $ CP \equal{} 1$. The ratio $ \frac{AP}{BP}$ can be represented in the form $ p \plus{} q\sqrt{r}$, where $ p,q,r$ are positive integers an... | 1. **Define the angles and segments:**
Let $\angle ACP = \theta$ and $\angle APC = 2\theta$. Also, let $AP = x$ and $PB = 4 - x$ because $AB = 4$. Let $AC = y$.
2. **Apply the Law of Sines in $\triangle ACP$:**
\[
\frac{\sin \theta}{x} = \frac{\sin 2\theta}{y}
\]
Since $\sin 2\theta = 2 \sin \theta \cos... | 7 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The $ 52$ cards in a deck are numbered $ 1, 2, \ldots, 52$. Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards from a team, and the two persons with higher numbered cards form another team. Let... | 1. **Define the problem and variables:**
- We have a deck of 52 cards numbered from 1 to 52.
- Alex and Dylan pick cards \(a\) and \(a+9\) respectively.
- We need to find the minimum value of \(p(a)\) such that \(p(a) \geq \frac{1}{2}\).
2. **Calculate the total number of ways to choose the remaining two card... | 263 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Let $ ABCDEF$ be a regular hexagon. Let $ G$, $ H$, $ I$, $ J$, $ K$, and $ L$ be the midpoints of sides $ AB$, $ BC$, $ CD$, $ DE$, $ EF$, and $ AF$, respectively. The segments $ AH$, $ BI$, $ CJ$, $ DK$, $ EL$, and $ FG$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of ... | 1. **Define the problem and setup:**
Let \( ABCDEF \) be a regular hexagon with side length \( s \). Let \( G, H, I, J, K, \) and \( L \) be the midpoints of sides \( AB, BC, CD, DE, EF, \) and \( FA \), respectively. The segments \( AH, BI, CJ, DK, EL, \) and \( FG \) bound a smaller regular hexagon inside \( ABCDE... | 7 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let $ N$ be the number of ordered pairs of nonempty sets $ \mathcal{A}$ and $ \mathcal{B}$ that have the following properties:
• $ \mathcal{A} \cup \mathcal{B} \equal{} \{1,2,3,4,5,6,7,8,9,10,11,12\}$,
• $ \mathcal{A} \cap \mathcal{B} \equal{} \emptyset$,
• The number of elements of $ \mathcal{A}$ is not an elemen... | To solve the problem, we need to count the number of ordered pairs of nonempty sets \(\mathcal{A}\) and \(\mathcal{B}\) that satisfy the given conditions. Let's break down the problem step by step.
1. **Understanding the Conditions:**
- \(\mathcal{A} \cup \mathcal{B} = \{1,2,3,4,5,6,7,8,9,10,11,12\}\)
- \(\mathc... | 4094 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Let $ P(z) \equal{} z^3 \plus{} az^2 \plus{} bz \plus{} c$, where $ a$, $ b$, and $ c$ are real. There exists a complex number $ w$ such that the three roots of $ P(z)$ are $ w \plus{} 3i$, $ w \plus{} 9i$, and $ 2w \minus{} 4$, where $ i^2 \equal{} \minus{} 1$. Find $ |a \plus{} b \plus{} c|$. | 1. Given the polynomial \( P(z) = z^3 + az^2 + bz + c \) with real coefficients, and the roots \( w + 3i \), \( w + 9i \), and \( 2w - 4 \), we need to find \( |a + b + c| \).
2. Let \( w = m + ni \), where \( m \) and \( n \) are real numbers. The roots of \( P(z) \) are then \( m + (n+3)i \), \( m + (n+9)i \), and \... | 136 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $ 100$ feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks ... | 1. **Identify the total number of potential gate assignments:**
- There are 12 gates, and Dave's initial gate and new gate are chosen randomly.
- The total number of potential gate assignments is \(12 \times 11 = 132\) because the new gate must be different from the initial gate.
2. **Determine the number of val... | 52 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $ 8: 7$. Find the minimum possible value of their common perimeter. | 1. Let the side lengths of the first isosceles triangle be \(a, b, b\) and the side lengths of the second isosceles triangle be \(c, d, d\). Given that the ratio of the lengths of the bases of the two triangles is \(8:7\), we have:
\[
\frac{a}{d} = \frac{8}{7} \implies 7a = 8d \implies d = \frac{7}{8}a
\]
2. ... | 676 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
There are $24$ different pencils, $4$ different colors, and $6$ pencils of each color. They were given to $6$ children in such a way that each got $4$ pencils. What is the least number of children that you can randomly choose so that you can guarantee that you have pencils of all colors.
P.S. for 10 grade gives same p... | 1. **Define the problem and variables:**
- We have 24 pencils, divided into 4 different colors: \( R_1, R_2, R_3, R_4 \).
- Each color has 6 pencils.
- There are 6 children, each receiving 4 pencils.
2. **Consider the worst-case distribution:**
- We need to find the minimum number of children to guarantee ... | 5 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Each of $1000$ elves has a hat, red on the inside and blue on the outside or vise versa. An elf with a hat that is red outside can only lie, and an elf with a hat that is blue outside can only tell the truth. One day every elf tells every other elf, “Your hat is red on the outside.” During that day, some of the elves t... | 1. **Initial Setup and Definitions**:
- Each elf has a hat that is either red on the outside and blue on the inside, or blue on the outside and red on the inside.
- An elf with a hat that is red on the outside can only lie.
- An elf with a hat that is blue on the outside can only tell the truth.
- Each elf ... | 998 | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
In the county some pairs of towns connected by two-way non-stop flight. From any town we can flight to any other (may be not on one flight). Gives, that if we consider any cyclic (i.e. beginning and finish towns match) route, consisting odd number of flights, and close all flights of this route, then we can found two t... | 1. **Lemma.** Let \( m, n \ge 2 \) be two positive integers. A graph \( G(V, E) \) is \( mn \)-colorable if and only if there exists an edge partition \( E = E_1 \cup E_2 \) such that \( G_1 = G(V, E_1) \) and \( G_2 = G(V, E_2) \) are \( m \)- and \( n \)-colorable, respectively.
2. **Proof of Lemma.**
- **If dir... | 4 | Logic and Puzzles | proof | Yes | Yes | aops_forum | false |
The polynomial $ x^3\minus{}ax^2\plus{}bx\minus{}2010$ has three positive integer zeros. What is the smallest possible value of $ a$?
$ \textbf{(A)}\ 78 \qquad
\textbf{(B)}\ 88 \qquad
\textbf{(C)}\ 98 \qquad
\textbf{(D)}\ 108 \qquad
\textbf{(E)}\ 118$ | 1. Let the polynomial be \( P(x) = x^3 - ax^2 + bx - 2010 \). Given that it has three positive integer zeros, let these zeros be \( \alpha, \beta, \gamma \).
2. By Vieta's formulas, we know:
\[
\alpha + \beta + \gamma = a
\]
\[
\alpha \beta + \beta \gamma + \gamma \alpha = b
\]
\[
\alpha \beta ... | 78 | Algebra | MCQ | Yes | Yes | aops_forum | false |
A drawer contains red, green, blue, and white socks with at least 2 of each color. What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?
$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$ | 1. **Identify the problem**: We need to determine the minimum number of socks to pull from a drawer containing red, green, blue, and white socks to guarantee a matching pair.
2. **Consider the worst-case scenario**: In the worst-case scenario, we would pull out one sock of each color without getting a matching pair. S... | 5 | Logic and Puzzles | MCQ | Yes | Yes | aops_forum | false |
A triangle has side lengths 10, 10, and 12. A rectangle has width 4 and area equal to the area of the triangle. What is the perimeter of this rectangle?
$ \textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 36$ | 1. **Calculate the area of the triangle using Heron's formula.**
Heron's formula states that the area \( A \) of a triangle with side lengths \( a \), \( b \), and \( c \) is given by:
\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \( s \) is the semi-perimeter of the triangle:
\[
s = \frac{a + b + c}{2}... | 32 | Geometry | MCQ | Yes | Yes | aops_forum | false |
What is the sum of all the solutions of $ x \equal{} |2x \minus{} |60\minus{}2x\parallel{}$?
$ \textbf{(A)}\ 32\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 92\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 124$ | To solve the equation \( x = |2x - |60 - 2x|| \), we need to consider different cases based on the values of the expressions inside the absolute values.
### Case 1: \( 60 - 2x \geq 0 \)
This implies \( x \leq 30 \).
#### Subcase 1.1: \( 2x - (60 - 2x) \geq 0 \)
This simplifies to:
\[
2x - 60 + 2x \geq 0 \implies 4x \... | 92 | Algebra | MCQ | Yes | Yes | aops_forum | false |
Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to $ 32$?
$ \textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 ... | To solve the problem, we need to count the number of different convex cyclic quadrilaterals with integer sides and a perimeter equal to 32. We will use the fact that a quadrilateral is cyclic if and only if the sum of its opposite angles is 180 degrees. This is equivalent to the condition that the sum of any three side... | 568 | Combinatorics | MCQ | Yes | Yes | aops_forum | false |
In $ \triangle ABC, \ \cos(2A \minus{} B) \plus{} \sin(A\plus{}B) \equal{} 2$ and $ AB\equal{}4.$ What is $ BC?$
$ \textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}$ | 1. Given the equation \(\cos(2A - B) + \sin(A + B) = 2\), we know that the maximum values of \(\cos x\) and \(\sin y\) are both 1. Therefore, for the sum to be 2, both \(\cos(2A - B)\) and \(\sin(A + B)\) must be equal to 1.
2. Since \(\cos(2A - B) = 1\), it implies that \(2A - B = 360^\circ k\) for some integer \(... | 2 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
For a positive integer $k$, let $d(k)$ denote the number of divisors of $k$ and let $s(k)$ denote the digit sum of $k$. A positive integer $n$ is said to be [i]amusing[/i] if there exists a positive integer $k$ such that $d(k)=s(k)=n$. What is the smallest amusing odd integer greater than $1$? | 1. **Understanding the Problem:**
We need to find the smallest amusing odd integer greater than 1. A positive integer \( n \) is said to be amusing if there exists a positive integer \( k \) such that \( d(k) = s(k) = n \), where \( d(k) \) is the number of divisors of \( k \) and \( s(k) \) is the digit sum of \( k... | 9 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Three speed skaters have a friendly "race" on a skating oval. They all start from the same point and skate in the same direction, but with different speeds that they maintain throughout the race. The slowest skater does $1$ lap per minute, the fastest one does $3.14$ laps per minute, and the middle one does $L$ laps a ... | 1. Let \( t \) be the total duration of the race (in minutes). The skaters will meet again when the fractional parts of their laps are the same, i.e., \(\{t\} = \{Lt\} = \{3.14t\}\).
2. The number of passings of the slowest skater by the fastest skater is given by:
\[
\lfloor 3.14t \rfloor - \lfloor t \rfloor - ... | 16 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
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