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1,606 | Xenia and Sergey play the following game. Xenia thinks of a positive integer $N$ not exceeding 5000. Then she fixes 20 distinct positive integers $a_{1}, a_{2}, \ldots, a_{20}$ such that, for each $k=1,2, \ldots, 20$, the numbers $N$ and $a_{k}$ are congruent modulo $k$. By a move, Sergey tells Xenia a set $S$ of posit... | [
"Sergey can determine Xenia's number in 2 but not fewer moves.\n\n\n\nWe first show that 2 moves are sufficient. Let Sergey provide the set $\\{17,18\\}$ on his first move, and the set $\\{18,19\\}$ on the second move. In Xenia's two responses, exactly one number occurs twice, namely, $a_{18}$. Thus, Sergey is able... | [
"2"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | Bot</im_start>></im_end></im_start></answer>}
</think>
To determine the number Xenia thought of, Sergey needs to make **2 moves**. Here's a step-by-step explanation:
1. **First Move:**
- Sergey asks for a set containing $\{17, 18\}$.
- The response is a set $\{a_{17}, a_{18}\}$.
- Sergey notes that $a_{17}$ ... |
1,610 | Given a positive integer $n$, determine the largest real number $\mu$ satisfying the following condition: for every $4 n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater ... | [
"The required maximum is $\\frac{1}{2 n+2}$. To show that the condition in the statement is not met if $\\mu>\\frac{1}{2 n+2}$, let $U=(0,1) \\times(0,1)$, choose a small enough positive $\\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\\left(\\frac{i}{n+1} \\pm... | [
"$\\frac{1}{2 n+2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | English |
</think>
Given a positive integer \( n \), we aim to determine the largest real number \( \mu \) such that for any \( 4n \)-point configuration \( C \) within the open unit square \( U = (0,1) \times (0,1) \), there exists an open rectangle containing exactly one point of \( C \) and with area at least \( \mu \).
To... |
1,612 |
Find (in closed form) the difference between the number of positive integers at most $2^{2017}$ with even weight and the number of positive integers at most $2^{2017}$ with odd weight. | [
"For every integer $M \\geq 0$, let $A_{M}=\\sum_{n=-2^{M}+1}^{0}(-1)^{w(n)}$ and let $B_{M}=$ $\\sum_{n=1}^{2^{M}}(-1)^{w(n)}$; thus, $B_{M}$ evaluates the difference of the number of even weight integers in the range 1 through $2^{M}$ and the number of odd weight integers in that range.\n\n\n\nNotice that\n\n\n\n... | [
"$2^{1009}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English | 's response.\n\n\n\nSolution Step by step:\n\n\n\n1. For every integer M ≥ 0,
A_M = sum_{n= -2^{M} + 1}^{0} (-1)^{w(n)}
and B_M = sum_{n=1}^{2^{M}} (-1)^{w(n)}
So, A_M = -B_{M-1} + A_{M-1}
and B_M = B_{M-1} + A_{M-1}
2. Therefore,
B_M = A_{M-1} + B_{M-1}
3. Then, starting from B_1:
A_1 = -B_0 + A_0
B_1 = A_0 + B_0
But... |
1,613 | Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k \leq n$, and $k+1$ distinct integers $x_{1}, x_{2}, \ldots, x_{k+1}$ such that
$$
P\left(x_{1}\right)+P\left(x_{2}\right)+\cdots+... | [
"There is only one such integer, namely, $n=2$. In this case, if $P$ is a constant polynomial, the required condition is clearly satisfied; if $P=X+c$, then $P(c-1)+P(c+1)=$ $P(3 c)$; and if $P=X^{2}+q X+r$, then $P(X)=P(-X-q)$.\n\n\n\nTo rule out all other values of $n$, it is sufficient to exhibit a monic polynom... | [
"2"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English | You:<think>
Alright, so I need to figure out all positive integers \( n \) that satisfy this condition. Let me break it down and think through it step by step.
First, the condition is that for every monic polynomial \( P \) of degree at most \( n \) with integer coefficients, there exists a positive integer \( k \leq... |
1,614 | Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \ldots, A_{k}$ of $X$ is tight if the union $A_{1} \cup \cdots \cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection ... | [
"The required maximum is $2 n-2$. To describe a $(2 n-2)$-element collection satisfying the required conditions, write $X=\\{1,2, \\ldots, n\\}$ and set $B_{k}=\\{1,2, \\ldots, k\\}$, $k=1,2, \\ldots, n-1$, and $B_{k}=\\{k-n+2, k-n+3, \\ldots, n\\}, k=n, n+1, \\ldots, 2 n-2$. To show that no subcollection of the $B... | [
"$2n-2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | English | thought process initiated</assistant thought process initiated>'The solution says the required maximum is $2n-2$. The example construction uses $B_k = \{1,2,…,k\}\} for k=1,…,n−1$, then for k=n,…,2n−2, B_k is defined as {k−n+2,…,n}\}. So for n=3, it's B₁={1}, B₂={1,2}, B₃={2,3}, B₄={3}, B₅={4} (Wait, n=3, 2n−2=4, so k... |
1,618 | Determine all prime numbers $p$ and all positive integers $x$ and $y$ satisfying $x^{3}+y^{3}=$ $p(x y+p)$. | [
"Up to a swap of the first two entries, the only solutions are $(x, y, p)=(1,8,19)$, $(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$. The verification is routine.\n\n\n\nSet $s=x+y$. Rewrite the equation in the form $s\\left(s^{2}-3 x y\\right)=p(p+x y)$, and express $x y$ :\n\n\n\n$$\n\nx y=\\frac{s^{3}-p^{2}}{3 s+p}... | [
"$(1,8,19), (2,7,13), (4,5,7)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Number Theory | Math | English | : I'm looking at the system's solution where it identifies three primes and their corresponding x, y pairs. The verification is mentioned but not done, so the user is to provide that.
I need to write a detailed thought process, as if explaining to a student, to understand how the solution arrives at these answers. I s... |
1,620 | Let $n \geqslant 2$ be an integer, and let $f$ be a $4 n$-variable polynomial with real coefficients. Assume that, for any $2 n$ points $\left(x_{1}, y_{1}\right), \ldots,\left(x_{2 n}, y_{2 n}\right)$ in the plane, $f\left(x_{1}, y_{1}, \ldots, x_{2 n}, y_{2 n}\right)=0$ if and only if the points form the vertices of ... | [
"The smallest possible degree is $2 n$. In what follows, we will frequently write $A_{i}=$ $\\left(x_{i}, y_{i}\\right)$, and abbreviate $P\\left(x_{1}, y_{1}, \\ldots, x_{2 n}, y_{2 n}\\right)$ to $P\\left(A_{1}, \\ldots, A_{2 n}\\right)$ or as a function of any $2 n$ points.\n\n\n\nSuppose that $f$ is valid. Firs... | [
"$2n$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | English | .'</im_start>><|im_end|>]\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\... |
1,631 | For a positive integer $a$, define a sequence of integers $x_{1}, x_{2}, \ldots$ by letting $x_{1}=a$ and $x_{n+1}=2 x_{n}+1$ for $n \geq 1$. Let $y_{n}=2^{x_{n}}-1$. Determine the largest possible $k$ such that, for some positive integer $a$, the numbers $y_{1}, \ldots, y_{k}$ are all prime. | [
"The largest such is $k=2$. Notice first that if $y_{i}$ is prime, then $x_{i}$ is prime as well. Actually, if $x_{i}=1$ then $y_{i}=1$ which is not prime, and if $x_{i}=m n$ for integer $m, n>1$ then $2^{m}-1 \\mid 2^{x_{i}}-1=y_{i}$, so $y_{i}$ is composite. In particular, if $y_{1}, y_{2}, \\ldots, y_{k}$ are pr... | [
"2"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English | You are a helpful math assistant.<|im_end|>
']
**Note:** You can assume that for a number $n$, the number of
prime factors of $n$ is the total count of prime factors with multiplicity. So, for example, 12 has 2 prime factors, 2 and 3.
Wait, but in the original problem, it's stated "y_n is a prime if and only if x_n i... |
1,641 | Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow... | [
"The required number is $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$. To prove this, trace the circumference counterclockwise to label the points $a_{1}, a_{2}, \\ldots, a_{2 n}$.\n\nLet $\\mathcal{C}$ be any good configuration and let $O(\\mathcal{C})$ be the set of all points from which arrows emerge.... | [
"$\\binom{2n}{n}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | English | .</im_start|>.
</im_end|>]
</think>
To determine the number of good configurations, we can use a recursive approach with the Catalan numbers.
**Solution:**
Let $D_n$ represent the number of good configurations for $2n$ points. The generating function for Catalan numbers is well-known:
$$
C(x) = \sum_{n=0}^{\infty}... |
1,644 | Given positive integers $m$ and $n \geq m$, determine the largest number of dominoes $(1 \times 2$ or $2 \times 1$ rectangles) that can be placed on a rectangular board with $m$ rows and $2 n$ columns consisting of cells $(1 \times 1$ squares $)$ so that:
(i) each domino covers exactly two adjacent cells of the boar... | [
"The required maximum is $m n-\\lfloor m / 2\\rfloor$ and is achieved by the brick-like vertically symmetric arrangement of blocks of $n$ and $n-1$ horizontal dominoes placed on alternate rows, so that the bottom row of the board is completely covered by $n$ dominoes.\n\n\n\nTo show that the number of dominoes in a... | [
"$m n-\\lfloor m / 2\\rfloor$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | English | \nThe required maximum is $mn - \lfloor m/2 \rfloor$ which is achieved by the brick-like vertically symmetric arrangement of blocks of $n$ and $n-1$ horizontal dominoes placed alternate rows so that the bottom row of the board is completely covered by $n$ dominoes. The number of dominoes in an arrangement satisfying th... |
1,645 | A cubic sequence is a sequence of integers given by $a_{n}=n^{3}+b n^{2}+c n+d$, where $b, c$ and $d$ are integer constants and $n$ ranges over all integers, including negative integers.
Determine the possible values of $a_{2015} \cdot a_{2016}$ for a cubic sequence satisfying the condition in part (a). | [
"The only possible value of $a_{2015} \\cdot a_{2016}$ is 0 . For simplicity, by performing a translation of the sequence (which may change the defining constants $b, c$ and $d$ ), we may instead concern ourselves with the values $a_{0}$ and $a_{1}$, rather than $a_{2015}$ and $a_{2016}$.\n\n\n\nSuppose now that we... | [
"0"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English | You think I have this right, but you tried to walk through the problem again, making sure that it's completely clear. So, in particular, you can simplify, expand, or reorder terms.</|im_start|>user replied>\n\n\n\nI think the solution is correct. I also noticed that in the original problem statement, for part (a), n ra... |
1,659 | Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$$
f(x+f(y))=f(x+y)+f(y)\tag{1}
$$
for all $x, y \in \mathbb{R}^{+}$. (Symbol $\mathbb{R}^{+}$denotes the set of all positive real numbers.) | [
"First we show that $f(y)>y$ for all $y \\in \\mathbb{R}^{+}$. Functional equation (1) yields $f(x+f(y))>f(x+y)$ and hence $f(y) \\neq y$ immediately. If $f(y)<y$ for some $y$, then setting $x=y-f(y)$ we get\n\n$$\nf(y)=f((y-f(y))+f(y))=f((y-f(y))+y)+f(y)>f(y),\n$$\n\ncontradiction. Therefore $f(y)>y$ for all $y \\... | [
"$f(x)=2 x$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | English | You're right, and $f(x)=2x$ is the only solution. To find this, I observe that $f(y)>y$ for all $y>0$. This allows defining $g(x)=f(x)-x$, which leads to $g(t+g(y))=g(t)+y$ for all $t>y>0$. By induction, I can show that $g(t + n g(y))=g(t) + n y$ for all $t > n y$ and positive integer $n$. Taking two variables, $y$ and... |
1,662 | Let $n>1$ be an integer. In the space, consider the set
$$
S=\{(x, y, z) \mid x, y, z \in\{0,1, \ldots, n\}, x+y+z>0\}
$$
Find the smallest number of planes that jointly contain all $(n+1)^{3}-1$ points of $S$ but none of them passes through the origin. | [
"It is easy to find $3 n$ such planes. For example, planes $x=i, y=i$ or $z=i$ $(i=1,2, \\ldots, n)$ cover the set $S$ but none of them contains the origin. Another such collection consists of all planes $x+y+z=k$ for $k=1,2, \\ldots, 3 n$.\n\nWe show that $3 n$ is the smallest possible number.\n\nLemma 1. Consider... | [
"$3 n$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | English | 's solution['It is easy to find $3 n$ such planes. For example, planes $x=i$, $y=i$, or $z=i$ $(i=1,2, \ldots, n)$ cover the set $S$ but none of them contains the origin. Another such collection consists of all planes $x+y+z=k$ for $k=1,2, \ldots, 3 n$. It is shown in Lemma 2 that a polynomial $P(x, y, z)$ of degree at... |
1,664 | Find all positive integers $n$, for which the numbers in the set $S=\{1,2, \ldots, n\}$ can be colored red and blue, with the following condition being satisfied: the set $S \times S \times S$ contains exactly 2007 ordered triples $(x, y, z)$ such that (i) $x, y, z$ are of the same color and (ii) $x+y+z$ is divisible b... | [
"Suppose that the numbers $1,2, \\ldots, n$ are colored red and blue. Denote by $R$ and $B$ the sets of red and blue numbers, respectively; let $|R|=r$ and $|B|=b=n-r$. Call a triple $(x, y, z) \\in S \\times S \\times S$ monochromatic if $x, y, z$ have the same color, and bichromatic otherwise. Call a triple $(x, ... | [
"$69$,$84$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Combinatorics | Math | English | You are a helpful math assistant.<|im_end|>
Answer
</think>
To solve the problem, we need to find all positive integers \( n \) such that the set \( S = \{1, 2, \ldots, n\} \) can be colored red and blue in a way that satisfies the given condition regarding the number of monochromatic ordered triples \( (x, y, z) \) w... |
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