Found problems: 5802
We consider graphs with vertices colored black or white. "Switching" a vertex means: coloring it black if it was formerly white, and coloring it white if it was formerly black.
Consider a finite graph with all vertices colored white. Now, we can do the following operation: Switch a vertex and simultaneously switch all of its neighbours (i. e. all vertices connected to this vertex by an edge). Can we, just by performing this operation several times, obtain a graph with all vertices colored black?
[It is assumed that our graph has no loops (a [i]loop[/i] means an edge connecting one vertex with itself) and no multiple edges (a [i]multiple edge[/i] means a pair of vertices connected by more than one edge).]
Find all function $ f: R^\plus{} \rightarrow R^\plus{}$ such that for any $ x,y,z \in R^\plus{}$ such that $ x\plus{}y \ge z$ , $ f(x\plus{}y\minus{}z) \plus{}f(2\sqrt{xz})\plus{}f(2\sqrt{yz}) \equal{} f(x\plus{}y\plus{}z)$
Let $n$ be a positive integer. For $i$ and $j$ in $\{1,2, \ldots, n\}$, let $s(i, j)$ be the number of pairs $(a, b)$ of nonnegative integers satisfying $a i+b j=n$. Let $S$ be the $n$-by-n matrix whose $(i, j)$-entry is $s(i, j)$.
For example, when $n=5$, we have $S=\left[\begin{array}{lllll}6 & 3 & 2 & 2 & 2 \\ 3 & 0 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 & 1 \\ 2 & 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 1 & 2\end{array}\right]$.
Compute the determinant of $S$.
A $k \times k$ array contains each of the numbers $1, 2, \dots, m$ exactly once, with the remaining entries all zero. Suppose that all the row sums and column sums are equal. What is the smallest possible value of $m$ if $k = 3^n$ ($n \in \mathbb{N}^+$)?
[i]Attila Sztranyák and Péter Erben[/i]
Let $\mathcal{S}$ be a set consisting of $n \ge 3$ positive integers, none of which is a sum of two other distinct members of $\mathcal{S}$. Prove that the elements of $\mathcal{S}$ may be ordered as $a_1, a_2, \dots, a_n$ so that $a_i$ does not divide $a_{i - 1} + a_{i + 1}$ for all $i = 2, 3, \dots, n - 1$.
In each square of a garden shaped like a $2022 \times 2022$ board, there is initially a tree of height $0$. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:
[list]
[*] The gardener chooses a square in the garden. Each tree on that square and all the surrounding squares (of which there are at most eight) then becomes one unit taller.
[*] The lumberjack then chooses four different squares on the board. Each tree of positive height on those squares then becomes one unit shorter.
[/list]
We say that a tree is [i]majestic[/i] if its height is at least $10^6$. Determine the largest $K$ such that the gardener can ensure there are eventually $K$ majestic trees on the board, no matter how the lumberjack plays.
Let $t(A)$ denote the sum of elements of a nonempty set $A$ of integers, and define $t(\emptyset)=0$. Find a set $X$ of positive integers such that for every integers $k$ there is a unique ordered pair of disjoint subsets $(A_{k},B_{k})$ of $X$ such that $t(A_{k})-t(B_{k}) = k$.
Determine all functions $ f$ from the set of positive integers to the set of positive integers such that, for all positive integers $ a$ and $ b$, there exists a non-degenerate triangle with sides of lengths
\[ a, f(b) \text{ and } f(b \plus{} f(a) \minus{} 1).\]
(A triangle is non-degenerate if its vertices are not collinear.)
[i]Proposed by Bruno Le Floch, France[/i]
Determine all integers $n\geqslant 2$ with the following property: every $n$ pairwise distinct integers whose sum is not divisible by $n$ can be arranged in some order $a_1,a_2,\ldots, a_n$ so that $n$ divides $1\cdot a_1+2\cdot a_2+\cdots+n\cdot a_n.$
[i]Arsenii Nikolaiev, Anton Trygub, Oleksii Masalitin, and Fedir Yudin[/i]
Denote the sequence $a_{i,j}$ in positive reals such that $a_{i,j}$.$a_{j,i}=1$. Let $c_i=\sum_{k=1}^{n}a_{k,i}$. Prove that $1\ge$$\sum_{i=1}^{n}\dfrac {1}{c_i}$
Let $n\ge 2$ be a positive integer. Find the positive integers $x$
\[\sqrt{x+\sqrt{x+\ldots +\sqrt{x}}}<n \]
for any number of radicals.
Define the sequences $ a_n, b_n, c_n$ as follows. $ a_0 \equal{} k, b_0 \equal{} 4, c_0 \equal{} 1$.
If $ a_n$ is even then $ a_{n \plus{} 1} \equal{} \frac {a_n}{2}$, $ b_{n \plus{} 1} \equal{} 2b_n$, $ c_{n \plus{} 1} \equal{} c_n$.
If $ a_n$ is odd, then $ a_{n \plus{} 1} \equal{} a_n \minus{} \frac {b_n}{2} \minus{} c_n$, $ b_{n \plus{} 1} \equal{} b_n$, $ c_{n \plus{} 1} \equal{} b_n \plus{} c_n$.
Find the number of positive integers $ k < 1995$ such that some $ a_n \equal{} 0$.
Ann and Bob play a game on the edges of an infinite square grid, playing in turns. Ann plays the first move. A move consists of orienting any edge that has not yet been given an orientation. Bob wins if at any point a cycle has been created. Does Bob have a winning strategy?
Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers $x$ and $y$ such that $x>y$ and $x$ is to the left of $y$, and replaces the pair $(x,y)$ by either $(y+1,x)$ or $(x-1,x)$. Prove that she can perform only finitely many such iterations.
[i]Proposed by Warut Suksompong, Thailand[/i]
Let $n$ be a positive integer. A [i]Japanese triangle[/i] consists of $1 + 2 + \dots + n$ circles arranged in an equilateral triangular shape such that for each $i = 1$, $2$, $\dots$, $n$, the $i^{th}$ row contains exactly $i$ circles, exactly one of which is coloured red. A [i]ninja path[/i] in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with $n = 6$, along with a ninja path in that triangle containing two red circles.
[asy]
// credit to vEnhance for the diagram (which was better than my original asy):
size(4cm);
pair X = dir(240); pair Y = dir(0);
path c = scale(0.5)*unitcircle;
int[] t = {0,0,2,2,3,0};
for (int i=0; i<=5; ++i) {
for (int j=0; j<=i; ++j) {
filldraw(shift(i*X+j*Y)*c, (t[i]==j) ? lightred : white);
draw(shift(i*X+j*Y)*c);
}
}
draw((0,0)--(X+Y)--(2*X+Y)--(3*X+2*Y)--(4*X+2*Y)--(5*X+2*Y),linewidth(1.5));
path q = (3,-3sqrt(3))--(-3,-3sqrt(3));
draw(q,Arrows(TeXHead, 1));
label("$n = 6$", q, S);
label("$n = 6$", q, S);
[/asy]
In terms of $n$, find the greatest $k$ such that in each Japanese triangle there is a ninja path containing at least $k$ red circles.
The sequence $S_0,S_1,S_2,\ldots$ is defined by[list][*]$S_n=1$ for $0\le n\le 2011$, and
[*]$S_{n+2012}=S_{n+2011}+S_n$ for $n\ge 0$.[/list]Prove that $S_{2011a}-S_a$ is a multiple of $2011$ for all nonnegative integers $a$.
There are $n$ balls numbered from $1$ to $n$, and $2n-1$ boxes numbered from $1$ to $2n-1$. For each $i$, ball number $i$ can only be put in the boxes with numbers from $1$ to $2i-1$. Let $k$ be an integer from $1$ to $n$. In how many ways we can choose $k$ balls, $k$ boxes and put these balls in the selected boxes so that each box has exactly one ball?
Choose positive integers $b_1, b_2, \dotsc$ satisfying
\[1=\frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb\]
and let $r$ denote the largest real number satisfying $\tfrac{b_n}{n^2} \geq r$ for all positive integers $n$. What are the possible values of $r$ across all possible choices of the sequence $(b_n)$?
[i]Carl Schildkraut and Milan Haiman[/i]
In each square of a garden shaped like a $2022 \times 2022$ board, there is initially a tree of height $0$. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:
[list]
[*] The gardener chooses a square in the garden. Each tree on that square and all the surrounding squares (of which there are at most eight) then becomes one unit taller.
[*] The lumberjack then chooses four different squares on the board. Each tree of positive height on those squares then becomes one unit shorter.
[/list]
We say that a tree is [i]majestic[/i] if its height is at least $10^6$. Determine the largest $K$ such that the gardener can ensure there are eventually $K$ majestic trees on the board, no matter how the lumberjack plays.
Prove that there are infinitely many triples $(a, b, p)$ of positive integers with $p$ prime, $a < p$, and $b < p$, such that $(a + b)^p - a^p - b^p$ is a multiple of $p^3$.
[i]Noam Elkies[/i]
Let $a, b$ be integers, and let $P(x) = ax^3+bx.$ For any positive integer $n$ we say that the pair $(a,b)$ is $n$-good if $n | P(m)-P(k)$ implies $n | m - k$ for all integers $m, k.$ We say that $(a,b)$ is $very \ good$ if $(a,b)$ is $n$-good for infinitely many positive integers $n.$
[list][*][b](a)[/b] Find a pair $(a,b)$ which is 51-good, but not very good.
[*][b](b)[/b] Show that all 2010-good pairs are very good.[/list]
[i]Proposed by Okan Tekman, Turkey[/i]
For a positive integer $k$, let $s(k)$ denote the number of $1$s in the binary representation of $k$. Prove that for any positive integer $n$,
\[\sum_{i=1}^{n}(-1)^{s(3i)} > 0.\]
[i]Holden Mui[/i]
The set $M$ consists of integers, the smallest of which is $1$ and the greatest $100$. Each member of $M$, except $1$, is the sum of two (possibly identical) numbers in $M$. Of all such sets, find one with the smallest possible number of elements.
Find all functions $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ such that
$$\frac{x+f(y)}{xf(y)}=f(\frac{1}{y}+f(\frac{1}{x}))$$
for all positive real numbers $x$ and $y$.
Find all functions $f : R\to R$ satisfying $xf(x + xy) = xf(x) + f(x^2)f(y)$ for all $x, y \in R$.