Let $n$ and $m$ be natural numbers such that $m+ i=a_ib_i^2$ for $i=1,2, \cdots n$ where $a_i$ and $b_i$ are natural numbers and $a_i$ is not divisible by a square of a prime number. Find all $n$ for which there exists an $m$ such that $\sum_{i=1}^{n}a_i=12$

Vanya has chosen two positive numbers, x and y. He wrote the numbers x+y, x-y, x/y, and xy, and has shown these numbers to Petya. However, he didn't say which of the numbers was obtained from which operation. Show that Petya can uniquely recover x and y.

Find all triples of integers ${(x, y, z)}$ satisfying ${x^3 + y^3 + z^3 - 3xyz = 2003}$

Ana chose some unit squares of a $50 \times 50$ board and placed a chip on each of them. Prove that Beto can always choose at most $99$ empty unit squares and place a chip on each so that each row and each column of the board contains an even number of chips.

Let $ABC$ be a triangle with $AC > BC,$ let $\omega$ be the circumcircle of $\triangle ABC,$ and let $r$ be its radius. Point $P$ is chosen on $\overline{AC}$ such taht $BC=CP,$ and point $S$ is the foot of the perpendicular from $P$ to $\overline{AB}$. Ray $BP$ mets $\omega$ again at $D$. Point $Q$ is chosen on line $SP$ such that $PQ = r$ and $S,P,Q$ lie on a line in that order. Finally, let $E$ be a point satisfying $\overline{AE} \perp \overline{CQ}$ and $\overline{BE} \perp \overline{DQ}$. Prove that $E$ lies on $\omega$.

Some cells of a $2n\times2n$ board are marked so that each cell has an even number of neighboring (i.e. sharing a side) marked cells. Find the number of such markings.

Compute the sum $$\sum_{k=0}^{n}\frac{(2n)!}{k!^2(n-k)!^2}.$$

Let $a,b,c,d \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ be real numbers such that $\sin{a}+\sin{b}+\sin{c}+\sin{d}=1$ and $\cos{2a}+\cos{2b}+\cos{2c}+\cos{2d}\geq \frac{10}{3}$. Prove that $a,b,c,d \in [0, \frac{\pi}{6}]$

Fill in each empty cell of the grid with a digit from 1 to 8 so that every row and every column contains each of these digits exactly once. Some diagonally adjacent cells have been joined together. For these pairs of joined cells, the same number must be written in both. [asy] filldraw((0,0)--(0,8)--(8,8)--(8,0)--cycle,white); path removex(pair p) { return ((p.x-0.5, p.y)--(p.x+0.5,p.y)); } path removey(pair p) { return ((p.x, p.y-0.5)--(p.x,p.y+0.5)); } unitsize(1cm); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle, linewidth(2)); for(int i = 0; i < 8; ++i){ draw((0,i)--(8,i)); } for(int j = 0 ; j<8; ++j){ draw((j,0)--(j,8)); } pair [] pointsa = {(1,2),(3,1),(5,7),(7,6)}; pair [] pointsb= {(1,5),(4,4),(2,7),(6,1),(7,3)}; for(int q = 0; q<4; ++q){ draw(removex(pointsa[q]), white+linewidth(2)); draw(removey(pointsa[q]),white+linewidth(2)); draw(arc(pointsa[q]+(0.5,-0.5),0.5,90,180)); draw(arc(pointsa[q]-(0.5,-0.5),0.5,270,0,CCW)); draw(pointsa[q]+(-0.5,0)--pointsa[q]+(-1,0)); draw(pointsa[q]+(0.5,0)--pointsa[q]+(1,0)); draw(pointsa[q]+(0,0.5)--pointsa[q]+(0,1)); draw(pointsa[q]+(0,-0.5)--pointsa[q]+(0,-1)); } for(int q = 0; q<5; ++q){ draw(removex(pointsb[q]), white+linewidth(2)); draw(removey(pointsb[q]),white+linewidth(2)); draw(arc(pointsb[q]+(0.5,0.5),0.5,180,270,CCW)); draw(arc(pointsb[q]-(0.5,0.5),0.5,0,90,CCW)); draw(pointsb[q]+(-0.5,0)--pointsb[q]+(-1,0)); draw(pointsb[q]+(0.5,0)--pointsb[q]+(1,0)); draw(pointsb[q]+(0,0.5)--pointsb[q]+(0,1)); draw(pointsb[q]+(0,-0.5)--pointsb[q]+(0,-1)); } int [][] x = { {1,0,0,0,0,0,0,0}, {2,3,0,0,0,0,0,0}, {0,4,5,0,0,0,0,0}, {0,0,6,0,1,0,0,0}, {0,0,0,7,0,1,0,0}, {0,0,0,0,0,3,4,0}, {0,0,0,0,0,0,2,8}, {0,0,0,0,0,0,0,5} }; for(int k = 0; k<8; ++k){ for(int l = 0; l<8; ++l){ if(x[k][l]!=0){ label(string(x[k][l]), (l+0.5,-k+7.5), fontsize(24pt)); } } } [/asy] There is a unique solution, but you do not need to prove that your answer is the only one possible. You merely need to find an answer that satisfies the constraints above. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justification acceptable.)

There's a large pile of cards. On each card a number from $1,2,\ldots n$ is written. It is known that sum of all numbers on all of the cards is equal to $k\cdot n!$ for some $k$. Prove that it is possible to arrange cards into $k$ stacks so that sum of numbers written on the cards in each stack is equal to $n!$.

For a finite set $A$ of positive integers, a partition of $A$ into two disjoint nonempty subsets $A_1$ and $A_2$ is $\textit{good}$ if the least common multiple of the elements in $A_1$ is equal to the greatest common divisor of the elements in $A_2$. Determine the minimum value of $n$ such that there exists a set of $n$ positive integers with exactly $2015$ good partitions.

Let $n$ be an even positive integer. We say that two different cells of a $n \times n$ board are [b]neighboring[/b] if they have a common side. Find the minimal number of cells on the $n \times n$ board that must be marked so that any cell (marked or not marked) has a marked neighboring cell.

$a_1,a_2,\cdots ,a_n,b_1,b_2,\cdots ,b_n$ are $2n$ positive real numbers such that $a_1,a_2,\cdots ,a_n$ aren't all equal. And assume that we can divide $a_1,a_2,\cdots ,a_n$ into two subsets with equal sums.similarly $b_1,b_2,\cdots ,b_n$ have these two conditions. Prove that there exist a simple $2n$-gon with sides $a_1,a_2,\cdots ,a_n,b_1,b_2,\cdots ,b_n$ and parallel to coordinate axises Such that the lengths of horizontal sides are among $a_1,a_2,\cdots ,a_n$ and the lengths of vertical sides are among $b_1,b_2,\cdots ,b_n$.(simple polygon is a polygon such that it doesn't intersect itself)

The function $f(x)$ is defined on the reals such that $$f\left(\frac{1-4x}{4-x}\right) = 4-xf(x)$$ for all $x \ne 4$. There exists two distinct real numbers $a, b \ne 4$ such that $f(a) = f(b) = \frac{5}{2}$. $a+b$ can be represented as $\frac{p}{q}$ where $p, q$ are relatively prime positive integers. Find $10p + q$.

Lucy starts by writing $s$ integer-valued $2022$-tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\mathbf{v}=(v_1,\ldots,v_{2022})$ and $\mathbf{w}=(w_1,\ldots,w_{2022})$ that she has already written, and apply one of the following operations to obtain a new tuple: \begin{align*} \mathbf{v}+\mathbf{w}&=(v_1+w_1,\ldots,v_{2022}+w_{2022}) \\ \mathbf{v} \lor \mathbf{w}&=(\max(v_1,w_1),\ldots,\max(v_{2022},w_{2022})) \end{align*} and then write this tuple on the blackboard. It turns out that, in this way, Lucy can write any integer-valued $2022$-tuple on the blackboard after finitely many steps. What is the smallest possible number $s$ of tuples that she initially wrote?

Determine all ordered pairs $(a,p)$ of positive integers, with $p$ prime, such that $p^a+a^4$ is a perfect square. [i]Proposed by Tahjib Hossain Khan, Bangladesh[/i]