An $n \times m$ table ( $n \le m$ ) is filled in accordance with the rules of the game "Minesweeper": mines are placed at some cells (not more than one mine at the cell) and in the remaining cells one writes the number of the mines in the neighboring (by side or by vertex) cells. Then the sum of allnumbers in the table is computed (this sum is equal to $9$ for the picture). What is the largest possible value of this sum? (V. Lebed) [img]https://cdn.artofproblemsolving.com/attachments/2/9/726ccdbc57807788a5f6e88a5acb42b10a6cc0.png[/img]

Prove that for all positive integers $n$, $$\frac{1}{1 \cdot 2}+\frac{1}{3 \cdot 4}+ ...+ \frac{1}{(2n - 1)2n}=\frac{1}{n + 1}\frac{1}{n + 2}+ ... +\frac{1}{2n}$$

Show that $$5^{2019} \mid \Sigma^{5^{2019}}_{k=1}3^{gcd (5^{2019},k)}$$

Prove the following equality for all positive integers $m,n$: $$\sum_{k=0}^{n} {m+k \choose k} 2^{n-k} +\sum_{k=0}^m {n+k \choose k}2^{m-k}= 2^{m+n+1}$$

Find positive reals $a, b, c$ which maximizes the value of $a+ 2b+ 3c$ subject to the constraint that $9a^2 + 4b^2 + c^2 = 91$

Calculate the sum $\sum_{n=1}^{1.000.000}[ \sqrt{n} ]$ . You may use the formula $\sum_{i=1}^{k} i^2=\frac{k(k +1)(2k +1)}{6}$ without a proof.

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

a) The product of $n$ integers equals $n$, and their sum is zero. Prove that $n$ is divisible by $4$. b) Let $n$ is divisible by $4$. Prove that there exist $n$ integers such, that their product equals $n$, and their sum is zero.

Determine the greatest natural number $n$, such that for each set $S$ of 2015 different integers there exist 2 subsets of $S$ (possible to be with 1 element and not necessarily non-intersecting) each of which has a sum of its elements divisible by $n$.

One pupil has $7$ cards of paper. He takes a few of them and tears each in $7$ pieces. Then, he choses a few of the pieces of paper that he has and tears it again in $7$ pieces. He continues the same procedure many times with the pieces he has every time. Will he be able to have sometime $2009$ pieces of paper?

The vertical side of a square is divided onto $n$ segments. The sum of the segments with even numbers lengths equals to the sum of the segments with odd numbers lengths. $n-1$ lines parallel to the horizontal sides are drawn from the segments ends, and, thus, $n$ strips are obtained. The diagonal is drawn from the lower left corner to the upper right one. This diagonal divides every strip onto left and right parts. Prove that the sum of the left parts of odd strips areas equals to the sum of the right parts of even strips areas.

For every four-digit number, we take the product of its four digits. Then we add all of these products together . What is the result? ( G Galperin)

Let $a_1, a_2, ..., a_n$ real numbers such that $a_1 + a_2 + ... + a_n = 0$ and $| a_1 | + | a_2 | + ... + | a_n | = 1$. Show that: $| a _ 1 + 2 a _ 2 + ... + n a _ n | \le \frac {n-1} {2}$.

Show the the sum of any three consecutive positive integers is a divisor of the sum of their cubes.

Let $n$ be any integer $\ge 2$. Prove that $\sum 1/pq = 1/2$, where the summation is over all integers$ p, q$ which satisfy $0 < p < q \le n$,$ p + q > n$, $(p, q) = 1$.

Determine a natural $n$ such that $$996 \le \sum_{k = 1}^{n}\frac{1}{k}$$

[i]Palindromic partitioning [/i] of the natural number $ A $ is called, when $ A $ is written as the sum of natural the terms $ A = a_1 + a_2 + \ ldots + a_ {n-1} + a_n $ ($ n \geq 1 $), in which $ a_1 = a_n , a_2 = a_ {n-1} $ and in general, $ a_i = a_ {n + 1 - i} $ with $ 1 \leq i \leq n $. For example, $ 16 = 16 $, $ 16 = 2 + 12 + 2 $ and $ 16 = 7 + 1 + 1 + 7 $ are [i]palindromic partitions[/i] of the number $16$. Find the number of all [i]palindromic partitions[/i] of the number $2006$.

Let $a_1,a_2,...,a_n$ be $n$ different positive integers where $n\ge 1$. Show that $$\sum_{i=1}^n a_i^3 \ge \left(\sum_{i=1}^n a_i\right)^2$$

Given an integer $N \geq 4$, determine the largest value the sum $$\sum_{i=1}^{\left \lfloor{\frac{k}{2}}\right \rfloor+1}\left( \left \lfloor{\frac{n_i}{2}}\right \rfloor+1\right)$$ may achieve, where $k, n_1, \ldots, n_k$ run through the integers subject to $k \geq 3$, $n_1 \geq \ldots\geq n_k\geq 1$ and $n_1 + \ldots + n_k = N$.

The sum of the reciprocals of three positive integers is equal to $1$. What are all the possible such triples?

The sequence of Fibonacci numbers $F_0, F_1, F_2, . . .$ is defined by $F_0 = F_1 = 1 $ and $F_{n+2} = F_n+F_{n+1}$ for all $n > 0$. For example, we have $F_2 = F_0 + F_1 = 2, F_3 = F_1 + F_2 = 3, F_4 = F_2 + F_3 = 5$, and $F_5 = F_3 + F_4 = 8$. The sequence $a_0, a_1, a_2, ...$ is defined by $a_n =\frac{1}{F_nF_{n+2}}$ for all $n \ge 0$. Prove that for all $m \ge 0$ we have: $a_0 + a_1 + a_2 + ... + a_m < 1$.

Suppose that $p> 2$ is a prime number. For each integer $k = 1, 2,..., p-1$, denote $r_k$ as the remainder of the division $k^p$ by $p^2$. Prove that $r_1+r_2+r_3+...+r_{p-1}=\frac{p^2(p-1)}{2}$

Five circles in a row are each labeled with a positive integer. As shown in the diagram, each circle is connected to its adjacent neighbor(s). The integers must be chosen such that the sum of the digits of the neighbor(s) of a given circle is equal to the number labeling that point. In the example, the second number $23 = (1+8)+(5+9)$, but the other four numbers do not have the needed value. [img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMi9lL2M2MzVkMmMyYTRlZjliNWEzYWNkOTM2OGVmY2NkOGZmOWVkN2VmLnBuZw==&rn=MjAxMSBCQU1PIDIucG5n[/img] What is the smallest possible sum of the five numbers? How many possible arrangements of the five numbers have this sum? Justify your answers.

Prove that for arbitrary fixed $a_1, a_2,.. , a_{31}$ the sum $\cos 32x + a_{31} \cos 31x +... + a_2 cos 2x + a_1 \cos x$ can take both positive and negative values as $x$ varies.

Since $24 = 3+5+7+9$, the number $24$ can be written as the sum of at least two consecutive odd positive integers. (a) Can $2005$ be written as the sum of at least two consecutive odd positive integers? If yes, give an example of how it can be done. If no, provide a proof why not. (b) Can $2006$ be written as the sum of at least two consecutive odd positive integers? If yes, give an example of how it can be done. If no, provide a proof why not.