Found problems: 700
Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed:
$$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$.
The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this.
Prove that Eduardo has a winning strategy.
[i]Proposed by Amine Natik, Morocco[/i]
$ a \minus{} )$ Find all prime $ p$ such that $ \dfrac{7^{p \minus{} 1} \minus{} 1}{p}$ is a perfect square
$ b \minus{} )$ Find all prime $ p$ such that $ \dfrac{11^{p \minus{} 1} \minus{} 1}{p}$ is a perfect square
Determine, with proof, if there exists an odd prime $p$ such that the following equation holds:
$$\sum_{n = 1}^{\frac{p-1}{2}} \cot\left(\frac{\pi n^2}{p}\right) = 69\sqrt{p}$$
[i]Proposed by Chris Bao[/i]
Let $m, n, a, k$ be positive integers and $k>1$ such that the equality $$5^m+63n+49=a^k$$
holds. Find the minimum value of $k$.
Find all triplets of positive integers $ (a,m,n)$ such that $ a^m \plus{} 1 \mid (a \plus{} 1)^n$.
We call a positive integer $n$ is $good$ , if there exist integers $a,b,c,x,y$ such that $n=ax^2+bxy+cy^2$ and $b^2-4ac=-20$. Prove that the product of any two good numbers is also a good number.
Find the smallest $x \in\mathbb{N}$ for which $\frac{7x^{25}-10}{83}$ is an integer.
S is the set of all ($a$, $b$, $c$, $d$, $e$, $f$) where $a$, $b$, $c$, $d$, $e$, $f$ are integers such that $a^2 + b^2 + c^2 + d^2 + e^2 = f^2$. Find the largest $k$ which divides abcdef for all members of $S$.
Find the total number of primes $p<100$ such that $\lfloor (2+\sqrt{5})^p \rfloor-2^{p+1}$ is divisible by $p$. Here $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.
Let $p$ and $q$ be prime numbers of the form $4k+3$. Suppose that there exist integers $x$ and $y$ such that $x^2-pqy^2=1$. Prove that there exist positive integers $a$ and $b$ such that $|pa^2-qb^2|=1$.
Determine all pairs $(x, y)$ of integers such that \[1+2^{x}+2^{2x+1}= y^{2}.\]
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]
Find all natural integers $n$ such that $(n^3 + 39n - 2)n! + 17\cdot 21^n + 5$ is a square.
Let $b \geqslant 2$ be a positive integer. Anu has an infinite collection of notes with exactly $b-1$ copies of a note worth $b^k-1$ rupees, for every integer $k\geqslant 1$. A positive integer $n$ is called payable if Anu can pay exactly $n^2+1$ rupees by using some collection of her notes. Prove that if there is a payable number, there are infinitely many payable numbers.
[i]Proposed by Shantanu Nene[/i]
We say that an integer $a$ is a quadratic, cubic, or quintic residue modulo $n$ if there exists an integer $x$ such that $x^2\equiv a \pmod n$, $x^3 \equiv a \pmod n$, or $x^5 \equiv a \pmod n$, respectively. Further, an integer $a$ is a primitive residue modulo $n$ if it is exactly one of these three types of residues modulo $n$.
How many integers $1 \le a \le 2015$ are primitive residues modulo $2015$?
[i] Proposed by Michael Ren [/i]
We call a positive integer $n\ge 4$[i] beautiful[/i] if there exists some permutation $$\{x_1,x_2,\dots ,x_{n-1}\}$$ of $\{1,2,\dots ,n-1\}$ such that $\{x^1_1,\ x^2_2,\ \dots,x^{n-1}_{n-1}\}$ gives all the residues $\{1,2,\dots, n-1\}$ modulo $n$. Prove that if $n$ is beautiful then $n=2p,$ for some prime number $p.$
The numbers from 1 to $ 2013^2 $ are written row by row into a table consisting of $ 2013 \times 2013 $ cells. Afterwards, all columns and all rows containing at least one of the perfect squares $ 1, 4, 9, \cdots, 2013^2 $ are simultaneously deleted.
How many cells remain?
Determine the number of all numbers which are represented as $x^2+y^2$ with $x, y \in \{1, 2, 3, \ldots, 1000\}$ and which are divisible by 121.
For any prime $p$, prove that there exists integer $x_0$ such that $p | (x^2_0 - x_0 + 3)$ $\Leftrightarrow$ there exists integer $y_0$ such that $p | (y^2_0 - y_0 + 25).$
Prove that there are infinite many positive integers $ n$ such that
$ n^2\plus{}1\mid n!$, and infinite many of those for which $ n^2\plus{}1 \nmid n!$.
Given odd integers $a,b,c$ prove that the equation $ax^2+bx+c=0$ cannot have a solution $x$ which is a rational number.
A subset $S$ of the set $M=\{1,2,.....,p-1\}$,where $p$ is a prime number of the kind
$12n+11$,is [i]essential[/i],if the product ${\Pi}_s$ of all elements of the subset
is not less than the product $\bar{{\Pi}_s}$ of all other elements of the set.The
[b]difference[/b] $\bigtriangleup_s=\Pi_s-\bar{{\Pi}_s}$ is called [i]the deviation[/i]
of the subset $S$.Define the least possible remainder of division by $p$ of the deviation of an essential subset,containing $\frac{p-1}{2}$ elements.
Find all positive integers $n$ such that the set $\{n,n+1,n+2,n+3,n+4,n+5\}$ can be partitioned into two subsets so that the product of the numbers in each subset is equal.
Let $p$ be an odd prime number. Show that the smallest positive quadratic nonresidue of $p$ is smaller than $\sqrt{p}+1$.
Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed:
$$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$.
The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this.
Prove that Eduardo has a winning strategy.
[i]Proposed by Amine Natik, Morocco[/i]