Found problems: 471
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]
Find all positive integer pairs $(a,n)$ such that $\frac{(a+1)^n-a^n}{n}$ is an integer.
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]
Suppose $a_0,a_1,\ldots, a_{2018}$ are integers such that \[(x^2-3x+1)^{1009} = \sum_{k=0}^{2018}a_kx^k\] for all real numbers $x$. Compute the remainder when $a_0^2 + a_1^2 + \cdots + a_{2018}^2$ is divided by $2017$.
For a nonnegative integer $n$ define $\operatorname{rad}(n)=1$ if $n=0$ or $n=1$, and $\operatorname{rad}(n)=p_1p_2\cdots p_k$ where $p_1<p_2<\cdots <p_k$ are all prime factors of $n$. Find all polynomials $f(x)$ with nonnegative integer coefficients such that $\operatorname{rad}(f(n))$ divides $\operatorname{rad}(f(n^{\operatorname{rad}(n)}))$ for every nonnegative integer $n$.
How many different remainders can result when the $100$th power of an integer is divided by $125$?
$
\textbf{(A) }1 \qquad
\textbf{(B) }2 \qquad
\textbf{(C) }5 \qquad
\textbf{(D) }25 \qquad
\textbf{(E) }125 \qquad
$
Show that there exists a set $ A$ of positive integers with the following property: for any infinite set $ S$ of primes, there exist [i]two[/i] positive integers $ m$ in $ A$ and $ n$ not in $ A$, each of which is a product of $ k$ distinct elements of $ S$ for some $ k \geq 2$.
Let $a, b, c, d$ be a permutation of the numbers $1, 9, 8,4$ and let $n = (10a + b)^{10c+d}$. Find the probability that $1984!$ is divisible by $n.$
Find all ordered triples of nonnegative integers $(a,b,c)$ satisfying $2^a \cdot 5^b - 3^c = 1.$
Consider the assertion that for each positive integer $n\geq2$, the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of $4$. Either prove the assertion or find (with proof) a counterexample.
Let $p$ be a prime number and let $A$ be a set of positive integers that satisfies the following conditions:
(i) the set of prime divisors of the elements in $A$ consists of $p-1$ elements;
(ii) for any nonempty subset of $A$, the product of its elements is not a perfect $p$-th power.
What is the largest possible number of elements in $A$ ?
Find all positive integers $n$ satisfying the following conditions simultaneously:
(a) the number of positive divisors of $n$ is not a multiple of $8$;
(b) for all integers $x$, we have
\[x^n \equiv x \mod n.\]
[i]
Proposed by usjl[/i]
Find the remainder when $2^{2019}$ is divided by $7$.
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 $k\le n$ and $k+1$ distinct integers $x_1,x_2,\cdots ,x_{k+1}$ such that \[P(x_1)+P(x_2)+\cdots +P(x_k)=P(x_{k+1})\].
[i]Note.[/i] A polynomial is [i]monic[/i] if the coefficient of the highest power is one.
Find the smallest positive integer solution to the equation $2^{2^k}\equiv k\pmod{29}$.
Find all natural numbers a, b such that $ a^{n}\plus{} b^{n} \equal{} c^{n\plus{}1}$ where c and n are naturals.
Let $n > 1$ be a given integer. Prove that infinitely many terms of the sequence $(a_k )_{k\ge 1}$, defined by \[a_k=\left\lfloor\frac{n^k}{k}\right\rfloor,\] are odd. (For a real number $x$, $\lfloor x\rfloor$ denotes the largest integer not exceeding $x$.)
[i]Proposed by Hong Kong[/i]
Solve the equation $p^2-pq-q^3=1$ in prime numbers.
[i]A. Golovanov[/i]
Find all triples of primes $(p,q,r)$ satisfying $3p^{4}-5q^{4}-4r^{2}=26$.
Find all positive integers $a$ such that for any positive integer $n\ge 5$ we have $2^n-n^2\mid a^n-n^a$.
Let $p$ be a prime and suppose $2^{2p} \equiv 1 (\text{mod}$ $ 2p+1)$ is prime. Prove that $2p+1$ is prime$^{1}$
[size=75]$^{1}$This is a special case of Pocklington's theorem. A proof of this special case is required.[/size]
Let $n{}$ be a positive integer. What is the smallest sum of digits that $5^n + 6^n + 2022^n$ can take?
Find all triples of primes $(p,q,r)$ satisfying $3p^{4}-5q^{4}-4r^{2}=26$.
Consider the assertion that for each positive integer $n\geq2$, the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of $4$. Either prove the assertion or find (with proof) a counterexample.
Let $m\in N$ and $E(x,y,m)=(\frac{72}x)^m+(\frac{72}y)^m-x^m-y^m$, where $x$ and $y$ are positive divisors of 72.
a) Prove that there exist infinitely many natural numbers $m$ so, that 2005 divides $E(3,12,m)$ and $E(9,6,m)$.
b) Find the smallest positive integer number $m_0$ so, that 2005 divides $E(3,12,m_0)$ and $E(9,6,m_0)$.