Olympiad problems

2007 Thailand Mathematical Olympiad, 18

Let $p_k$ be the $k$-th prime number. Find the remainder when $\sum_{k=2}^{2550}p_k^{p_k^4-1}$ is divided by $2550$.

2007 Thailand Mathematical Olympiad, 15

Tags: number theory , remainder

Compute the remainder when $222!^{111} + 111^{222!} + 111!^{222} + 222^{111!}$ is divided by $2007$.

2017 Finnish National High School Mathematics Comp, 1

Tags: number theory , remainder , euclidean algorithm , division

By dividing the integer $m$ by the integer $n, 22$ is the quotient and $5$ the remainder. As the division of the remainder with $n$ continues, the new quotient is $0.4$ and the new remainder is $0.2$. Find $m$ and $n$.

2019 Saudi Arabia Pre-TST + Training Tests, 2.1

Tags: number theory , remainder

Let be given a positive integer $n \ge 3$. Consider integers $a_1,a_2,...,a_n >1$ with the product equals to $A$ such that: for each $k \in \{1, 2,..., n\}$ then the remainder when $\frac{A}{a_k}$ divided by $a_k$ are all equal to $r$. Prove that $r \le n- 2$

2016 Indonesia MO, 7

Tags: number theory , remainder , sum

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 $r_1+r_2+r_3+...+r_{p-1}=\frac{p^2(p-1)}{2}$

2010 Junior Balkan Team Selection Tests - Romania, 3

Tags: factorial , remainder , number theory

Determine the integers $n, n \ge 2$, with the property that the numbers $1! , 2 ! , 3 ! , ..., (n- 1)!$ give different remainders when dividing by $n $.

2024 AMC 12/AHSME, 14

Tags: remainder

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 $

2020 Estonia Team Selection Test, 3

Tags: number theory , remainder

The prime numbers $p$ and $q$ and the integer $a$ are chosen such that $p> 2$ and $a \not\equiv 1$ (mod $q$), but $a^p \equiv 1$ (mod $q$). Prove that $(1 + a^1)(1 + a^2)...(1 + a^{p - 1})\equiv 1$ (mod $q$) .

2020 Tournament Of Towns, 7

Tags: combinatorics , game , remainder

Gleb picked positive integers $N$ and $a$ ($a < N$). He wrote the number $a$ on a blackboard. Then each turn he did the following: he took the last number on the blackboard, divided the number $N$ by this last number with remainder and wrote the remainder onto the board. When he wrote the number $0$ onto the board, he stopped. Could he pick $N$ and $a$ such that the sum of the numbers on the blackboard would become greater than $100N$ ? Ivan Mitrofanov

2013 Saudi Arabia IMO TST, 4

Tags: number theory , combinatorics , remainder

Determine whether it is possible to place the integers $1, 2,...,2012$ in a circle in such a way that the $2012$ products of adjacent pairs of numbers leave pairwise distinct remainders when divided by $2013$.

2018 Mathematical Talent Reward Programme, SAQ: P 5

Tags: fibonacci sequence , remainder , periodic sequence , polynomial , algebra

[list=1] [*] Prove that, the sequence of remainders obtained when the Fibonacci numbers are divided by $n$ is periodic, where $n$ is a natural number. [*] There exists no such non-constant polynomial with integer coefficients such that for every Fibonacci number $n,$ $ P(n)$ is a prime. [/list]

2022 JBMO Shortlist, N5

Tags: number theory , remainder , prime , shortlist

Find all pairs $(a, p)$ of positive integers, where $p$ is a prime, such that for any pair of positive integers $m$ and $n$ the remainder obtained when $a^{2^n}$ is divided by $p^n$ is non-zero and equals the remainder obtained when $a^{2^m}$ is divided by $p^m$.

2011 Saudi Arabia BMO TST, 4

Tags: number theory , remainder

Let $p \ge 3$ be a prime. For $j = 1,2 ,... ,p - 1$, let $r_j$ be the remainder when the integer $\frac{j^{p-1}-1}{p}$ is divided by $p$. Prove that $$r_1 + 2r_2 + ... + (p - 1)r_{p-1} \equiv \frac{p+1}{2} (\mod p)$$

1939 Moscow Mathematical Olympiad, 051

Tags: number theory , remainder

Find the remainder after division of $10^{10} + 10^{10^2} + 10^{10^3} + ... + 10^{10^{10}}$ by $7$.