Olympiad problems

1990 Romania Team Selection Test, 2

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

2013 QEDMO 13th or 12th, 2

Tags: binomial , number theory , divisible , divide

Let $p$ be a prime number and $n, k$ and $q$ natural numbers, where $q\le \frac{n -1}{p-1}$ should be. Let $M$ be the set of all integers $m$ from $0$ to $n$, for which $m-k$ is divisible by $p$. Show that $$\sum_{m \in M} (-1) ^m {n \choose m}$$ is divisible by $p^q$.

1996 VJIMC, Problem 2

Tags: binomial coefficient , summation , binomial , number theory , algebra

Let $\{x_n\}^\infty_{n=0}$ be the sequence such that $x_0=2$, $x_1=1$ and $x_{n+2}$ is the remainder of the number $x_{n+1}+x_n$ divided by $7$. Prove that $x_n$ is the remainder of the number $$4^n\sum_{k=0}^{\left\lfloor\frac n2\right\rfloor}2\binom n{2k}5^k$$

1996 VJIMC, Problem 2

Tags: recurrence relation , summation , binomial , algebra , sequence

Let $\{a_n\}^\infty_{n=0}$ be the sequence of integers such that $a_0=1$, $a_1=1$, $a_{n+2}=2a_{n+1}-2a_n$. Decide whether $$a_n=\sum_{k=0}^{\left\lfloor\frac n2\right\rfloor}\binom n{2k}(-1)^k.$$

2020 Czech and Slovak Olympiad III A, 6

Tags: inequalities , binomial , number theory , divisible

For each positive integer $k$, denote by $P (k)$ the number of all positive integers $4k$-digit numbers which can be composed of the digits $2, 0$ and which are divisible by the number $2 020$. Prove the inequality $$P (k) \ge \binom{2k - 1}{k}^2$$ and determine all $k$ for which equality occurs. (Note: A positive integer cannot begin with a digit of $0$.) (Jaromir Simsa)