This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 110

Give are the integers $a_{1}=11 , a_{2}=1111, a_{3}=111111, ... , a_{n}= 1111...111$( with $2n$ digits) with $n > 8$ . Let $q_{i}= \frac{a_{i}}{11} , i= 1,2,3, ... , n$ the remainder of the division of $a_{i}$ by$ 11$ . Prove that the sum of nine consecutive quotients: $s_{i}=q_{i}+q_{i+1}+q_{i+2}+ ... +q_{i+8}$ is a multiple of $9$ for any $i= 1,2,3, ... , (n-8)$
What is the largest number $N$ for which there exist $N$ consecutive positive integers such that the sum of the digits in the $k$-th integer is divisible by $k$ for $1 \le k \le N$ ? (S Tokarev)
An integer $N$ is the product of two consecutive integers. (a) Prove that we can add two digits to the right of this number and obtain a perfect square. (b) Prove that this can be done in only one way if $N > 12$
If $m$ is a positive integer, prove that $2^m$ cannot be written as a sum of two or more consecutive natural numbers.
Which positive integers cannot be represented as a sum of (two or more) consecutive integers?
Prove that there exists a sequence of $100$ different integers such that the sum of the squares of any two consecutive terms is a perfect square. (S Tokarev)
Sequence $(G_n)$ is defined by $G_0 = 0, G_1 = 1$ and $G_n = G_{n-1} + G_{n-2} + 1$ for every $n \ge2$. Prove that for every positive integer $m$ there exist two consecutive terms in the sequence that are both divisible by $m$.
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.
Prove that the product of five consecutive positive integers cannot be a perfect square.
Show the the sum of any three consecutive positive integers is a divisor of the sum of their cubes.