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Source: 2002 British Mathematical Olympiad Round 2

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Math-Problem-Solving

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Math-Problem-Solving

#1 h Apr 3, 2025, 10:13 AM • 1 Y

Y by cubres

Solve this.

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60 posts

XAN4

#2 h Apr 4, 2025, 12:31 PM • 1 Y

Y by teomihai

Now we prove that this is only possible for all $n=2^k,k\in\mathbb N$ or $n=2^k\times m, k\in\mathbb N$ and $m=2^k\pm1$ is an odd prime .
The problem is equivalent to asking if for the numbers $1,2,\dots,n$ , there exist $1\leq p\leq q\leq n$ such that $\sum_{i=p}^qi\equiv0(mod$ $n)$ . If there exists $n$ such that $n$ cannot be expressed as $2^k, k\in\mathbb N$ , let’s say that $n=2^k\times m, k\in\mathbb N$ and $m$ is odd. $\sum_{i=p}^qi=\frac{(q+p)(q-p-1)}2$ , and obviously either $q+p$ or $q-p-1$ is even. If the difference between $2^{k+1}$ and $m$ is at least $3$ , then clearly making the larger one $p+q$ and the smaller one $p-q$ solves it. However, if the difference is $1$ , then it’s clear that there are no solutions unless $m$ is not a prime (in which case there’s obviously s solution).

This post has been edited 2 times. Last edited by XAN4, Apr 4, 2025, 12:38 PM

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