2001 IMO Shortlist Problems/C2

Problem

Let $n$ be an odd integer greater than 1 and let $c_1, c_2, \ldots, c_n$ be integers. For each permutation $a = (a_1, a_2, \ldots, a_n)$ of $\{1,2,\ldots,n\}$ , define $S(a) = \sum_{i = 1}^n c_i a_i$ . Prove that there exist permutations $a \neq b$ of $\{1,2,\ldots,n\}$ such that $n!$ is a divisor of $S(a) - S(b)$ .

Solution

We shall prove this by contradiction. Assume that for some $n$ -tuple of $c_i$ there does not exist two permutations $a$ and $b$ of $s_n=\{ 1, 2, \ldots,n\}$ such that $n!|S(a)-S(b)$ . Note that there are $n!$ distinct permutations of $s_n$ , which means there are $n!$ distinct sums $S(a)$ . Because no two of them are congruent modulo $n!$ , we have that for some $n$ -tuple of $c_i$ there exists a unique permutation $x$ of $s_n$ such that $S(x)\equiv y\bmod{n}$ for all integers $0\leq y\leq n-1$ . This means that the sum of every $S(a)$ is congruent to $\sum_{i=0}^{n!-1}=\frac{(n!-1)n!}{2}$ . However, that same sum is congruent to

$\sum_{i=1}^{n}\left(c_i\sum_{j=1}^{n} (n-1)!*a_j\right)=\sum_{i=1}^{n} c_i\cdot \frac{n(n+1)(n-1)!}{2}=\frac{c_i(n+1)!}{2}\bmod{n!}$

Note that $2|n+1$ , so $n!|\frac{(n+1)!}{2}$ , so $n!$ divides the sum. However, the sum is also congruent to $\frac{(n!-1)n!}{2}$ modulo $n!$ , and $n!-1$ is odd, so $n!$ couldn't possibly divide the sum. This leads to a contradiction, so our previous assumption must be false. This proves the problem statement.

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2001 IMO Shortlist Problems/C2

Problem

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