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Problem 1333

AIME late

Algebra Difficulty 5.5 Prove it Singapore Mathematical Olympiad (SMO) · Singapore

Let $n$ be a positive integer and $a_1, a_2, \dots, a_{2n}$ be $2n$ distinct integers. Given that the equation
$|x - a_1| |x - a_2| \dots |x - a_{2n}| = (n!)^2$
has an integer solution $x = m$ , find $m$ in terms of $a_1, \dots, a_{2n}$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We have
$|m - a_1| |m - a_2| \cdots |m - a_{2n}| = (n!)^2.$
First we show that we cannot have distinct $i$ , $j$ , $k$ so that $|m - a_i| = |m - a_j| = |m - a_k|$ . If so, then two of $(m - a_i)$ , $(m - a_j)$ , $(m - a_k)$ must be of the same sign, say $(m - a_i)$ , $(m - a_j)$ . Then $a_i = a_j$ , a contradiction. Thus the values of $|m - a_1|$ , $|m - a_2|$ , ..., $|m - a_{2n}|$ are $1$ , $1$ , $2$ , $2$ , ..., $n$ , $n$ . Also if $|m - a_i| = |m - a_j|$ , then we must have $m - a_i = -(m - a_j)$ otherwise $a_i = a_j$ . Therefore $\sum (m - a_i) = 0$ and $m = \frac{a_1 + \cdots + a_{2n}}{2n}$ .

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