IMO 1985 SL 7

Problem

1a.(CZS 3) The positive integers x1, . . . , xn, n \geq3, satisfy x1 < x2 < \cdot \cdot \cdot < xn < 2x1. Set P = x1x2 \cdot \cdot \cdot xn. Prove that if p is a prime number, k a positive integer, and P is divisible by pk, then P pk \geqn!.

Solution

Let ki \geq0 be the largest integer such that pki | xi, i = 1, . . . , n, and yi = xi/pki. We may assume that k = k1 + \cdot \cdot \cdot + kn. All the yi must be

distinct. Indeed, if yi = yj and ki > kj, then xi \geqpxj \geq2xi \geq2x1, which is impossible. Thus y1y2 . . . yn = P/pk \geqn!. If equality holds, we must have yi = 1, yj = 2 and yk = 3 for some i, j, k. Thus p \geq5, which implies that either yi/yj \leq1/2 or yi/yj \geq5/2, which is impossible. Hence the inequality is strict.