IMO 2001 SL A5

Origin: BUL | Category: Algebra

Problem

Find all positive integers a1, a2, . . . , an such that 100 = a0 a1

• a1 a2
• \cdot \cdot \cdot + an−1 an , where a0 = 1 and (ak+1 −1)ak−1 \geqa2 k(ak −1) for k = 1, 2, . . ., n −1.

Solution

Let a1, a2, . . . , an satisfy the conditions of the problem. Then ak > ak−1, and hence ak \geq2 for k = 1, . . . , n. The inequality (ak+1 −1)ak−1 \geq a2 k(ak −1) can be rewritten as ak−1 ak + ak ak+1 −1 \leqak−1 ak −1. Summing these inequalities for k = i + 1, . . . , n −1 and using the obvious inequality an−1 an < an−1 an−1, we obtain ai ai+1 + \cdot \cdot \cdot + an−1 an < ai ai+1−1. Therefore ai ai+1 \leq99 100 −a0 a1 −\cdot \cdot \cdot −ai−1 ai < ai ai+1 −1 for i = 1, 2, . . ., n −1. (1) Consequently, given a0, a1, . . . , ai, there is at most one possibility for ai+1. In our case, (1) yields a1 = 2, a2 = 5, a3 = 56, a4 = 2802 = 78400. These values satisfy the conditions of the problem, so that this is a unique solution.