IMO 2001 SL N1

Origin: AUS | Category: Number Theory

Problem

Prove that there is no positive integer n such that for k = 1, 2, . . . , 9, the leftmost digit (in decimal notation) of (n + k)! equals k.

Solution

For each positive integer x, define \alpha(x) = x/10r if r is the positive integer satisfying 10r \leqx < 10r+1. Observe that if \alpha(x)\alpha(y) < 10 for some x, y \inN, then \alpha(xy) = \alpha(x)\alpha(y). If, as usual, [t] means the integer part of t, then [\alpha(x)] is actually the leftmost digit of x. Now suppose that n is a positive integer such that k \leq\alpha((n+k)!) < k +1 for k = 1, 2, . . ., 9. We have 1 < \alpha(n + k) = \alpha((n + k)!) \alpha((n + k −1)!) < k + 1 k −1 \leq3 for 2 \leqk \leq9, from which we obtain \alpha(n+k +1) > \alpha(n+k) (the opposite can hold only if \alpha(n + k) \geq9). Therefore 1 < \alpha(n + 2) < \cdot \cdot \cdot < \alpha(n + 9) \leq5 4. On the other hand, this implies that \alpha((n+4)!) = \alpha((n+1)!)\alpha(n+2)\alpha(n+ 3)\alpha(n + 4) < (5/4)3\alpha((n + 1)!) < 4, contradicting the assumption that the leftmost digit of (n + 4)! is 4.