IMO 1978 SL 3

Origin: CUB

Problem

Let n > m \geq1 be natural numbers such that the groups of the last three digits in the decimal representation of 1978m, 1978n coincide. Find the ordered pair (m, n) of such m, n for which m + n is minimal.

Solution

What we need are m, n for which 1978m(1978n−m −1) is divisible by 1000 = 8 \cdot125. Since 1978n−m −1 is odd, it follows that 1978m is divisible by 8, so m \geq3. Also, 1978n−m −1 is divisible by 125, i.e., 1978n−m \equiv1 (mod 125). Note that 1978 \equiv−2 (mod 5), and consequently also −2n−m \equiv1. Hence 4 | n −m = 4k, k \geq1. It remains to find the least k such that 19784k \equiv1 (mod 125). Since 19784 \equiv(−22)4 = 4842 \equiv(−16)2 = 256 \equiv6, we reduce it to 6k \equiv1. Now 6k = (1 + 5)k \equiv1 + 5k + 25
k  (mod 125), which

reduces to 125 | 5k(5k −3). But 5k −3 is not divisible by 5, and so 25 | k. Therefore 100 | n −m, and the desired values are m = 3, n = 103.