21st APMC 1998 Problem 7 Find positive integers m, n with the smallest possible product mn such that the number m m n n ends in exactly 98 zeros.
{m,n} = {75,98} Solution Let 2 a be the highest power of 2 dividing m and 2 b be the highest power of 2 dividing n. Let 5 A be the highest power of 5 dividing m, and 5 B be the highest power of 5 dividing n. We need min(ma + nb, mA + nB) = 98 and mn as small as possible. If A = 0, then mA is a multiple of 5. If A > 0, then m is a multiple of 5, so mA is a multiple of 5. Similarly nB is always a multiple of 5, so mA + nB is a multiple of 5. Hence ma + nb = 98 and mA + nB > 98. Now 98 is not a multiple of 5, so m and n cannot both be divisible by 5. Assume m is a multiple of 5 and n is not. If m is not a multiple of 2, then n has to supply all the power of 2. Since nb = 98, and 98 is not a multiple of 4, n is not a multiple of 4, so b = 1 and n = 98. Now mA > 98. If A = 1, then the smallest possible value of m is 105. If A = 2, then m > 49, but m is an odd multiple of 25, so the smallest possible value is 75. If A > 2, then 125 divides m, so m > 75. Thus the smallest solution is m = 75, n = 98. If m is a multiple of 2, then we must have mA > 98 ≥ ma, so A > a. So A ≥ 2. So m must be a multiple of 50. But we cannot have m ≥ 100, for then ma > 98, so m = 50. Hence nb = 48. If b = 1, then n = 48 and so b ≠ 1. Contradiction. If b = 2, then n = 24, so b = 3. Contradiction. If b = 3, then n = 16, so b = 4. Contradiction. If b ≥ 4, then 16 ≤ 2b ≤ n = 48/b ≤ 12. Contradiction. 21st APMC 1998 © John Scholes jscholes@kalva.demon.co.uk 28 Nov 2003 Last corrected/updated 28 Nov 03