11th Irish 1998 Problem B1 Find all positive integers n with exactly 16 positive divisors 1 = d 1 < d 2 < ... < d 16 = n such that d 6 = 18 and d 9 - d 8 = 17.
n = 2·3 3 37, 2·3 3 71 Solution d 6 = 18, so n is also divisible by 1, 2, 3, 6, 9. Hence d 1 = 1, d 2 = 2, d 3 = 3, d 4 = 6, d 5 = 9. So n is not divisible by 4, nor by any prime p  17. If n = 2 1 3 7 , then d 7 = 27, d 8 = 54, d 9 = 81 and d 9 - d 8 ≠ 17. So we must have n = 2 1 3 3 p for some prime p > 17. If p = 23, then d 7 = 23, d 8 = 27, d 9 = 46, which does not work. If 27  54, then d 7 = 27, d 8 = 54, d 9 = p. Hence p = 54 + 17 = 71, which is also prime. 11th Irish 1998 © John Scholes jscholes@kalva.demon.co.uk 17 Dec 2003 Last updated/corrected 17 Dec 03