26th IMO 1985 shortlist Problem 26 Find the smallest positive integer n such that n has exactly 144 positive divisors including 10 consecutive integers. Solution
: 110880 = 2 5 3 2 5.7.11. Solution by Demetres Christofides If n = p a q b ... , then n has exactly (a + 1)(b + 1) ... distinct positive divisors. 144 = 2·2·2·2·3·3, so n can have at most 6 prime divisors. It is divisible by 10 consecutive integers. These must include a multiple of 5, a multiple of 7, a multiple of 9 and a multiple of 8. So n must be divisible by 2, 3, 5, 7. Also if n the power of 2 dividing n is 2 m , then m >= 3, so (m + 1) ≥ 4, which must account for at least two of the factors 2·2·2·2·3·3. So n has either 4 or 5 prime factors. If it has 5, then clearly we must take the fifth to be as small as possible and hence to be 11. Thus we must have n = 2 a 3 b 5 c 7 d 11 e , with a ≥ 3, b ≥ 2, c ≥ 1, d ≥ 1, e ≥ 0 and (a + 1)(b + 1)(c + 1)(d + 1)(e + 1) = 144. So (a+1, b+1, c+1, d+1, e+1) must be a permutation of one of the following: (12, 3, 2, 2, 1), (9, 4, 2, 2, 1), (8, 3, 3, 2, 1), (6, 6, 2, 2, 1), (6, 4, 3, 2, 1), (6, 3, 2, 2, 2), (4, 3, 3, 2, 2). Where possible the larger exponents should go with the smaller primes, so that gives the following possibilities: 2 11 3 2 5.7, 2 8 3 3 5.7, 2 7 3 2 5 2 7, 2 5 3 5 5.7, 2 5 3 3 5 2 7, 2 5 3 2 5.7.11, 2 3 3 2 5 2 7.11. Taking out the common factor 2 3 3 2 5·7 we are comparing: 2 8 = 256, 2 5 3 = 96, 2 4 3 3 = 432, 2 2 3·5 = 60, 2 2 11 = 44, 5·11 = 55. So the smallest is 2 5 3 2 5·7·11 (corresponding to 44). 26th IMO shortlist 1985 © John Scholes jscholes@kalva.demon.co.uk 11 Sep 2002