IMO shortlist 1985/26 solution

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

Answer: 110880 = 2⁵3²5.7.11.

Solution by Demetres Christofides

If n = p^aq^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^a3^b5^c7^d11^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: 2¹¹3²5.7, 2⁸3³5.7, 2⁷3²5²7, 2⁵3⁵5.7, 2⁵3³5²7, 2⁵3²5.7.11, 2³3²5²7.11.

Taking out the common factor 2³3²5·7 we are comparing: 2⁸ = 256, 2⁵3 = 96, 2⁴3³ = 432, 2²3·5 = 60, 2²11 = 44, 5·11 = 55. So the smallest is 2⁵3²5·7·11 (corresponding to 44).

26th IMO shortlist 1985