IMO 1990 SL 1

Origin: AUS

Problem

The integer 9 can be written as a sum of two consecutive integers: 9 = 4+5. Moreover, it can be written as a sum of (more than one) consecutive positive integers in exactly two ways: 9 = 4 + 5 = 2 + 3 + 4. Is there an integer that can be written as a sum of 1990 consecutive integers and that can be written as a sum of (more than one) consecutive positive integers in exactly 1990 ways?

Solution

Let N be a number that can be written as a sum of 1990 consecutive integers and as a sum of consecutive positive integers in exactly 1990 ways. The former requirement gives us N = m + (m + 1) + \cdot \cdot \cdot + (m + 1989) = 995(2m + 1989) for some m. Thus 2 ∤N, 5 | N, and 199 | N. The latter requirement tells us that there are exactly 1990 ways to express N as n + (n + 1) + \cdot \cdot \cdot + (n + k), or equivalently, express 2N as (k + 1)(2n + k). Since N is odd, it follows that one of the factors k + 1 and 2n + k is odd and the other is divisible by 2, but not by 4. Evidently k +1 < 2n+k. On the other hand, every factorization 2N = ab, 1 < a < b, corresponds to a single pair (n, k), where n = b−a+1 (which is an integer) and k = a −1. The number of such factorizations is equal to d(2N)/2 −1 because a = b is impossible (here d(x) denotes the number of positive divisors of an x \inN). Hence we must have d(2N) = 2 \cdot 1991 = 3982. Now let 2N = 2\cdot5e1 \cdot199e2 \cdotpe3 3 \cdot \cdot \cdot per r be a factorization of 2N into prime numbers, where p3, . . . , pr are distinct primes other than 2, 5, and 199 and e1, \cdot \cdot \cdot , er are positive integers. Then d(2N) = 2(e1 + 1)(e2 + 1) \cdot \cdot \cdot (er + 1), from which we deduce (e1 + 1)(e2 + 1) \cdot \cdot \cdot (er + 1) = 1991 = 11 \cdot 181. We thus get {e1, e2} = {10, 180} and e3 = \cdot \cdot \cdot = er = 0. Hence N = 510 \cdot 199180 and N = 5180 \cdot 19910 are the only possible solutions. These numbers indeed satisfy the desired properties.