### 43rd Putnam 1982

**Problem A5**

a, b, c, d are positive integers satisfying a + c ≤ 1982 and a/b + c/d < 1. Prove that 1 - a/b - c/d > 1/1983^{3}.

**Solution**

Let us consider the general case with 1982 replaced by N. We have a < N, so 1 - a/b > 1/N. So there is no benefit in making d too large. Certainly, for example, we must have d < 2Nc < 2N^{2}. Similarly for b. So there are only finitely many candidates for a, b, c, d. Hence there is a set of values which minimises 1 - a/b - c/d. Let us adopt these values.

Evidently 1 - a/b - c/d = k/bd for some positive integer k. We may assume that b ≥ d. The fact that a, b, c, d is an optimal set means that k cannot be too large. Multipying across, k = bd - ad - bc. So if k > d, we could increase a by 1. That would reduce 1 - a/b - c/d to (k-d)/bd, contradicting minimality. So k ≤ d.

Now bd = k + ad + bc <= k + ab + bc = k + b(a + c) <= k + bN. Hence d <= N + k/b <= N + 1.

But (k/bd + a/b) = 1 - c/d ≥ 1/d ≥ 1/(N+1). We have (k/bd + a/b) = 1/b (a + k/d) ≤ 1/b (a + 1) <= 1/b N. So 1/b ≥ 1/(N(N+1)) > 1/(N+1)^{2}. Hence 1 - a/b - c/d = k/bd > k/(N+1)^{3} ≥ 1/(N+1)^{3}.

43rd Putnam 1982

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001