### 34th IMO 1993 shortlist

**Problem 23**
Show that for any finite set S of distinct positive integers, we can find a set T ⊇ S such that every member of T divides the sum of all the members of T.

**Solution**

Let S = {a_{1}, a_{2}, ... , a_{r}}. Put s = ∑ a_{i}, m = lcm(S). Suppose m = 2^{k}n, where n is odd. Put n = b_{t}...b_{1}b_{0} in binary. Adjoin 2^{i}s for i>1 and b_{i}=1. The adjoined numbers have sum (n-1)s, so the enlarged set has sum ns. Now adjoin ns, 2ns, 2^{2}ns, ... , 2^{h-1}ns, where h = max(k,t). The adjoined numbers have sum (2^{h}-1)ns, so the enlarged set T has sum 2^{h}ns. All the members of T divide the sum.

34th IMO shortlist 1993

© John Scholes

jscholes@kalva.demon.co.uk

8 Nov 2003

Last corrected/updated 8 Nov 03