### 12th Putnam 1952

**Problem A6**

Prove that there are only finitely many cuboidal blocks with integer sides a x b x c, such that if the block is painted on the outside and then cut into unit cubes, exactly half the cubes have no face painted.

**Solution**

It is sufficient to show that abc = 2(a - 2)(b -2)(c - 2) has only finitely many solutions in integers c ≤ b ≤ a. If c ≤ 4, then c ≥ 2(c - 2), but ab > (a - 2)(b - 2), so there are no solutions. If c ≥ 10, then c/(c - 2) ≤ 10/8, so D ≤ 125/64 < 2, where for convenience we have written abc/( (a - 2)(b - 2)(c - 2) ) as D, so there are no solutions. Hence c = 5, 6, 7, 8, or 9.

If c = 5 and b ≥ 24, then D ≤ 5/3 (24/22)^{2} < 2, so there are no solutions. Now a/(a - 2) is strictly monotonic decreasing, so there is at most one solution for given b, c. Hence there are only finitely many solutions for c = 5.

Similarly, for c = 6, we find that are only solutions for b < 16; for c = 7, we find b < 14; for c = 8, b < 12 and for c = 9, b < 12. Hence there are only finitely many solutions in each case.

12th Putnam 1952

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002