### 30th Putnam 1969

**Problem B1**

The positive integer n is divisible by 24. Show that the sum of all the positive divisors of n - 1 (including 1 and n - 1) is also divisible by 24.

**Solution**

Let n = 24m. We show first that if d is a divisor of n - 1 = 24m - 1, then d^{2} - 1 is divisible by 24. Clearly d is not a multiple of 3 (because n is), so 3 must divide d^{2} - 1. Also d must be odd (like n - 1), so d - 1 and d + 1 are consecutive even numbers, so one must be a multiple of 4, and there product d^{2} - 1 must be a multiple of 8.

Now 24m - 1 cannot be a square (because squares are congruent to 0 or 1 mod 4), so its divisors come in pairs d, (24m - 1)/d. But d + (24m - 1)/d is divisible by 24 (because d^{2} - 1 and 24m are and no factor of 24 can divide d). Hence the sum of all the divisors of n - 1 is divisible by 24.

30th Putnam 1969

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002