3rd Centromerican 2001 Problem A3 Find all squares which have only two non-zero digits, one of them 3.
36, 3600, 360000, ... Solution A square must end in 0, 1, 4, 5, 6, or 9. So the 3 must be the first digit. If a square ends in 0, then it must end in an even number of 0s and removing these must give a square. Thus we need only consider numbers which do not end in 0. The number cannot end in 9, for then it would be divisible by 3 but not 9 (using the sum of digits test). It cannot end in 5, because squares ending in 5 must end in 25. So it remains to consider 1, 4, 6. 36 is a square. But if there are one or more 0s between the 3 and the 6, then the number is divisible by 2 but not 4, so 36 is the only solution ending in 6. Suppose 3·10 n + 1 = m 2 , so 3·2 n 5 n = (m-1)(m+1). But m+1 and m-1 cannot both be divisible by 5, so one must be a multiple of 5 n . But 5 n &t; 3·2 n + 2 for n > 1, so that is impossible for n > 1. For n = 1, we have 31, which is not a square. Thus there are no squares 3·10 n + 1. A similar argument works for 3·10 n + 4, because if 3·10 n + 4 = m 2 , then 5 cannot divide m-2 and m+2, so 5 n must divide one of them, which is then too big, since 5 n > 3·2 n + 4 for n > 1. For n = 1 we have 34, which is not a square. 3rd OMCC 2001 © John Scholes jscholes@kalva.demon.co.uk 30 January 2004 Last corrected/updated 30 Jan 04