3rd ASU 1969 Problem 8 Find four different three-digit numbers (in base 10) starting with the same digit, such that their sum is divisible by three of the numbers. Solution
: 108, 117, 135, 180. Sum 540 = 108·5 = 135·4 = 180·3. Try looking for a number of the form 3·4·5·n. We want 12n, 15n and 20n to have the same first digit. If the first digit is 1, this requires n = 9. We must now check that the fourth number which must be 60n - 12n - 15n - 20n = 13n also has three digits starting with 1. It does, so we are home. [In fact, in this case the first digit must be 1, since 20n > 3/2 12n.] 3rd ASU 1969 © John Scholes jscholes@kalva.demon.co.uk 6 Nov 1998