31st Eötvös 1927 Problem 2 Find the sum of all four digit numbers (written in base 10) which contain only the digits 1 - 5 and contain no digit twice.
Denote the digits as d 3 d 2 d 1 d 0 . If we fix d 0 = k, then there are 4·3·2 ways of choosing the other digits, so there are 24 such numbers. The sum of their d 0 digits is 24k. Thus the sum of the d 0 digits for all the numbers is 24(1+2+3+4+5) = 24·15 = 360. A similar argument works for the other digits, so the sum of the numbers is 360(1+10+100+1000) = 399960. 31st Eötvös 1927 © John Scholes jscholes@kalva.demon.co.uk 1 Nov 2003 Last corrected/updated 1 Nov 03