Find the sum of all four digit numbers (written in base 10) which contain only the digits 1 - 5 and contain no digit twice.
Solution
Denote the digits as d3d2d1d0. If we fix d0 = k, then there are 4·3·2 ways of choosing the other digits, so there are 24 such numbers. The sum of their d0 digits is 24k. Thus the sum of the d0 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.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03