6 digit divisibility by 666

Question

Let n, n+1, n+2 be digits (where n is different from 0). Using them, it is possible to create 6 different 3 digit numbers. Prove that the sum of the 6 3 digit numbers is divisible by 666

Answer 1

the six numbers are...

100n+10(n+1)+(n+2)

100n+10(n+2)+(n+1)

100(n+1)+10n+(n+2)

100(n+1)+10(n+2)+n

100(n+2)+10n+(n+1)

100(n+2)+10(n+1)+n

add up the n's and add up the numbers...

600n+60n+6n+600+60+6=666n+666=666(n+1) which is divisible by 666