Please read Digit Math: Introduction before you continue.
The problem
It seems I have developed a fascination for the number 5, so here we go again. Here I would be using Digit Math to prove that all integers which end with digit 5 are always divisible by 5.
The proof
Case 1:
Take a two digit number xω5, i.e. x has only one digit.
xω5 is divisible by 5, if xω5×15 (i.e. xω5×0.2) leave no remainder.
So, xω5×2 must end with one zero.
xω5×0ω2=0ω2xω10=(2x+1)ω0
Since (1) ends with zero so it is proved.
Case 2:
xω5 is a number where x has n digits.
Multiplying xω5 by 10(n−1) to make both sides of ω equal in number of digits.
So,
xω(5×10n−1)×0ω(2×10n−1)=0ω(2x×10n−1)ω(10×10n−1)=(2x×10n−1)ω10n=(2x×10n−1+1)ω]0[n
In the above equation it is easy to see that the result ends with 10 after we divide the 10n−1. Therefore, this is not going to leave any remainder. (Proved)
Case 3:
55=1. (Proved)
Yeah, this is a trivial case, but for the sake of completeness.