Prove that , if the difference between a three digit number in the form of a0b that is digit at ten's place is 0 and the sum of it's digits is divided by 99, we always get the number equal to a?

Answer 1

a0b has the value 100a+0+b

100a+b - (a+b) = 99a

divide by 99, and you always get a

Answer 2

Uncompleted