If you take a 3 digit number and reverse its numbers like 651 to 156 and then subtract it from the first number like 651 minus 156 the answer is always divisible by 99 why?

Answer 1

Given a three digit number n = "d1 d2 d3" (i.e. a number 0 <= n <= 999, where d1, d2 and d3 are it's digits), we can express n as 100d1 + 10d2 + 1d3. The number n' with the digits of nin reverse order would be: n' = "d3 d2 d1", which could be expressed as n' = 100d3 + 10d2 + 1d1. Subtracting n from n' (or vice versa) we get the following equation:

n - n' = (100d1 + 10d2 + 1d3) - (100d3 + 10d2 + 1d1)

resolving and rearranging we obtain:

n - n' = (100d1 - 1d1) + (10d2 - 10d2) + (100d3 - 1d3) = 99d1 - 99d3 = 99(d1 - d3)

Since d1 and d3 are integers (and d2 cancels out), we see that the result 99(d1 - d3) is always divisible by 99.

(The equation holds for all integers values for d1, d2 and d3, not just for the digits 0, 1, ... , 9, but we can no longer write it as a "three digit number". )

Example:

651 - 156 = (6*100 + 5*10 + 1*1) - (1*100 + 5*10 + 6*1) = 500 - 0 - 5 = 495 = 99*5