Question

show that if the difference between a three digit number in the form of a0b and the sum of its digit is divided by 99 we always get the number equal to a

Answer 1

The three-digit number is in the form of a0b. 

Its expanded form is 100a + 10(0) + b = 100a + b. 

Sum of its digits is a + b. 

Now the difference between the three digit number and the sum of its digits (say d) is,

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

d = 100a + b - a - b

d = 99a

Dividing this difference by 99,

d/99 = (99a)/99

d/99 = a

I.e., when this difference is divided by 99 we always get the number equal to a.

Hence Proved!