Question

Aarush asked Soham to take any 2 or 3 digit number and reverse the digits, then subtract the larger number from the smaller number. Soham did the same and concluded that the result is exactly divisible by 9. What Soham concluded is:

• A. True
• B. False

Answer 1

The correct option is A

True

Let's consider a two digit number ab.Its general form will be $10\times a+1\times b$.
If we reverse the digits, its general form will be $10\times b+1\times a$. If we subtract both of the numbers, we get:
$(10\times a+1\times b)-(10\times b+1\times a)$
= (10a + b) - (10b + a)
= 10a + b -10b - a
= 9a - 9b
= 9 (a-b)
Observe that the result is a multiple of 9 and hence is exactly divisible by 9.
Similarly, for 3 digit numbers, the difference of the number and number obtained by reversing the digits (if abc is the number) will be given by:
$(100\times a+10\times b+c)-(100\times c+10\times b+a)$
=100a + 10b + c - (100c + 10b + a)
= 100a + 10b + c - 100c - 10b - a
= 99a - 99c
= 99 (a-c)
= $9\times 11(a-c)$
which is also divisible by 9.

Note that this happens with any 2 or 3 number.