Aarush asked Soham to take a two digit number and reverse it, 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:
Not divisible by 9
Let's consider a two digit number ab.Its general form will be 10×a+1×b.
If we reverse it, then the number will become ba. Its general form will be 10×b+1×a. If we subtract both of the numbers, we get:
(10×a+1×b)−(10×b+1×a)=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×a+10×b+c)−(100×c+10×b+a)=99(a−c)
which is divisible by 9.
Note that this happens with a number of any number of digits