Question

Difference between 3 digit number and number formed by reversing the digits, is always divisible by

Answer 1

Let's assume that x is a three digit number which can be written as (100a+10b+c).

So x has:
a in the hundreds place
b in the tens place
c in the ones place

if we reverse the digits, we will get (100c+10b+a)

Difference = (100a+10b+c) - (100c+10b+a)

=> D = 99a - 99c

=> D = 99(a-c)

Since a and c can be natural numbers from 1 to 9, (a-c) is always a whole number.

So the difference can be written as 99m where m is a whole number.

Now, 99m is always divisible by: 1, 3, 9, 11, 33, 99

So it is always divisible by 1, 3, 9, 11, 33 and 99

Answer 2

1. Let xyz be 3 digit number
2. if 123 is a 3 digit number it can always be written as (3*1)+(2*10)+(1*100)
3. In similar way if xyz is a 3 digit number it can be written as (z*1)+(y*10)+(x*100)
4. If we reverse xyz we get zyx and it can be written as (z*100)+(y*10)+(x*1)
5. now if we calculate difference we get [(x*100)+(y*10)+x]-[(z*100)+(y*10)+x]
6. On calculating difference we get result as 99(x-z)
7. hence it will be divisible by 1,3,9,11,99,33