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Question

The difference between a 3 digit number and a number formed by reversing its digits is divisible by
  1. 99

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Solution

The correct option is A 99
Let 100a+10b+c be the original number.
Then, the number obtained by reversing its digits is 100c+10b+a.

Assuming a>c, subtracting smaller number from larger
(100a+10b+c)(100c+10b+a)
=(1001)a+(1010)b+(1100)c
=99a99c=99(ac)
So, it is always divisible by 99.

Now, 99=11×9
Therefore, 99(ac)=11×9(ac)
Hence, the difference obtained is always divisible by 99, 9 and 11.

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