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Question

The difference between a 3 digit number and the number formed by reversing its digits is not always divisible by


A
9
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B
11
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C
99
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D
5
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Solution

The correct option is D 5

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.

So, the difference obtained is not divisible by 5 unless (a - c) is divisible by 5.


Mathematics

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