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

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Solution

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

Assuming a>c, subtracting smaller number from larger
$(100 a + 10 b + c) - (100 c + 10 b + a)$
$= (100 - 1) a + (10 - 10) b + (1 - 100) c$
$= 99 a - 99 c = 99 (a - c)$
So, it is always divisible by $99$ .

Now, $99 = 11 \times 9$
Therefore, $99 (a - c) = 11 \times 9 (a - c)$
Hence, the difference obtained is always divisible by 99, 9 and 11.

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The difference between a 3 digit number and a number formed by reversing its digits is divisible by99

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