Take a 3-digit number 'abc', reverse this number and divide the difference of the original number and the reverse by 99, the remainder is:

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Solution

The correct option is D
0

abc can be written as $100 a + 10 b + c$
Similarly, $c b a = 100 c + 10 b + a$
The difference is:
$a b c - c b a = (100 a + 10 b + c) - (100 c + 10 b + a)$
$= 99 a - 99 c$
$= 99 (a - c)$
On dividing by $99$ ,
$D i v i d e n d = D i v i s o r \times Q u o t i e n t + R e m a i n d e r$
$99 (a - c) = 99 \times (a - c) + 0$
$R e m a i n d e r = 0$

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Take a 3-digit number 'abc', reverse this number and divide the difference of the original number and the reverse by 99, the remainder is:

The correct option is D 0 abc can be written as 100a+10b+c Similarly, cba=100c+10b+a The difference is: abc−cba=(100a+10b+c)−(100c+10b+a) =99a−99c =99(a−c) On dividing by 99, Dividend=Divisor×Quotient+Remainder 99(a−c)=99×(a−c)+0 Remainder=0