If the number $(100 \times e + 10 \times g + f)$ is reversed and the difference of reversed and original number is divided by $33$ , then the quotient and remainder are:

(f - g); 3

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99 (e - f); 0

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33 (e - g); 3

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3 (e - f); 0

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Solution

The correct option is D
3 (e - f); 0

$100 e + 10 g + f = e g f$
Reversing the digits
$f g e = 100 f + 10 g + e$
Taking the difference of $e g f - f g e$
$= (100 e + 10 g + f) - (100 f + 10 g + e)$
$= 100 e + 10 g + f - 100 f - 10 g - e$
$= 99 e - 99 f = 99 (e - f)$
When divided by 33:
$D i v i d e n d = D i v i s o r \times Q u o t i e n t + R e m a i n d e r$
$99 (e - f) = 33 \times 3 (e - f) + 0$
$R e m a i n d e r = 0$
$Q u o t i e n t = 3 (e - f)$

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If the number (100×e+10×g+f) is reversed and the difference of reversed and original number is divided by 33, then the quotient and remainder are:

If the number $(100 \times e + 10 \times g + f)$ is reversed and the difference of reversed and original number is divided by $33$ , then the quotient and remainder are: