$T w o n u m b e r s, x a n d y a r e s u c h t h a t w h e n$
$d i v i d e d b y 6, t h e y l e a v e r e m a i n d e r 4 a n d 5$
$r e s p e c t i v e l y . F i n d t h e r e m a i n d e r w h e n$ $x^{3} + y^{3} i s d i v i d e d b y 6 ?$

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Solution

The correct option is B 3
$R e m \frac{[x]}{6} = 4, R e m \frac{[y]}{6} = 5$

$R e m = \frac{[x^{3} + y^{3}]}{6} = \frac{[4^{3} + 5^{3}]}{6}$

$= \frac{[64 + 125]}{6}$

$\Rightarrow R e m [\frac{4 + 5}{6}] = R e m (\frac{9}{6}) = 3$

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