# Find the remainder obtained when 10! + 111 is divisible by 143. Option 1) Option 2) Option 3) Option 4) Option 5)

$Remainder\: of \:\frac{ \left ( 10!+11 \right )}{143}$

$\Rightarrow \: \: \: \:\frac{ \left ( 10! \right )}{143}+\frac{\left ( 111 \right )}{143}$

$\Rightarrow \: \: \: \: \frac{\left ( 720\times 5040 \right )}{143}+111= \frac{\left ( 32+111 \right )}{143}$

0 remainder

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0

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