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Tejas

asked 2016-10-21 11:18:08

What will be the remainder when $10!+111$ is divided by $143?$

Raj

answered 2016-10-23 18:47:52

Ans: $0$

$143=11×13$ where $11,13$ are co-prime.

Using Wilson's theorem, $10! \equiv -1 \pmod{11}$

Therefore,

$10!+111 \equiv -1+1\equiv 0 \pmod{11}~~\cdots(1)$

$10!=10×9×8×7×6×5×4×3×2\\

=(2×7)(3×8)(4×6)(5×10)(9)\\

\equiv 1(-2)(-2)(-2)(-4) \equiv 32 \equiv 6 \pmod{13}$

Therefore,

$10!+111 \equiv 6+7\equiv 13\equiv 0 \pmod{13}~~\cdots(2)$

From $(1)$ and $(2)$,

$10!+111$ is divisible by both $11$ and $13$.

Since $11$ and $13$ are co-prime, $10!+111$ is also divisible by their product $143$

$143=11×13$ where $11,13$ are co-prime.

Using Wilson's theorem, $10! \equiv -1 \pmod{11}$

Therefore,

$10!+111 \equiv -1+1\equiv 0 \pmod{11}~~\cdots(1)$

$10!=10×9×8×7×6×5×4×3×2\\

=(2×7)(3×8)(4×6)(5×10)(9)\\

\equiv 1(-2)(-2)(-2)(-4) \equiv 32 \equiv 6 \pmod{13}$

Therefore,

$10!+111 \equiv 6+7\equiv 13\equiv 0 \pmod{13}~~\cdots(2)$

From $(1)$ and $(2)$,

$10!+111$ is divisible by both $11$ and $13$.

Since $11$ and $13$ are co-prime, $10!+111$ is also divisible by their product $143$

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