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Question
Find the remainder when $110!$ Is divided by $107^2$

numbers
2016-10-21 11:19:23
Tejas
Ans: $10807$

Required remainder
$=\text{rem}\left(\dfrac{110!}{107^2}\right)\\ =\text{rem}\left(\dfrac{110×109×108×107×106!}{107^2}\right)$

But, $\text{rem}\left(\dfrac{kx}{ky}\right)=k×\text{rem}\left(\dfrac{x}{y}\right)$

Therefore, required remainder
$=107×\text{rem}\left(\dfrac{110×109×108×106!}{107}\right)\\ =107×\text{rem}\left(\dfrac{3×2×1×106!}{107}\right)\\ =107×\text{rem}\left(\dfrac{6×106!}{107}\right)$

If $p$ is a prime number, by Wilson's theorem,
$\text{rem}\left(\dfrac{(p-1)!}{p}\right)=p-1$
Hence, $\text{rem}\left(\dfrac{106!}{107}\right)=106$

Therefore, required remainder
$=107×\text{rem}\left(\dfrac{6×106}{107}\right)\\ =107×\text{rem}\left(\dfrac{636}{107}\right)\\ =107×101\\ =10807$

Note: Using congruence relation, we can write Wilson's Theorem as
$(p-1)!\equiv(p-1)\equiv -1 \pmod p$
when $p$ is prime.

Hence,
$(6×106!) \equiv (6×-1) \equiv -6 \equiv 101 \pmod{107}$

therefore, required remainder
$=107×101=10807$

(2) (0)
2016-10-22 21:53:24
Raj (Senior Maths Expert, careerbless.com)