Find the remainder when $3(12!)-15(11!)$ divides $22(11!)+22!$

Question

Find the remainder when $3(12!)-15(11!)$ divides $22(11!)+22!$. Options A) 11! B) 10! C) 12! D) 9!

My Approach: ${11!(22+22\cdot 21\cdot 20*\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12)\over 11!(3\cdot 12-15)} \text{ then }{(22+22\cdot 21\cdot 20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12)\over 21} \text{ then }{22\over21}$ gives remainder of 1 and the other term gives 0. So the final remainder I'm getting is 1.

These are the options given. I tried to take the 11! common from both numerator and denominator and cancelled it. Then in denominator I'm only left with 21. So As per my calculation I'm getting 1 as remainder but it's not present in the options. Please help.

Answer 1

The answer is 11!. I read the rule somewhere that if I cancel common factors from divisor and dividend then in the end I need to multiply it to the answer. In my case I cancelled common factor 11! and Remainder I got is 1 so 1*11!= 11! is the answer.

Answer 2

You could just do: $$36(11!)-15(11!)=21(11!) \quad\text{(divisor)}$$ $$ 22(11!)+22!= ({22!\over 11!} +22)(11!) \quad\text{(dividend)}$$ then $$21\mid {22!\over11!}\text{ and } 22\equiv 1 \bmod {21}\implies 11! \text{ is the answer}$$

Answer 3

We can factor out $\ \color{}{c=11!}\ $ using $\ \ ca\bmod cn\, =\, c(a\bmod n),\ $ the mod Distributive Law.

$$\begin{align} &\ \ \ \ \ \ (22(11!)\!+\! 22!\bmod\, 3(12!)\!-\!\!15(11!)\\[.4em] =\ & 11!\:(22\!+\!22!/11!\ \bmod\ \color{#c00}{3(12)\!-\!15})\\[.4em] =\ & 11!\ \ {\rm by}\ \ 22\!+\!22!/11! \equiv 1\! +\! 0\!\!\!\pmod{\!\color{#c00}{21}} \end{align}$$