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.

score 3 · Answer 1 · edited Oct 23 '17 at 14:46

3

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 $$21\mid {22!\over11!}\text{ and } 22\equiv 1 \bmod {21}\implies 11! \text{ is the answer}$$

edited Oct 23 '17 at 14:46

Simply Beautiful Art

74,685

answered Jul 26 '17 at 12:06

You are implicitly applying the mod Distributive Law - see my answer. – Bill Dubuque Jul 26 '17 at 13:42

Bill Dubuque · Answer 2 · 2017-07-26T13:40:59.613

2

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

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

edited Jul 26 '17 at 13:40

answered Jul 26 '17 at 13:19

Bill Dubuque

272,048

score 0 · Answer 3 · answered Jul 26 '17 at 11:25

0

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.

answered Jul 26 '17 at 11:25

Rohit

37

I explain how to operationally solve it in this manner in my answer. Using the explicit formula in the mod Distributive law helps up keep track of all the relevant info (including the factor $c$ which you forgot to bring back when your separated it into two steps in your question). – Bill Dubuque Jul 26 '17 at 17:45
Thank you Bill. – Rohit Jul 27 '17 at 12:59

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

3 Answers3