Question

remainder when (16! + 91 ) is divided by 323? Please explain the approach :)
Thank you !

Answer 1

familiar with wilson's theorem?

Answer 2

My approach is to use wolfram alpha..

Ha ha ha ha..

Just kidding..

Answer 3

323 =17 * 19
well I am.. and it made sense for the first part.. ((17-1) ! + 1 + 85 )/ 16
But not for the second part.. i.e. with 19

Answer 4

could you explain wilsons theorem with 19 ?

Answer 5

lol......i thought its 13*17 let me work on 19..

Answer 6

I did not get how you got 1 + 85 there ??

Answer 7

by wilson's theorem

\(\mod 19: \ \ \ \ −1 ≡ 18! ≡ 18⋅17⋅16! ≡ (−1) (−2)⋅16!\)

so we can write \(2.16! ≡-1≡18 \ \ \mod 19\) and \(16! ≡9 \ \ \mod 19\)

since we know that \(\gcd(2,19)=1\)

Answer 8

ok ! but how does that lead to a remainder of 5 ! :)
sorry .. but im a little new to this

Answer 9

so till here we found out

\(16! + 91≡5 \ \ \ \mod 17\)
\(16! + 91≡5 \ \ \ \mod 19\)

Answer 10

am i right?

Answer 11

no..

Answer 12

@waterineyes why?....lol

Answer 13

Because I am not getting what you both are doing here..

It is passing approximately 2 meters from my head..

Ha ha ha..

Answer 14

16!+91≡5 mod19 .. how exactly.. :(

Answer 15

did u get how 16!≡9 mod19 ?

Answer 16

Asking me ??

Answer 17

yes!

Answer 18

loooool...what happend here? @Ajax12

did u get how 16!≡9 mod19 ?

Answer 19

wilson's says

18!≡18 mod 19 right?

Answer 20

yes! got that too ! :D

Answer 21

so last stage of solving problrm

If \(\gcd(p, q) = 1\), then for all integers \(x\) and \(a\), \(x ≡ a \ \ \mod p\)

and \(x ≡ a \mod q\) then \(x ≡ a \ \ \mod pq\).

This is a corollary of a famous theorem known as the Chinese Remainder Therorem.


so we deduce that \(16!+91≡5 \ \ \mod \ 323\) since we proved that

\(16!+91≡5 \ \ \mod \ 17\)
\(16!+91≡5 \ \ \mod \ 19\)

Answer 22

gracias! :D

Answer 23

yw....:)