Question

if

\(\large\tt \begin{align} \color{black}{p=1!+(2\times 2!)+(3\times 3!)+.......+(10\times 10!),
}\end{align}\)

then \(\large p+2 \) when divided by \(\large 11!\) leaves remainder of

a.)10
b.)0
c.)7
d.)1

Answer 1

i think its \(\Large 1\)

Answer 2

\[p=\sum_{1}^{10}n*n!\]
\[a _{n}=n*n!=(n+1-1)*n!=(n+1)!-n!\]
put n=1,2,...,10 and add vertically downwards
solve

Answer 3

@ganeshie8 is my solution correct?

Answer 4

@mathmath333

Answer 5

looks good to me, i think surjithayer also has the same idea(telescoping) :

\[\sum\limits_{n=1}^{10}\left[f(n+1) - f(n)\right] = f(10+1) -f(1) \]

Answer 6

can i give another idea xD

Answer 7

you should be preparing for regression analysis exam

Answer 8

you're thinking of wilson's thm right ? ;)

Answer 9

1!=1 mod 11
2!=2 mod 11
3!=6 mod 11
4!= 3! mod 11* 4 mod 11 ( goes like this for the rest )
4!=2 mod 11
5!=-1 mod 11
6!=5 mod 11
7!=2 mod 11
8!=5 mod 11
9!=1 mod 11
10!=-1 mod 11

thus u would have
P mod 11 =1+(2*1)+(3*6)+(4*2)+(5*-1)+(6*5)+(7*2)+(8*5)+(9*1)+(10*-1) mod 11

reduce more
P mod 11 =1+(2)+(7)+(8)+(6)+(8)+(3)+(7)+(9)+(1) mod 11
reduce more xD
p mod 11=52 mod 11=8 mod 11
p+2=10 mod 11 =-1 mod 11

i guess i made a mistack somewhere xD using my lazy method lol

Answer 10

wilson only work for
10! =-1 mod 11 part xD
i used lazy analysis but see i got -1 hmm

Answer 11

which is the same as 10 lol
so its ovc something wrong :O

Answer 12

answer should be 1 hmm
i''ll check my sol again -.-

Answer 13

lol ok i know what i have done lil mistake second term should be
2*2 instead of 2*1

Answer 14

thus u would have
P mod 11 =1+(2*2)+(3*6)+(4*2)+(5*-1)+(6*5)+(7*2)+(8*5)+(9*1)+(10*-1) mod 11

reduce more
P mod 11 =1+(4)+(7)+(8)+(6)+(8)+(3)+(7)+(9)+(1) mod 11
reduce more xD
p mod 11=54 mod 11=-1 mod 11
p+2=1 mod 11
wew ! done :3

Answer 15

i used the idea that \((1!+2!+3!+4!.....10*10+2)/11! \\
\implies (1!+2*2!+3*3!+4*4!.....10*10!+2)/10\\
\implies (1!+2*2!+...5*5!+2)/10---\text{as after 5*5! all will}\\
\text{be divisible by 10}\\
\implies remainder ~~1\)

but i like surjithiyers and pricejones ideas more

Answer 16

Ohk, you are just reducing the congruences nice :)

Answer 17

thats for ikram ^

Answer 18

\(
o-0\)

Answer 19

nice catch !

Answer 20

Clever !!

Answer 21

was that foe me?

Answer 22

Yes :)

Answer 23

*jealous* :P

Answer 24

yay

Answer 25

btw i never know that
(k-1)n!=k!-n! ( or its only special case and k need to be n+1 ? )
so its something new to look for !

Answer 26

oh sorry i misread, one sec

Answer 27

(n+1)n! = (n+1)!

Answer 28

i got it now lol
so
((n+1)-1 )n!=(n+1)n!-n!=(n+1)!-n!
so yeah it need to be n+1 not interesting as i thought

Answer 29

was out of my mind for few second xD

Answer 30

thnx everyone!