What is the remainder when 1! + 2*2! + 3*3! + 4*4! + … + 12*12! is div

Question

What is the remainder when  1! + 2*2! + 3*3! + 4*4! + … + 12*12!  is divided by 13?

(A) 12
(B) 11
(C) 10
(D) 1
(E) 0

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firas92 · Accepted Answer

n*n!= *n!=(n+1)n!-n!

That is, n*n!=(n+1)!-n!

So the given equation can be rewritten as

2!-1!+3!-2!+4!-3!+5!-4!+....+13!-12!

After cancelling out the equal opposites, we are left with

-1!+13! = 13!-1

13! is a multiple of 13 so 1 less than 13 factorial must leave a remainder of 13-1=12

Answer is (A)

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Raksat · Answer

I think the answer should be A. My Approach:
we can write 1!+2∗2!+3∗3!+4∗4!+…+12∗12!1!+2∗2!+3∗3!+4∗4!+…+12∗12! 
as 1+(3-1)2!+(4-1)*3!+.....................+(13-1)*12!
or, 1+3*2!-2!+4*3!-3!+................+13*12!-12!
or, 1+3!-2!+4!-3!+........................+13!-12!
or, 1-2!+13!
or, 13!-1
so when 13!-1 is divided by 13 the reminder should be 12

NitishJain · Answer

IMO  A 
n * n! can be written as (n+1 - 1)* n!

So, 1!+2∗2!+3∗3!+4∗4!+…+12∗12!1!+2∗2!+3∗3!+4∗4!+…+12∗12! = (2! -1!)+ (3!-2!) + (4!-3!) + (5!-4!) ........(13! - 12!)

All the terms will be cancelled and left will be: 13! -1

so when divided be 13: 13! will be completely divisible.

-1 willbe left which is equal to 12 as a remainer

bumpbot · Answer

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