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

ANS: $12$.

Let $S=1!+2\times2!+3\times3!…+12\times12!$.
It can be seen that $T_{n}=n\times n!=(n+1-1)\times n!$.
or $T_{n}=(n+1)n!-n!=(n+1)!-n!$
Substitute $n=1,2,3…12$, we get
$S=(2!-1!)+(3!-2!)…+(13!-12!)$$=13!-1!=13!-1$
So,
$rem(\frac{S}{13})$$=rem(\frac{13!-1}{13})=rem(\frac{13!}{13})$$-rem(\frac{1}{13})=0-1=-1$
or rem=$-1+13=12$.

## 10 Replies to “What is the remainder when 1! +2*2! +3*3! +4*4! +… +12*12! Is divided by 13?”

1. Shubham Illuminati Vij says:

You can write n*n! as (n+1-1)*n! =  (n+1)! – n!
now
1*1! +2*2! +3*3! +4*4! +… +n*n!      for n=12
=  2! – 1! + 3!-2! + 4! – 3! +  …. (n+1)! – n!
You can cancel alternate terms, which gives
=  (n+1)! – 1!                           Now put n=12
= 13! -1
= -1 (mod 13)                          13! = 0 (mod 13)
= 12 (mod 13)    Ans ðŸ™‚

2. Himanshu Bhardwaj says:

## 12

The trick in these type of questions is often observing the pattern

1! + 2*2! = 1 + 4 = 5 = 3! – 1
1! + 2*2! + 3*3! = 1 + 4 + 18 = 23 = 4! – 1
1! + 2*2! + 3*3! + 4*4! = 1 + 4 + 18 + 96 = 119 = 5! – 1
1! + 2*2! + 3*3! + 4*4! + 5*5! = 1 + 4 + 18 + 96 + 600 = 719 = 6! – 1
So, we can say
1! +2*2! +3*3! +4*4! +… +12*12! = 13! – 1
=> Rem [(1! +2*2! +3*3! +4*4! +… +12*12!) / 13] = -1 = 12

I have answered a bunch of very similar questions on remainders. You can get the complete list here: Remainder Theorem and related concepts for CAT Preparation by Ravi Handa on CAT Preparation

3. Ajay Sharma says:

We can solve this question easily by redistributing the multipliers of the terms.

$R[\frac {1! + 2 \times 2! + 3 \times 3! + 4 \times 4! +… + 12 \times 12!}{13}]$

$=R[\frac {(2â€“1)1! + (3â€“1) \times 2! + (4â€“1) \times 3! +… +(13â€“1) \times 12!}{13}]$

$=R[\frac {(2! – 1!) + (3! – 2!) + (4! – 3!) + â€¦â€¦ + (13! – 12!)}{13}]$

$=R[\frac {(13! – 1)}{13}]$

$=R[\frac {13!}{13}] – R[\frac {1}{13}]$

=0 – 1

4. Deepak Rathee says:

={1.1! + 2.2! + 3.3! + 4.4! +…………+ 12.12!}/13
={(2-1).1! + (3-1).2! + (4-1).3! + (5-1).4! +…………+ (13-1).12!}/13
={2.1!-1.1! + 3.2!-1.2! + 4.3!-1.3! + 5.4!-1.4! +…………+ 13.12!-1.12!}/13
={2!-1! + 3!-2! + 4!-3! + 5!-4! +…………+ 13!-12!}/13
=(13! – 1)/13
=(13! -13+13 – 1)/13
=(13! -13 + 12)/13
=(12!-1) + 12/13

remainder = 12

5. Mohamed Rameez says:

it forms pattern….
lets take
(1!+2*2!)/3=2
(1!+2*2!+3*3!)/4=3
(1!+2*2!+3*3!+4*4!)/5=4

so remainder is l less than the divider
so in this case remainder-12….

6. Ajay Sharma says:

n*n! can be written as (n+1)! – n!
For n=12
now 2!-1! + 3!-2! + 4! -3! ……. (n+1)! – n!.
cancel alternate terms so we get, 13! – 1! 13!(mod 13) =0  therfore 13-1= 12.

7. Nitish Grover says:

Step 1 : Write the T(n)th Term for the pattern .
T(n)th Term for this pattern is n*n! which is equalent to (n+1 – 1)*n!=(n+1)*n! – 1*n! = (n+1)! – n!

Step 2 : Add them from n = 1 to n = n , in this case n = 12 so imagine adding 1st 2 terms :
[ 2! – 1! ] + [ 3! – 2! ] …we note something here , terms cancel out…if you carry on this pattern you'll find it will give you 13! – 1! (for any n it will give you :  n+1 ! – 1 ! ).

Step 3: Calculate the remainder of 13! – 1 by 13 , you can say it is -1 as 13! is div by 13, but -1 is nothing but means 13 -1 = 12, hence the ans.

The trick was :
1. To note the T(n) th term
2. Deduce the pattern , in this case it was of the form T(n+1)-T(n)
3. Rest is simple computation.

8. Sarthak Dash says:

1! +2*2! = 5 = 3!-1
1! + 2*2! + 3*3! = 23 = 4!-1
Similarly
1! + 2*2! + 3*3! +….+12*12! = 13!-1
So
1! + 2*2!+…+12*12! / 13
= 13! – 1 / 13
= 13! / 13 – 1/13
= 0 – 1 / 13
= -1/13
= -1+13
= 12 —> remainder

9. Nitish Grover says:

The sum of  n terms in the series is of the form (n+1)!-1!.Hence the sum of 12 terms in the series is 13!-1! which when divided by 13 will leave remainder as -1 or in the positive sense 12.I reckon this is a CAT question.

10. Rajendra Rajput says:

tn= n.n!
tn=(n+1-1)n!
tn=(n+1)!-n!
Sn=13!-1!
remainder= 12