ANS: [math]12[/math].

Let [math]S=1!+2\times2!+3\times3!…+12\times12![/math].

It can be seen that [math]T_{n}=n\times n!=(n+1-1)\times n![/math].

or [math]T_{n}=(n+1)n!-n!=(n+1)!-n![/math]

Substitute [math]n=1,2,3…12[/math], we get

[math]S=(2!-1!)+(3!-2!)…+(13!-12!)[/math][math]=13!-1!=13!-1[/math]

So,

[math]rem(\frac{S}{13})[/math][math]=rem(\frac{13!-1}{13})=rem(\frac{13!}{13})[/math][math]-rem(\frac{1}{13})=0-1=-1[/math]

or rem=[math]-1+13=12[/math].

You can write n*n! as (n+1-1)*n! = (n+1)! – n!

1*1! +2*2! +3*3! +4*4! +… +n*n! for n=12

now

= 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™‚## 12

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

1! + 2*2! = 1 + 4 = 5 =

3! – 11! + 2*2! + 3*3! = 1 + 4 + 18 = 23 =

4! – 11! + 2*2! + 3*3! + 4*4! = 1 + 4 + 18 + 96 = 119 =

5! – 11! + 2*2! + 3*3! + 4*4! + 5*5! = 1 + 4 + 18 + 96 + 600 = 719 =

6! – 1So, 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

= 12I 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 PreparationWe can solve this question easily by

redistributing the multipliersof the terms.[math]R[\frac {1! + 2 \times 2! + 3 \times 3! + 4 \times 4! +… + 12 \times 12!}{13}][/math]

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

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

[math]=R[\frac {(13! – 1)}{13}][/math]

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

=0 – 1

=

(-1) or 12 (Answer)={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

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….

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.

The answer is 12.

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.

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

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.tn= n.n!

tn=(n+1-1)n!

tn=(n+1)!-n!

Sn=13!-1!

remainder= 12