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

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].

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

  1. 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 🙂

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

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

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

    [math]=R[\frac {(2! – 1!) + (3! – 2!) + (4! – 3!) + …… + (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)

  4. ={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. 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. 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.

  7. 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. 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. 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.

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