Latest Elitmus Aptitude Question SOLUTION: 1! * 2^2 + 2! 3^2 + 3! * 4^2 + .....+ 20! 21^2 ) / 21 what is the remainder....?

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(#M40157788) ELITMUS Numerical Ability question Keep an EYE

1! * 2^2 + 2! 3^2 + 3! * 4^2 + .....+ 20! 21^2 ) / 21 what is the remainder....?

Asked In Elitmus Sniper (7 years ago)

Unsolved Read Solution (7)

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pooja kumari  (7 years ago)

quesn is in the form( 2!*2+3!*3+4!*4+5!*5+...........+21!*21)/21
so remainder will be
(2*2+6*3+3*4+12+15+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0)/21=61/21
hence remainder will be 19

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Mallikarjuna Reddy  (7 years ago)

from 6!*7^2 every number is divided by 21
so the required remainder is (1! * 2^2 + 2! 3^2 + 3! * 4^2 + 4!*5^2 + 5!*6^2 )/21
ans is 19

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rajat  (7 years ago)

(1!+2!+3!+.......20!)(2^2+3^2.......21^2)
as after 7 each term is divisibel by 21 . so rem(1!+2!+3!+4!+5!+6!)/21=12 rem.
as sum of square=n(2n+1)(n+1)/2=21*(21*2+1)*(21+1)/2=3311.
as 1 is not there subtract 1 in 3311=3310.
3310/21=13 rem.so
rem=13*12/21=9
9 ans

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abdullah imran  (7 years ago)

From 6th term onward remainder will be zero..
So remainder = R[(2+18+4+4+5+0+0+...)/21]=R[33/21]=12 -ans

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Aakash kesarwani  (7 years ago)

it is squre proportional of its area . So T=time to file
(4+16/9+1)T=4*122.
T =72 minutes.

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Manisha Choudhary  (7 years ago)

question is in the form 1! +2*2! +3*3! +4*4! +.... 20!*20 / 21
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, till 20!*20= 21! -1/ 21
Now, apply wilson theorem = -1 = 19

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Manisha Choudhary  (7 years ago)

oups!! the final answer is 20. As, it satisfy the wilson theorem .

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