Latest Elitmus Aptitude Question SOLUTION: 4(4!)+5(5!)+6(6!)+7(7!)+.......+19(19!)

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(#M40154086) ELITMUS Numerical Ability question Keep an EYE

4(4!)+5(5!)+6(6!)+7(7!)+.......+19(19!)

Asked In Elitmus dytort (8 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

gagan  (8 years ago)

correct Q: find remainder of (4(4!)+5(5!)+6(6!)+......+19(19!)) /64.
sol:
5-1(4!)+6-1(5!)+7-1(6!)+.....+20-1(19!)/64
5.4!-4!+6.5!-5!+7.6!-6!+....20.19!-19!/64
5!-4!+6!-5!+7!-6!+...........+20!-19!/64
(-4!+20!)/64

20! is divisible by 64, therefore remainder is 64-4!=40

Is this solution Helpfull? Yes (12) | No   

Akshay  (8 years ago)

8*8!+9*9!....... will be completely divisible by 64 so we have to concern about first 4 terms so
4(4!)+5(5!)+6(6!)+7(7!) is 40296 when divided by 64 will gives 40 so answer is 40

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sumit payasi  (8 years ago)

20(20!) - 1 -4 -18

Is this solution Helpfull? Yes | No (1)  

sumit payasi  (8 years ago)

n*n! ka summation (n+1)! - 1
20!-1 -(4!-1) = 20!-4! leave the first ans submitted by me

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akshay kumar  (8 years ago)

option kya tha???????

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arti kumari  (8 years ago)

here we have to find remainder if it will divide by 64.

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Abhishek  (8 years ago)

=(4.4! + 5.5! + 6.6! +7.7!)/64
take 4! common
4!(4 + 5.5 + 6.6.5 + 7.7.6.5)/64
=4!(4+25+180+1470)/64
=4!*1679/64
=4*3*2*1679/64
=3*1679/8
=5037/8
5032 is divisible by 8 so 5 would be the remainder .

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