Latest TCS Aptitude Question SOLUTION: 1!+2!+3!...+50! when divided by 5!, the remainder is?

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(#M40026212) TCS question Keep an EYE

1!+2!+3!...+50! when divided by 5!, the remainder is?

Asked In TCS litun (10 years ago)

Unsolved Read Solution (8)

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Triloki Shaw  (10 years ago)

5!=120 now look at the sum it is clear that
after 5! all are divisible by 5!(120) so
take the sum of first four terms i.e.,
1!+2!+3!+4!=33
therefore rem=33.

Is this solution Helpfull? Yes (55) | No (1)  

Chandra Bhatt  (10 years ago)

33 will be the remainder.

Is this solution Helpfull? Yes (7) | No (1)  

SAI SARAN  (10 years ago)

33
no need of finding remainder after 5!
(1!+2!+3!+4!)%5!=33%120=33

Is this solution Helpfull? Yes (4) | No (1)  

manasa  (10 years ago)

remainder is 33

Is this solution Helpfull? Yes (4) | No   

Rishabh Prakash Sharma  (10 years ago)

5!/5! + 6!/5! + .... gives no remainders, so you only have to worry about the first 4 terms.

(1! + 2! + 3! +... 50!)/5!
= 1!/5! + 2!/5! + 3!/5! + 4!/5! + ... 50!/5!
= (1! + 2! + 3! + 4!)/5! + .... 50!/5!

(1! + 2! + 3! + 4!)/5!
= (1 + 2*1 + 3*2*1 + 4*3*2*1)/5!
= (1+2+6+24)/5!
= (33)/5!
= (33)/(5*4*3*2*1)
= (11)/(5*4*2*1)
= 11/(40)

Remainder 11

Is this solution Helpfull? Yes (3) | No (18)  

jyothi Am  (10 years ago)

33 how..?can anyone explain?

Is this solution Helpfull? Yes (1) | No (5)  

santanu  (10 years ago)

rem is 33%5=3

Is this solution Helpfull? Yes (1) | No (4)  

Guru Pranav  (10 years ago)

1!+2!+3!+4!=33
rem=33.

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