Latest TCS Aptitude Question SOLUTION: 1!+2!+3!...+50! Divided by 5! Remainder will be ?

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1!+2!+3!...+50! Divided by 5! Remainder will be ?

Asked In TCS IQBAL MEHDI (11 years ago)

Unsolved Read Solution (5)

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karpagam  (11 years ago)

Clearly 5! onwards each is divisible by 5!.
SO the remainder will come from the division (1!+2!+3!+4!) divided by 5!
Now (1!+2!+3!+4!) = 1+2+6+24 = 33 and 5! = 120 so the remainder is 33.

Is this solution Helpfull? Yes (22) | No (4)  

chandramani singh  (11 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)
=33/120=33

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rohan nimbalkar  (11 years ago)

Anyone can send me tcs (latest this year)placement papers urgent

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Mohamed Yasin  (11 years ago)

1!+2!+3!+4!=33 after this we get a zero in all the factorials. so the remainder is (no end with 0+ 33)/5 = 3

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snehil  (11 years ago)

only !1,!2,!3,!4 will not be divisible by 5 rest have 5 with them all will divisible by 5
so !1+!2+!3+!4=33
now 33%5=3
ans will 3

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