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 Shivendra Bhardwaj (11 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

Amar  (11 years ago)

1!+2!+3!...+50! divided by 5!
remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.

so remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120).

so remainder is 33.

Is this solution Helpfull? Yes (44) | No (7)  

RAVI KIRAN  (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 (19) | No (2)  

Garima  (11 years ago)

33 is correct.

when 33 is divided by 5! ( 120) , u will get 33 as remainder.

@ Vivek,
I think u r considering the case when 33 is divided by 5 .

Is this solution Helpfull? Yes (6) | No (2)  

vivek chauhan  (11 years ago)

@ amar 33 is not possible dude it will be 3.

Is this solution Helpfull? Yes (4) | No (30)  

Manan Saxena  (11 years ago)

answer will be zero.........as the total for this sum is 3.1035x10^64 which is easily divisible by 5

Is this solution Helpfull? Yes (3) | No (15)  

Akshay Prakash  (11 years ago)

till 4! last digit will be non zero.
after 4! all terms will have last digit as 0 so it doesn't matter how many terms are there last digit is always zero and sum is divisible by 5.
Now, if we see 1!+...+4!=33
Therefore last digit of 1!+......+50! will be 3
hence remainder is 3.

Is this solution Helpfull? Yes (2) | No (1)  

Amish Choudhary  (11 years ago)

let us see it in simple way here we are taking 33/120 i.e. (1!+2!+3!+4!)/5!
now, take an example divide(12/6) is cancelled by (2/1) similarly when we cancel 5 with any digit it get cancelled if digit is greater than 5.

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

abhishek mishra  (11 years ago)

1!+2!+3!+.............+50!
1! is the smallest and 50! is the highest
now the half of 50! is 25! and if we increase 1! in 50! it will give 51!
Now 25! * 51!=1275!
So,1!+2!+3!+...........+50!=1275!
now divide 1275! with 5!=255 and 1275 is completely divisible by 5 .
Hence, (1!+2!+3!+.................+50!)/5 then the remainder will be zero because it is completely divisible by 5.

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

Navin  (7 years ago)

1!+2!+3!...+50! divided by 5!
remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.



so remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120).

so remainder is 33

can anyone explain me this what to divide with whom?

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SHANU MAHAJAN  (11 years ago)

pls can anyone explain me the complete solution. concept behind taking only 1! to 4!

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Nikhil kumar jha  (8 years ago)

1!+2!+3!...+50! divided by 5!
remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.

so remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120).

so remainder is 33

Is this solution Helpfull? Yes | No   

neha upadhyay  (7 years ago)

The remainder when the terms greater than 5! are divided by 5! becomes 0 so we need to consider the terms upto 4!.
So remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.
So remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120)
So remainder is 33.

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Bindu sree  (6 years ago)

Reminder 3

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AMIT KUMAR  (4 years ago)

1! + 2! +3! +4! +5! ......+50!
after 5! eery term will be divisible by 5! beacuse 5! is the factor
now , 1 + 2+6+24=33
33 will be the reminder

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