Latest TCS Aptitude Question SOLUTION: what will be the remainder if 1!+2!+3!+..+100! is divided by 7

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what will be the remainder if 1!+2!+3!+..+100! is divided by 7

Asked In TCS Shishir kumar (11 years ago)

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Sourangsu Bandyopadhyay  (11 years ago)

7! onwards all terms are divisible by 7 as 7 is one of the factor. So there is no remainder left for those terms i.e. remainder left after dividing 7! + 8! + 9! + ... +100! is 0.
The only part to be consider is
= 1! + 2! + 3! + 4! + 5! + 6!
= 1 + 2 + 6 + 24 + 120 + 720
= 873
The remainder left after dividing 873 by 7 is 5
Hence, the remainder is 5.

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Sumit Bhattacharyya  (11 years ago)

7! onward terms are divisible by 7...
Summation till 6!=873
hence remainder = 873%7 = 5

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Arindam Dhara  (11 years ago)

all terms from 7 will have the factor 7...so we consider the 1st 6 terms: 1+2+6+24+120+720 thats 873
which leaves remainder 5 when divided by 7

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sami wardhan karn  (11 years ago)

after 6! all terms divisible by 7.so we hve to cnsdr 1! +2!+3!+4!+5!+6!=873%7=5

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Raviranjan Pandey  (11 years ago)

(1!+3!)+(2!+4!)+(5!+6!)+(7!+8!+9!+..........+100!)
=(7)+(26)+5!(1+6)+7!(1+8+8.9+............+8.9.10......100)...WE FIND THAT EXCEPT 26 THE SECOND TERM, ALL THE OTHER TERMS ARE DIVISIBLE BY 7.HENCE THE REMAINDER WE GET ON DIVIDING WITH 7 IS THE REMAINDER WE GET WHEN WE DO 26/7...THAT IS ...5 IS THE REMAINDER .

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Aditya Kushwaha  (11 years ago)

rearranging the series we get.
(1!+3!)+(2!+4!)+(5!+6!)+(7!+8!+9!+..........+100!)
=(7)+(26)+5!(1+6)+7!(1+8+8.9+............+8.9.10......100)...WE FIND THAT EXCEPT 26 THE SECOND TERM, ALL THE OTHER TERMS ARE DIVISIBLE BY 7.HENCE THE REMAINDER WE GET ON DIVIDING WITH 7 IS THE REMAINDER WE GET WHEN WE DO 26/7...THAT IS ...5 IS THE REMAINDER .

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Shashi nath jha  (11 years ago)

after 7! we don't need to calculate as remainder is 0.
720+120+24+6+2+1/7=5 as remainder...

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poonam  (11 years ago)

1+2+6+24+120+720 ;;
873/7 =>
remainder: 5

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Guru Pranav  (11 years ago)

remainder will be 5.

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