Latest TCS Aptitude Question SOLUTION: when (1!)^1!+(2!)^2!+(3!)^3!+...............(100!)^100! is divided by 5,the remainder obtained is? a. 2 b. 0 c. 4 d. none of these

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(#M40039060) TCS Numerical Ability question Keep an EYE

when (1!)^1!+(2!)^2!+(3!)^3!+...............(100!)^100! is divided by 5,the remainder obtained is?

a. 2
b. 0
c. 4
d. none of these

Asked In TCS bhagwandas kushwaha (9 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

RAKESH  (9 years ago)

a) 2
(1!)^1!+(2!)^2!+(3!)^3!+...............(100!)^100! / 5

= 1 + 2^2 + 6^6 + 24^24 + [(5!)^5! + (6!)^6! + .....+ (100!)^100!] / 5

from (5!)^5! to 100!^100! each term is divisible by 5 & gives rem = 0

reqd rem = 1 + 2^2 + 6^6 + 24^24 / 5 => rem= 7/5 = 2
[unit digit of 1 + 2^2 + 6^6 + 24^24 = 1+4 +6+6 = 7]

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sarika  (9 years ago)

1 4 6^6 24^24 120^120....
from 120 the last digit is 0
for 6^6 last digit is 6 and for 24^24 odd powers will give 4 and even powers will give 6
so 1+4+6+6=17 gives remainder 2 when divided by 5
ans is 2

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Pranoy Bej  (9 years ago)

ANSWER==> d)none of this
as (1!)^1!+(2!)^2!+(3!)^3!+...............(100!)^100! = 1+4+6^6+....
the last digit of the summation will be 1. so the remainder will be 1

Is this solution Helpfull? Yes (2) | No (7)  

Rajadurai  (9 years ago)

the ans is (d)

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mithun  (9 years ago)

(5!+.......100!)/5, rem is 0.
then 1!^1!=1,2!^2!=4,3!^3!=6,4!^4!=6
then (1+4+6+6)/5
=17/5
=2
ans:a)2

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JAGAN  (9 years ago)

last digit will be zero
so it is divisible by 5
rem=0

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lakshmi narayanan  (9 years ago)

option b 0

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Pranoy Bej  (9 years ago)

@Rakesh: thanks for correcting me!

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Navdeep Singh  (9 years ago)

remainder 0 , since last digit is zero

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mohdzuhiab  (9 years ago)

after 5! oll the no. will be divisible by 5., as they contain 5 in every term., so add up 1!+2!..+4!/5 ans.(2)

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nagesh  (9 years ago)

plz send any of 1 tcs qustn pepr...

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pampanagouda  (9 years ago)

CAN ANY ONE SEND ME THE TCS PAPER TO MA MAIL ID IS pampanagouda032@gmail.com

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ARNAB MUKHERJEE  (6 years ago)

1!(mod 5) = 1 (mod 5)
2!(mod 5) = 2 (mod 5)
3!(mod 5) = 6 (mod 5) =1(mod 5)
4!(mod 5) = 24 (mod 5)= 4 (mod 5)
5!(mod 5) = 120 (mod 5)= 0(mod 5)
when n>=5,then mod value become 0.
hence, (1!+2!+3!+.......+100!) (mod 5) = (1+2+1+4+0+0+.....+0)(mod 5)
=8(mod 5) =3(mod 5)
so,required remainder is 3...
ANS:d. -----> none of these...

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rupam bumba  (5 years ago)

i do not under stand one thing.... why more then one people giving same explanation

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