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(#M40012999) TCS question Keep an EYE

(1!^1!+2!^2!+3!^3!+4!^4!+......+50!^50!)/5
The remainder for the above is--------?

Asked In TCS Madhu (11 years ago)

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Pallavi Barjatiya  (11 years ago)

remainder for above is 2.
no. ahead 5!^5 will be completely divisable by 5.
rest 1!*1+2!^2+3!^3+4!^4 = 331997
when divided by 5 gives remailder 2. ans :)

Is this solution Helpfull? Yes (24) | No (3)  

Prashant  (11 years ago)

5! ^ 5! and above terms will be divisible by 5.
so take the terms before that.
1 + 2^2 + 6^6 + 24^24 /5 (check for the remainders , dont calculate this.)
4+1+1+1/5 (1 leaves 4 as remainder , 2^2 leaves 1 , 6 wil always giv 1 , 24^24 wil also give 1 as 4 as unit digits in even powers as 6)
7/5 leaves 2 as remainder.

Is this solution Helpfull? Yes (11) | No (3)  

kalpana srivastava  (11 years ago)

the anwser is 2

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manjushree hossur  (11 years ago)

@Mr.prasanth, 1 leaves 1 as remainder not 4

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saurabh yadav  (11 years ago)

ater 5!^5! the remainder will be 0 so,
(1^1!+2^2!+6^3!+24^4!)/5
no calculate for rtemainder as for
1st term remainder=1
2nd term remainder=-1
3rd term remainder=1
4th term remainder=-1^(4!)=1
so answer =2

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Swarnajit Adhikary  (10 years ago)

after 5!^5! the remainder will be 0 so,
(1^1!+2^2!+6^3!+24^4!)/5=331997/5
so remainder =2

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