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Let N=(1!)^(3)+(2!)^(3)+(4!)^(3)+(8!)^(3...

Let `N=(1!)^(3)+(2!)^(3)+(4!)^(3)+(8!)^(3)+….` upto 20 terms, then

A

digit at unit's place is 3

B

digit at thousand's place is 5.

C

digit at ten's place is 3

D

digit at thousand's place is 6

Text Solution

Verified by Experts

AC
`N=(1!)^(3)+(2!)^(3)+(4!)^(3)+…..`
`N=1+2^(3)+24^(3)+…….`
=1+8+13824+(1000x some natural number )
`:.` unit digit is 3
tens digit is 3
thousands digit is 8.
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