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Sum of the last 2 digit in sum(r=0)^(101...

Sum of the last `2` digit in `sum_(r=0)^(101)(101-r)`

A

`4`

B

`8`

C

`0`

D

`5`

Text Solution

Verified by Experts

The correct Answer is:
D
Sum of the last ……….
`sum_(t=0)^(101)(101-r)`!
`= 101!+100!+99!+….+7!+6!+5!+4!+3!+2!+1!+0!`
We observe that `10!, 11!, 12!.......101!` each have more than 2 zero in the end hence no contribution in the last 2 digits of resulting "sum"
`:.` last 2 digits in `sum_(r=0)^(101)(101-r)!`
= last 2 digits in `9!+8!+7!+6!+5!+4!+3!+2!+1!+0!`
`= 362880+40320+5040+720+120+24+6+2+1+1 = 409114`
`:` Last 2 digits `= 14 , "sum" = 5`
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