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Sum of the last `2` digit in `sum_(r=0)^(101)(101-r)`A. `4`B. `8`C. `0`D. `5` |
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Answer» Correct Answer - 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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