Calculate the last 2 digits of the sum : S=1!+2!+3!+...+2015! (n!=123...n) (?)

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Answer 1 · 2017-06-17T23:26:33

#13#

#1!+2!+3!+4!+5!+6!+7!+8!+9!#

#=1+2+6+24+120+720+5040+40320+362880#

#=409113#

Any larger factorial will have #2# trailing #0#'s, so will not affect the last two digits #13#

Calculate the last 2 digits of the sum : S=1!+2!+3!+...+2015! (n!=1*2*3*...*n) (?)