Remainder when divisor is 11!
Number Theory
Level
3
Let n!=1x2x3x4x...xn for integer n>=1. If p =1!+(2+2!)+(3+3!)+...+(10+10!), then p+2 when divided by 11! leaves a remainder of?
The answer is 1.
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1 solution
P = r = 1 ∑ 1 0 r • r ! = r = 1 ∑ 1 0 ( ( r + 1 ) − 1 ) • r ! = r = 1 ∑ 1 0 ( r + 1 ) ! − r ! = 1 1 ! − 1 ! ( w h y ) Hence remainder when P + 2 = 1 1 ! + 1 is divided by 1 1 ! is obviously 1