modactorial !!!!

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What is the value of

( 1 ! + 2 ! + 3 ! . . . . . . . . . . + n ! ) m o d 5 (1!+2!+3!..........+n!)\mod5 for n > 4 ? n>4?


The answer is 3.

Rishabh Jain by Rishabh Jain , 23, India

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1 solution

Milind Prabhu
Sep 14, 2014

Just have to find ( 1 ! + 2 ! + 3 ! + 4 ! ) m o d 5 (1!+2!+3!+4!) mod 5 because all the terms after that will be divisible by 5.

This gives 33 33 m o d mod 5 5 = 3 = \boxed { 3 }

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