Question

N= 1!+2!+3!.....+10!. what is the last digit of N^N?

Answer 1

answer is 1523775454569

Answer 2

Answer:

Summing up the factorials gives,

N=1+2+6+24+120+720+5040+40320+362880+3628800=4037913

Now we have to find the last digit of N^N

N^N=(4037913)^4037913

for finding out the last digit, we can write only the last digit of

4037913 which is 3, hence

(3)^4037913

[(3)^(4*1009478)]*3 .using cyclicity of 3

1*3=3