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Q.
The last digit of $(1 !+2 !+\ldots \ldots+2009 !)^{500}$ is
Permutations and Combinations
Solution:
As $1 !+2 !+3 !+4 !=33$ and last digit of $n !$, for $n \geq 5$ is 0 , last digit of $(1 !+2 !+\cdots+2009 !)^{500}$ is same as that of last digit of $3^{500}$
But $3^{500}=9^{250}=(10-1)^{250}$, and last digit of $(10-$ $1)^{250}$ is 1.