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Q.
The last digit of $(1 !+2 !+\cdots+2005 !)^{500}$ is
Permutations and Combinations
Solution:
$N =1 !+2 !+\cdots+2005 !$
$=(1 !+2 !+3 !+4 !)+(5 !+\cdots+2005 !)$
$=33+$ and integer having $0$ in its unit's place
$=$ an integer having $3$ in its unit's place
Hence, $N ^{500}$ is an integer having 1 in its unit's place,