First slide

Combinations

Question

The last digit of $(1! + 2! + \dots + 2005!)^{500}$ is

Moderate

A

9
B

2
C

7
D

1

Solution

$\begin{matrix} N = 1! + 2! + \dots + 2005! \\ = (1! + 2! + 3! + 4!) + (5! + \dots + 2005!) \end{matrix}$
= 33 + an integer having 0 in its unit's place
= an integer having 3 in its unit's place
Hence, N⁵⁰⁰ is an integer having 1 in its unit's place.

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