First slide

Permutations

Question

The last digit of ${(1! + 2! + . . . . . . . + 2009!)}^{500}$ is

Moderate

A

1
B

2
C

7
D

9

Solution

As $1! + 2! + 3! + 4! = 33$ and last digit of n!, for
$n \geq 5$ is 0, last digit of $(1! + 2! + \dots + 2009!)^{500}$ 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.

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