The last digit of (1 !+2 !+ ldots ldots+2009 !)500 is

Question

The last digit of (1 !+2 !+ ldots ldots+2009 !)500 is (A) 1 (B) 2 (C) 7 (D) 9

Tardigrade Admin · Accepted Answer

As 1 !+2 !+3 !+4 !=33 and last digit of n !, for n ≥ 5 is 0 , last digit of (1 !+2 !+⋅s+2009 !)500 is same as that of last digit of 3500 But 3500=9250=(10-1)250, and last digit of (10- 1)250 is 1.

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

1

2

7

9

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.

Q. The last digit of (1!+2!+……+2009!)500 is

1

2

7

9

As 1!+2!+3!+4!=33 and last digit of n!, for n≥5 is 0 , last digit of (1!+2!+⋯+2009!)500 is same as that of last digit of 3500 But 3500=9250=(10−1)250, and last digit of (10− 1)250 is 1.

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

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.