The last digit of (1 !+2 !+⋅s+2005 !)500 is

Question

The last digit of (1 !+2 !+⋅s+2005 !)500 is (A) 9 (B) 2 (C) 7 (D) 1

Tardigrade Admin · Accepted Answer

N =1 !+2 !+⋅s+2005 ! =(1 !+2 !+3 !+4 !)+(5 !+⋅s+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,

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

9

2

7

1

$N = 1! + 2! + \dots + 2005!$
$= (1! + 2! + 3! + 4!) + (5! + \dots + 2005!)$
$= 33 +$ 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,

Q. The last digit of (1!+2!+⋯+2005!)500 is

9

2

7

1

N=1!+2!+⋯+2005! =(1!+2!+3!+4!)+(5!+⋯+2005!) =33+ and 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,

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

$N = 1! + 2! + \dots + 2005!$
$= (1! + 2! + 3! + 4!) + (5! + \dots + 2005!)$
$= 33 +$ 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,