Question

The digit in the unit's place of $7^{171} +(177)!$ is

• A. 3
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (C)

1

Answer 2

Given expression
$(7)^{171}+(177) !$
We know that the cycle of 7 is 4, i.e.,$(7)^{3}=1$
and in case of more the 4 number in factorial notation gives always 0 at unit place.
$\Rightarrow \left(7^{3}\right)^{57}+(177) !=1+0=1$