A) 3
B) 2
C) 1
D) 0
Correct Answer: C
Solution :
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. \[\Rightarrow \] \[{{({{7}^{3}})}^{57}}+(177)!=1+0=1\]You need to login to perform this action.
You will be redirected in
3 sec