Question

The digit in the unit's place of 7^171 + (177)! is
(a) 3
(b) 2
(c) 1
(d) 0

Answer 1

177! is going to have a bunch of zeros at its end, so we just need to figure out what the ones digit of 7^171 is. Notice that:
\[7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, 7^5 = 16807, \ldots \]
The ones place repeats after the 4th term. What we have is a sequence that goes like:
7, 9, 3, 1, 7, 9, 3, 1,...
This sequence repeats every 4 terms, so lets see what the 171th (the power) term is:
171/4 = 42 with a remainder of 3. That means going out 171 terms will cycle through the 4 terms 42 times, then stop on the 3rd term, which is 3. So 7^171 will end with a 3, and likewise:
\[7^{171}+(177!)\]
will also end in a 3.

Answer 2

Right boss

You are a genius