Problem:
The units digit of 3100171002131003 is
Answer Choices:
A. 1
B. 3
C. 5
D. 7
E. 9
Solution:
Consider the first few powers of 3,7 and 13:
392781243749343….1….713169…7….1…..3
Clearly, the units digits in each case go through a cycle of length 4, with the units digit being 1 if the power is a multiple of 4. Let u(n) be the units digit of n. Since 4 divides 1000,
u(31001)u(71002)u(131003)=u(31)=3,=u(72)=9,=u(133)=7.
So u(3100171002131003)=u(3⋅9⋅7)=9.
(This solution can be expressed much more briefly using congruences.)
Alternate solution. Any power of either 7⋅13=91 or 34=81 has a units digit of 1. Thus 3100171002131003=3⋅13⋅81250911002 which clearly has a units digit of 9.