When integers are multiplied, the only digits that affect the units digit of a product are the units digits of the numbers being multiplied.
For instance, the units digit of 9 x 9 is the same as the units digit of 209 x 139. The units digits of both are 1.
So in this case, the only digits that matter are the 7 from 17 and the 3 from 1973.
So essentially the problem becomes this. What is the units digit of (7³)�, or 7¹², and what is the units digit of 3^3², or 3�?
The other thing you need to realize to do this problem is that when an integer is raised to consecutive powers, the resulting units digits repeat in a cyclical pattern.
So to get the answer here, we need the cyclical pattern for 7 and the cyclical pattern for 3.
7¹ = 7 (units digit 7)
7² = 49 (units digit 9)
7³ = _ _ 3 (units digit 3 - I am not going to bother multiplying out 7 x 49 since I only need the units digit. Since 7 x 9 = 63, I know the units digit of 7 x 49 is 3.)
7� = _ _ 1 units digit of 1. (Since 7 x 3 = 21 I know that the units digit of 7 x _ _ 3 is 1.)
Now that I am back to 1 as the units digit, when I multiply by 7 again, I will get a units digit of 7 and the pattern will repeat.
7 9 3 1 7 9 3 1 7 9 3 1
So 7¹² has a units digit of 1.
The pattern for 3 is the following.
3¹ = 3 (units digit 3)
3² = 9 (units digit 9)
3³ = 27 (units digit 7)
3� = 81 (units digit 1)
Then, since the last digit of 81 is 1, the pattern starts over, and so the pattern for 3 is the following.
3 9 7 1 3 9 7 1 3 9 7 1
Now since the units digit of 17¹² is 1 and the units digit of 1973� is 3, I was tempted to subtract 3 from 11 and get 8.
Then I figured out that 1973� > 17¹². So the answer is going to be a negative number and we have to reverse the subtraction to get the last digit.
So 3 - 1 = 2 and the correct answer is B.