Looking at the question. I misread it initially. This question actually would never be on the test, but in case you’re wondering it’s going off the pattern of 3^x.
Knowing that the unit digit repeats every 4 powers. So if you can take X/4 and turn it into x/4 = 4n + k then you can solve the question by figuring out what k is since it will be between 0-3, inclusively.
In this situation, x= 7^11
So (7^11/4) is what we need to put into 4n+k format. To do this it’s easier to view 7 as (8-1). To then break this down further, you need to understand the concept of negative remainders… which will never show up on the actual exam.
Originally I thought the question was ((3^7)^11) which would be 3^77
This is a question that could be tested since it’s quite easy to see how to turn 77/4 into 4n + k
It would be 4(19) + 1
So you would look for the unit digit of 3^1 and then see what the remainder would be if dividing by 5