Just how many digits does this number have anyway? When asking that question, the tool to use is the log function. Typing log(3^2153) would hypothetically give you the answer (you'll have to round up to the nearest whole number) but even that is impossibly big for the calculator. I can use a property of logarithms to rewrite it as 2153* log (3) which is calculable: . Practically, this means that the number 3*2153 has 1028 digits in it, and I'm interested in just the last one. Interestingly, this subquestion of how many digits a huge number has been able to be answered for far longer than we have had calculators, by using log tables, but that's an entirely different story.
Back to the question at hand. If we aren't going to just cheat and look at all the digits of a supercalculator, how can we answer the question? Someone always suggests just multiplying it out by hand. As ridiculous as that would be (simply multiplying the last step would probably take 5 minutes, and that's after getting there in the first place!) the approach actually reveals something very important that can apply, so lets try it.
Notice, this cycle of numbers repeats every four digits. Specifically, every multiple of four (3^4, 3^8, and 3^12) ends in a 1. The closest power of four to 2153 is 2152, and so:
I encourage you to try to answer the rest of these questions, which can all be answered using similar techniques:
Last digit of 2^2153
Last digit of 4^2153
Last digit of 5^2153
Last digit of 6^2153
Last digit of 7^2153
Last digit of 8^2153
Last digit of 9^2153