Calculate the last decimal digit of
111333 +333111
Have you tried exploring much smaller powers of each of those numbers?
111^1 = 111
111^2 = 12321
111^3 = 1367631
...
111^6 = 1870414552161
Seems relatively safe to say that 111^n (where n is an integer) ends in a 1. Makes sense, considering the ones digit is always going to be a product of previous 1s digits, which are all 1. No matter how many 1s you multiply together, you still end up with a 1s digit of 1.
Now, 333 is a little trickier, since multiplying a bunch of 3s together gets you a number that ends in:
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729
3^7 = 2187
3^8 = 6561
...
Check out the last digits there - they end in 3, 9, 7, 1, 3, 9, 7, 1 - looks like a pattern that repeats after each group of four numbers.
I'll leave it to you to determine, using the pattern, where 333^111 might fall.