I've read this PDF where it explains how to find the last digit of a number.
If I were to find the last digit of $9^{1500}$ I would simply write it as $(3^{2})^{1500}$ and then use the patterns in the PDF for $3^{4n}$.
The problem here is that I'm asked to find the last $2$ digits. I think I could try to find patterns for the last 2 digits of $3^x$ or for $9^x$ but this would waste a lot of time, and since this problem was supposed to be solved by hand, I think this is not the best method. I'm also having problems to find literature about these problems of finding last digits of large exponents. Can somebody recommend be some?
Best Answer
$(10-1)^{1500} = 10^{1500} - \binom{1500}{1}10^{1499}{1^1} + ... + \binom{1500}{1498}10^21^{1498} - \binom{1500}{1499}10^11^{1499} + 1^{1500}$ Only the last term is not divisible by hundred. Thus, $9^{1500} \equiv 1 \mod 100$ $i.e.$ the last two digits are 01.