elementary number theory - Last two digits of $9^{1500}$

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?

Answer

Calculating the last two digits is like taking the number $\mod 100$.

$$9 \equiv 9\mod 100\\

9^2 \equiv 81\mod 100\\
9^3=729\equiv 29\mod 100\\
9^4=9\cdot 9^3 \equiv 9\cdot 29 = 261\equiv 61\mod 100\\
9^5\equiv 9\cdot 61 = 549\equiv 49\mod 100\\
9^6 \equiv 9\cdot 49 = 441\equiv 41\mod 100\\
9^7\equiv 9\cdot 41 = 369\equiv 69\mod 100\\
9^8\equiv 9\cdot 69 = 721\equiv 21\mod 100\\
9^9\equiv 9\cdot 21 = 189\equiv 89\mod 100\\
9^{10}\equiv 9\cdot 89 = 801\equiv 1\mod 100\\
$$

Now, knowing that $9^{10}\equiv 1\mod 100$, your remainging task should be easier.