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?
 A: 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.
A: $(10-1)^{1500} = 10^{1500} - \binom{1500}{1}10^{1499}{1^1} + ... + \binom{1500}{1498}10^21^{1498} - \binom{1500}{1499}10^11^{1499} + 1^{1500}$
Only the last term is not divisible by hundred. Thus, $9^{1500} \equiv 1 \mod 100$ $i.e.$ the last two digits are 01.
A: HINT:
$$9^{2n}=(10-1)^{2n}=(1-10)^{2n}\equiv1-\binom{2n}110^1\pmod{100}\equiv1-20n$$
$\implies9^{2n}\equiv1\pmod{100}$ if $5|n$
Else $1-20n\equiv1+80n\pmod{100}$
A: $$\phi(100) = 40 \therefore 9^{1500} \equiv (9^{40})^{37}9^{20} \equiv 9^{20} \equiv \pm1, \pm51 \pmod{100}$$
as these all square to $1$ mod 100. To find which of the four is the right one: $$9^{1500} \equiv 81^{750} \equiv 1^{750}\equiv 1 \pmod{20} $$
So $01$.
A: $$9^2\equiv -19\pmod{100}$$
$$\implies 9^4\equiv 361\pmod{100}$$
$$\implies9^4\equiv 61\pmod{100}$$
$$\implies9^8\equiv 3721\pmod{100}$$
$$\implies 9^8\equiv21\pmod{100}$$
$$\implies9^8.9^{2}\equiv(21.81)\pmod{100}$$
$$\implies9^{10}\equiv1701\pmod{100}$$
$$\implies9^{10}\equiv 01\pmod{100}$$
$$\implies9^{1500}\equiv 01\pmod{100}$$
A: Since $\phi(100)=\phi(2^2\cdot5^2)=1\cdot2^1\cdot4\cdot5^1=40$ (where $\phi$ is Euler's Totient), we know that
$$
x^{40}\equiv1\pmod{100}
$$
Therefore, since $1500\equiv20\pmod{40}$
$$
9^{1500}\equiv9^{20}\pmod{100}
$$
and since $20=10100_\text{two}$, we can compute by squaring and multiplying
$$
\begin{align}
9^2&\equiv81\\
9^4&\equiv61\\
9^5&\equiv49\\
9^{10}&\equiv1\\
9^{20}&\equiv1\pmod{100}
\end{align}
$$
Thus, the last two digits of $9^{1500}$ are '$01$'.
A: Hint $\ 9\equiv -1\pmod{\!10}\,\Rightarrow\,9^{\large 10}\equiv (-1)^{\large 10}\,\pmod{\!10^{\large 2}}\ $
since  $\,\ \    a\,\equiv\ b\, \pmod{\!n}\ \Rightarrow\,\ a^{\large n}\,\equiv\  b^{\large n}\ \ \pmod{\!n^{\large 2}},\ $ [simple proof here]
