Find the last 2 digits of $13^{204}$. Find the last $2$ digits of $13^{204}$. 
Please help, I don't understand where to start. 
 A: To find the last two digits of a number, look at it mod $100$ (in general, to find the last $k$ digits, look at it mod $10^k$).
With this, we find that:
\begin{align*}
13^{204}\mod 100 
\end{align*}
We can use Euler's theorem, which says:
$$a^{\varphi(k)}\equiv a\mod k$$
Where $\varphi(x)$ is Euler's Totient function.
We have that $\varphi(100) = \varphi(4)\varphi(25)$ as it's multiplicative on prime powers (meaning that $\varphi(p^kq^j) = \varphi(p^k)\varphi(q^j)$ for $p\neq q$ that are primes).
It's known that $\varphi(p^k) = p^k - p^{k-1}$, so we get that:
$$\varphi(100) = \varphi(4)\varphi(25) = \varphi(2^2)\varphi(5^2) = (4-2)(25-5) = 40$$
So, we know that:
$$a^{40}\equiv a\mod 100\implies a^{39}\equiv 1\mod 100$$
Now, we can write:
$$204 = 195 + 9 = 39\times 5 + 9$$
So, we have that:
$$13^{204}\mod 100 \equiv (13^{39}\mod 100)^5(13^9\mod 100)$$
So, we just need to compute $13^9\mod 100$.
Hopefully the numerator is now small enough that you can tackle this on your own.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&13^{204} = 28561^{51}\implies 13^{204}\
\overbrace{\mbox{Last Digit}}^{\ds{\mrm{LD}}}\,\,\,\,\, \mbox{is equal to}\  \color{#f00}{\large1}
\\[5mm]
&\mrm{LD}\pars{{28561^{51} - \color{#f00}{1} \over 10}} =
\mrm{LD}\pars{28560\sum_{n = 0}^{50}28561^{n} \over 10} =
\mrm{LD}\pars{2856 \times 51} = \color{#f00}{\large 6}
\end{align}

The last two digits are $\ds{\color{#f00}{\large 61}}$.

