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Find the last $2$ digits of $13^{204}$. Please help, I don't understand where to start.

Chinnapparaj R
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Pragna
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  • Related : https://math.stackexchange.com/questions/607829/get-the-last-two-digits-of-16100-and-17100, https://math.stackexchange.com/questions/362012/find-the-last-two-digits-of-781, https://math.stackexchange.com/questions/1844558/how-to-find-last-two-digits-of-22016 https://math.stackexchange.com/questions/1337375/what-are-the-last-two-digits-of-7717 https://math.stackexchange.com/questions/1385396/last-two-digits-of-145532 https://math.stackexchange.com/questions/679842/find-the-last-two-digits-of-345 – lab bhattacharjee Jun 25 '17 at 06:05

2 Answers2

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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.

Mark Schultz-Wu
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{\color{#f00}{\large 61}}$.

Felix Marin
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