Exercise 2.3.36 (Last two digits of $2^{1000}$, of $3^{1000}$)

Question

What are the last two digits, that is, the tens and unit digits, of $2^{1000}$? of $3^{1000}$?

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} First, $\phi(25) = 25\left( 1 - \frac{1}{5} ight) = 20$. By Euler's theorem,
$$2^{20} \equiv 1 \pmod {25}.$$
Therefore,  
$$x = 2^{1000} = (2^{20})^{50} \equiv 1 \pmod{25}.$$
Moreover, $2^2 \equiv 0 \pmod 4$, therefore
$$x = 2^{1000} = (2^2)^{500} \equiv 0 \pmod 4.$$
So $x$ is solution of the system of congruences
$$
\left\{
\begin{array}{ll}
x \equiv 1 \pmod {25},\
x \equiv 0 \pmod 4,
\end{array}
ight.
$$
equivalent to
$$
\left\{
\begin{array}{ll}
x \equiv 76 \pmod {25},\
x \equiv 76 \pmod 4,
\end{array}
ight.
$$
The solutions are $x = 76 + k \, 100,\ k\in \Z$, so $2^{1000} = 76 + k \, 100$. The last two digits of $2^{1000}$ are $76$.

\bigskip
Similarly, $3^{40} \equiv 1 \pmod{25},$ thus
$$y = 3^{1000} \equiv 1 \pmod{25}.$$
Moreover, $3^2 \equiv 1 \pmod 4$, thus
$$y = 3^{1000} \equiv 1 \pmod{4}.$$
So $y$ is solution of the system of congruences
$$
\left\{
\begin{array}{ll}
y \equiv 1 \pmod {25},\
y \equiv 1 \pmod 4.
\end{array}
ight.
$$
The solutions are $y = 1 + k 100$. Therefore the two last digits of $3^{1000}$ are $01$.
\end{proof}
\end{document}

Exercise 2.3.36 (Last two digits of $2^{1000}$, of $3^{1000}$)

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Exercise 2.3.36 (Last two digits of $2^{1000}$, of $3^{1000}$)

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