Exercise 2.1.29 (Last digit of $2^{400}$)

Question

What is the last digit in the ordinary decimal representation of $2^{400}$?

Richard Ganaye · Accepted Answer

\begin{proof} 
Note that $2^4 \equiv 1 \pmod 5$, thus $2^{400} = (2^4)^{100} \equiv 1 \pmod 5$.

Moreover $2^{400} \equiv 0 \pmod 2$.

Write $n = 2^{400}$. We know that 
\begin{align*}
&n \equiv 1 \pmod 5,\
&n\equiv 0 \pmod 2.
\end{align*}
Therefore
\begin{align*}
&n \equiv 6 \pmod 5,\
&n\equiv 6 \pmod 2.
\end{align*}
Since $5 \mid n-6$ and $2 \mid n-6$, where $5 \wedge 2 = 1$, we obtain $10 \mid n-6$, so
$$n \equiv 6 \pmod{10}.$$
The last digit of $2^{400}$ is $6$.

\bigskip
($2^{400} =  2582249878086908589655919172003011874329705792829223512830659356540647\

622016841194629645353280137831435903171972747493376$.

Thanks to ipython!
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Exercise 2.1.29 (Last digit of $2^{400}$)

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Exercise 2.1.29 (Last digit of $2^{400}$)

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