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QUESTION
Use Euler's theorem to find
\begin{description}
\item[(i)]
the units digit of $3^{250}$.
\item[(ii)]
the last two digits of $3^{250}$.
\end{description}
ANSWER
The unit digit of an integer is given by its congruence class mod
10, and the final two digits by its congruence class mod 100.
Since 10=2.5 and $100=2^2.5^5$ we have
$\phi(10)=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=4$
and
$\phi(100)=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=40$.
Since gcd(3,10)=gcd(3,100)=1, we may use Eulers theorem to
evaluate (i) $3^{250}$ mod 10 and (ii)$3^{250}$ mod 100. (of
course we could just do (ii) and deduce (i) from it- but doing
both is useful for practice!)
\begin{description}
\item[(i)]
$3^{250}\equiv(3^4)^{62}.3^2\equiv1^{62}.3^2\equiv9$ mod 10. Thus
the unit digit of $3^{250}$ is 9.
\item[(ii)]
$3^{250}\equiv(3^{40})^6.3^{10}\equiv1^6.3^{10}\equiv 3^{10}$ mod
100. Now $3^4\equiv81\equiv-19$ mod 100, so
$3^{10}\equiv(-19).(-19).9\equiv19.171\equiv19.-29\equiv-551\equiv49$
mod 100. Thus the last two digits of $3^{250}$ are 49.
[Comment: The above solution evaluates $3^{10}$ mod 100 by direct
calculation. You might like to investigate the following rather
quicker procedure:-
(a) Calculate $3^{10}$ mod 4. (b) Calculate $3^{10}$ mod 25. (c)
Use the Chinese Remainder theorem to find a unique simultaneous
solution of the congruences (a) and (b) mod 100.]
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