the unit digit for $3^{100}\cdot 37^{98}$ What is the best  way to know the unit digit for $3^{100}\cdot 37^{98}$
 A: First find the unit digit pattern of numbers being multiplied by $3$, starting with $3$.
$$ 3, 9, 7, 1, 3, 9, ... $$
Now do the same for $7$
$$ 7, 9, 3, 1, 7, 9, ... $$
Now find the $100^{th}$ and $98^{th}$ element respectively of each of those sequences and multiply them together.
A: We essentially need $3^{100}\cdot37^{98}\pmod{10}$
$3^4=81\equiv1\pmod{10}$ or
using Euler's theorem,  $3^4\equiv1\pmod{10}$ as $\phi(10)=\phi(2)\phi(5)=4$
$\implies 3^{4n}\equiv1$
Also, $37\equiv7\pmod{10}\implies 37^{98}\equiv7^{98}\pmod{10}$
Again, using Fermat's Little Theorem  $7^4\equiv1\pmod{10}\implies 7^{4n}\equiv1$
So,  $3^{100}\cdot37^{98}=(3^4)^{25}\cdot(7^4)^{24}\cdot7^2\equiv1\cdot1\cdot9\pmod{10}$
A: Hint $\rm\ mod\ 10\!:\,\ 37\equiv 7\equiv 3^{-1}\!\Rightarrow\: 3^{100}37^{98}\!\equiv 3^{100} 3^{-98}\!\equiv 3^2$
A: $$
\begin{align}
37 * 3 & \equiv 1 \text{ (mod 10)}\\
(37 * 3)^n & \equiv 1 \text{ (mod 10)}\\
(37 * 3)^{98} & \equiv 1 \text{ (mod 10)}\\
(37 * 3)^{98} * 3^2 & \equiv 1 * 3^2 \text{ (mod 10)}\\
or, 37^{98} * 3^{100} & \equiv 9 \text{ (mod 10)}\\
\end{align}
$$
