Is the number $3^{444}+4^{333}$ divisible by 5?

## List 5
There is a pattern for the unit digit of a power of 3: it has a period of 4, as it repeats every 4 times.

$$
\begin{aligned}
& 3 \\
& 3^{2}=9 \\
& 3^{3}=27 \\
& 3^{4}=81 \\
& 3^{5}=243 \\
& 3^{6}=\ldots 9 \\
& 3^{7}=\ldots 7 \\
& 3^{8}=\ldots 1
\end{aligned}
$$

Since 444 is a multiple of 4, the unit digit of $3^{444}$ is 1.

Similarly, the unit digit of powers of 4 has a period of 2. Indeed, we have:

$$
\begin{array}{ll}
4^{1}=4 & ; \quad 4^{3}=64 \\
4^{2}=16 \quad & ; \quad 4^{4}=256
\end{array}
$$

Since 333 is odd, the unit digit of $4^{333}$ is 4. Therefore, the unit digit of $3^{444}+4^{333}$ is $1+4=5$, and thus it is divisible by 5.

REMEMBER: Numbers divisible by 5 end in 0 or 5.

## List 5