Please help <3

We can find a pattern in the exponents. Because we are only looking for units, let's find a pattern in 3s.

3^1, 3^2, 3^3, 3^4, 3^5

3, 9, 7, 1, 3

So every 4th multiple will have a units digit of 1. 120 is divisible by 4, meaning 123^120 has a units digit of 1, so 123^123 will have a units digit of 7.

thanks!

Too bad I can't use \equiv on begin{align*} enviorments.

Let $f(x)$ equal units digit of $x.$

$f(3^1) = 3$

$f(3^2) = 9 $

$\begin{align*} f(3^3) &= 27 \pmod{10} \\ &= 7 \pmod{10} \\ &= 7 \end{align*}$

$\begin{align*} f(3^4) &= 81 \pmod{10} \\ &= 1 \pmod{10} \\ &= 1 \end{align*}$

$\begin{align*} f(3^5) &= 243 \pmod{10} \\ &= 3 \pmod{10} \\ &= 3 \end{align*}$

Thus, given $3^n,$

$n \equiv 0 \pmod 4, f(n) = 1$

$n \equiv 1 \pmod 4, f(n) = 3$

$n \equiv 2 \pmod 4, f(n) = 9$

$n \equiv 3 \pmod 4, f(n) = 7$

$123 \equiv 3 \pmod 4, $ so $f(n) = \boxed{7}$

You could have also noted that the units digit is simply $\begin{align*} 3^3 &= 27 \pmod{10} \\ &= 7 \pmod{10} \\ &= \boxed{7} \end{align*}$