$3^{123} mod 100$

The OP began by looking for a pattern but stated that

...despite much painful inspection did not yield an obvious useful pattern.

You can use some light theory to actually predict the form and structure of the pattern.

Observe that if $a in {0,2,4,6,8}$ and $b in {1,3,7,9}$ and

$quad 3 times (10 a + b) equiv 10 ,a' + b' pmod{100} text{ with } a',b' in {0,1,2,3,4,5,6,7,8,9}$

then in fact $a' in {0,2,4,6,8}$ and $b' in {1,3,7,9}$.

This is our main (theoretical) pattern and

$quad 3^1 equiv 03 pmod{100}$
$quad 3^2 equiv 09 pmod{100}$
$quad 3^3 equiv 27 pmod{100}$
$quad 3^4 equiv 81 pmod{100}$
$quadtext{-------------------------}$
$quad 3^5 equiv 43 pmod{100}$

It easy to verify that the units digit will move

$quad 3 mapsto 9 mapsto 7 mapsto 1$

inside each of these four cycles.

Considering that $3$ is a unit, we can argue that one of these $4$-cycles will end on

$$quad 01 quad text{the multiplicative identify}$$

and that no repetition is possible until the identify is reached.

Since the tens digit can only cycle over the set ${0,2,4,6,8}$, there are at most five of these $4$-cycles that have to be calculated.

Calculating the $2^{nd}$ $4$-cycle:

$quad 3^5 equiv 43 pmod{100}$
$quad 3^6 equiv 29 pmod{100}$
$quad 3^7 equiv 87 pmod{100}$
$quad 3^8 equiv 61 pmod{100}$
$quadtext{-------------------------}$

Calculating the $3^{rd}$ $4$-cycle:

$quad 3^9 equiv 83 pmod{100}$
$quad 3^{10} equiv 49 pmod{100}$
$quad 3^{11} equiv 47 pmod{100}$
$quad 3^{12} equiv 41 pmod{100}$
$quadtext{-------------------------}$

Calculating the $4^{th}$ $4$-cycle:

$quad 3^{13} equiv 23 pmod{100}$
$quad 3^{14} equiv 69 pmod{100}$
$quad 3^{15} equiv 07 pmod{100}$
$quad 3^{16} equiv 21 pmod{100}$
$quadtext{-------------------------}$

At this point we really don't have to calculate the $5^{th}$ $4$-cycle since we know it has to be the last one.

We can now use the fact that

$tag 1 3^{20} equiv 1 pmod{100}$

and work out the remaining details for the OP's question.

You are indeed correct. There is, however, one minor improvement. Using the Carmichael function, you can argue that a smaller power of $3$, namely $3^{lambda(100)}=3^{20}equiv 1bmod 100$. The Carmichael function of divides half the Euler totient function when the argument is even and the Euler totient is a multiple of $4$, which is true for $lambda(100)$; thus $3^{20}$ can replace $3^{40}$ in the argument.

At a more elementary level, you can render $3^4=80+1$ and raise both sides to the fifth power, thus $3^{20}equiv1bmod 100$ as the Binomial Theorem for $(80+1)^5$ gives multiples of $100$ plus $1$.

Correct, an alternative solution:

$$ begin{align} 3^{123}&=left(3^{2}right)^{61}cdot 3\ &=left(10-1right)^{61}cdot 3\ &equivleft(binom{61}{1}10^{1}left(-1right)^{60}-1right)cdot 3 &mod{100}\ &equiv 27 &mod100 end{align} $$