Last 2 digits????

$100=2^2\cdot 5^2$ . $\,(-1)^{40}\equiv 1\pmod{\! 4}$ , so

$A:=1^{40}+2^{40}+3^{40}+\cdots+123^{40}\equiv 1+0+1+0+\cdots +0+1$

$\equiv 62\equiv 2\pmod{\! 4}\,\Rightarrow\, A=4k+2$ with $k\in\Bbb Z$ .

$a^{40}\equiv a^{40\pmod{\! 20}}\equiv a^0\equiv 1$ mod $25$ for $5\nmid a$ by Euler's theorem.

mod $25$ : $\ 4k+2\equiv 1+1+1+1+0+\cdots\equiv 99\equiv -1$

\iff 4k\equiv -3\equiv -28\stackrel{:4}\iff k\equiv -7\equiv 18

$\,\Rightarrow\, k=25m+18$ with $m\in\Bbb Z$ .

$A=4(25m+18)+2=100m+74$ .