Digit in the ten's place of an expression

Note that $25^2=625\equiv 25\bmod 100$, so that in fact $25^n\equiv 25\bmod 100$ for any $n$. Because $\phi(100)=40$, by Euler's theorem we have that $a^{40}\equiv1\bmod 100$ for any $a$ relatively prime to 100 (as 23 is). Thus $23^{41}\equiv 23\bmod 100$. Now put these results together to find $$23^{41}\cdot 25^{40}\bmod 100.$$