Digit in the ten's place of an expression What is the digit in the ten's place of $23^{41}* 25^{40}$ ? How do you calculate this? The usual method for this kind of problem is using the Binomial theorem, but I couldn't solve it.
 A: 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 
$$23^{41}\cdot 25^{40}\bmod 100.$$
A: $$(24+1)^{40} \times (24-1)^{41}$$ $$= 23\times (24^2-1)^{40}$$ $$=23\times (575)^{40}$$
You can extract a pattern for powers of $5$. The last digits will always be $5$ and for this case $575^n$ the last two digits will alternate between $25$ (even n) and $75$ for odd n.
So $$\cdots 25$$ $$\underline{\qquad \times 23}$$  $$\quad \cdots 75$$ $$\underline{\;\cdots 50\times}$$ $$=\cdots 75$$
I have tried to emulate the multiplication procedure to show it.
A: On applying $\,\ ab\,\bmod\, ac\, =\ a\,(b\bmod c)\, =\, $ mod Distributive Law we get
$\, 25^{\large 1+J} 23^{\large 1+2K}\bmod 100\, =\, 25\,(\!\!\!\!\underbrace{25^{\large J} 23^{\large 1+2K}}_{\Large \equiv\ 1^{\Large J}(-1)^{\Large 1+2K} \equiv\ \color{#c00}3\!\!\!\!\!}\!\!\!\!\!\bmod 4)\, =\, 25(\color{#c00}3)$
