# 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.

What is the digit in the ten's place of ${23}^{41}\ast {25}^{40}$? How do you calculate this? The usual method for this kind of problem is using the Binomial theorem.
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DoryErrofbi
Note that ${25}^{2}=625\equiv 25mod100$, so that in fact ${25}^{n}\equiv 25mod100$ for any $n$. Because $\varphi \left(100\right)=40$, by Euler's theorem we have that ${a}^{40}\equiv 1mod100$ for any $a$ relatively prime to $100$ (as $23$ is). Thus ${23}^{41}\equiv 23mod100$. Now put these results together to find
${23}^{41}\cdot {25}^{40}mod100.$
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Deborah Proctor
On applying mod Distributive Law we get