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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.

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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 $$23^{41}\cdot 25^{40}\bmod 100.$$

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$$(24+1)^{40} \times (24-1)^{41}$$ $$= 23\times (24^2-1)^{40}$$ $$=23\times (575)^{40}$$

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.

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    $\begingroup$ More simply, $23^{41} 25^{40} = (23 \cdot 25)^{40} 23 = 575^{40} 23$. $\endgroup$ – Shai Covo Jun 20 '11 at 6:40
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    $\begingroup$ @shai you're right. When I saw the problem, somehow $24^2 =576$ came to my mind.. $\endgroup$ – kuch nahi Jun 20 '11 at 6:48
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On applying $\,\ ab\,\bmod\, ac\, =\ a\,(b\bmod c)\, =\, $ mod Distributive Law we get

$\, 25^{\large 1+J} 23^{\large 1+2K}\bmod 100\, =\, 25\,(\!\!\!\!\underbrace{25^{\large J} 23^{\large 1+2K}}_{\Large \equiv\ 1^{\Large J}(-1)^{\Large 1+2K} \equiv\ \color{#c00}3\!\!\!\!\!}\!\!\!\!\!\bmod 4)\, =\, 25(\color{#c00}3)$

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