Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 6 down vote accepted

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

share|cite|improve this answer

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

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.

share|cite|improve this answer
More simply, $23^{41} 25^{40} = (23 \cdot 25)^{40} 23 = 575^{40} 23$. – Shai Covo Jun 20 '11 at 6:40
@shai you're right. When I saw the problem, somehow $24^2 =576$ came to my mind.. – kuch nahi Jun 20 '11 at 6:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.