$\displaystyle 9^{100}-1$ is divisible by 10.

How can I do this problem? My book doesn't give adequate info on how to do it.

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- Jun 26th 2010, 05:34 PMdwsmithT or F: 9^{100}-1 is divisible by 10
$\displaystyle 9^{100}-1$ is divisible by 10.

How can I do this problem? My book doesn't give adequate info on how to do it. - Jun 26th 2010, 05:40 PMAckbeet
Off the top of my head, could you show it's divisible by 2? How about 5? If you can show both of those, you'd be done. The first one seems to me straightforward. Perhaps the second one is a bit trickier. Any ideas?

- Jun 26th 2010, 05:43 PMAckbeet
Or you could look at patterns. Does $\displaystyle 10|(9^{2}-1)$? How about $\displaystyle 9^{4}-1$? And $\displaystyle 9^{6}-1$?

- Jun 26th 2010, 05:44 PMundefined
Use Euler's theorem. Coincidentally, I just mentioned this in the other thread you started not long ago.

$\displaystyle \varphi(10)=4$

$\displaystyle \gcd(9,10)=1$

$\displaystyle 9^{100}-1\equiv 9^{25\cdot4}-1\equiv \left(9^4\right)^{25}-1\equiv1^{25}-1\equiv1-1\equiv0\ (\text{mod}\ 10)$ - Jun 27th 2010, 12:33 AMalexmahone
This is how I would do it:

$\displaystyle 9^{100}-1=(10-1)^{100}-1$

On expanding the RHS with the binomial theorem, we see that it is indeed divisible by 10. - Jun 27th 2010, 05:26 AMmelese
Using the the following rule: If $\displaystyle a\equiv b (mod\ n)$, then $\displaystyle a^k\equiv b^k(mod\ n)$.

Now, notice $\displaystyle 9\equiv -1(mod\ 10)$, then $\displaystyle 9^{100}\equiv (-1)^{100}\equiv 1(mod\ 10)$.

Then $\displaystyle 9^{100}-1\equiv 0 (mod\ 10)$.