T or F: 9^{100}-1 is divisible by 10

**dwsmith** · June 26th 2010, 05:34 PM

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

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

**Ackbeet** · June 26th 2010, 05:40 PM

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?

**Ackbeet** · June 26th 2010, 05:43 PM

Or you could look at patterns. Does $10|(9^{2}-1)$ ? How about $9^{4}-1$ ? And $9^{6}-1$ ?

**undefined** · June 26th 2010, 05:44 PM

Originally Posted by dwsmith

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

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

Use Euler's theorem. Coincidentally, I just mentioned this in the other thread you started not long ago.

$\varphi(10)=4$

$\gcd(9,10)=1$

$9^{100}-1\equiv 9^{25\cdot4}-1\equiv \left(9^4\right)^{25}-1\equiv1^{25}-1\equiv1-1\equiv0\ (\text{mod}\ 10)$

**alexmahone** · June 27th 2010, 12:33 AM

This is how I would do it:

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

On expanding the RHS with the binomial theorem, we see that it is indeed divisible by 10.

**melese** · June 27th 2010, 05:26 AM

Originally Posted by dwsmith

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

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

Using the the following rule: If $a\equiv b (mod\ n)$ , then $a^k\equiv b^k(mod\ n)$ .

Now, notice $9\equiv -1(mod\ 10)$ , then $9^{100}\equiv (-1)^{100}\equiv 1(mod\ 10)$ .
Then $9^{100}-1\equiv 0 (mod\ 10)$ .

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