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

• Jun 26th 2010, 05:34 PM
dwsmith
T or F: 9^{100}-1 is divisible by 10
$9^{100}-1$ 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 PM
Ackbeet
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 PM
Ackbeet
Or you could look at patterns. Does $10|(9^{2}-1)$? How about $9^{4}-1$? And $9^{6}-1$?
• Jun 26th 2010, 05:44 PM
undefined
Quote:

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.

$\varphi(10)=4$

$\gcd(9,10)=1$

$9^{100}-1\equiv 9^{25\cdot4}-1\equiv \left(9^4\right)^{25}-1\equiv1^{25}-1\equiv1-1\equiv0\ (\text{mod}\ 10)$
• Jun 27th 2010, 12:33 AM
alexmahone
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.
• Jun 27th 2010, 05:26 AM
melese
Quote:

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