Show that 7^n - 1 is divisible by 6 for any positive integer n.
I'll post the solution on Wednesday.
Simple induction... is this a homework assignment question?
(http://latex.sidoh.org/?render64=JCQgJFxcKg0KXGxlZnQoN15uLTFccmlnaHQpXG1vZHs2fT0wIFxcDQo3Xm5cZXF1aXYgMVxsZWZ0KFxtb2R7Nn1ccmlnaHQpIFxcDQooN14xLTE9NilcZXF1aXYgMChcbW9kezZ9KSQNCg0KJFxcJEFzc3VtZTogJDdeay0xXGVxdWl2IDAoXG1vZHs2fSlcXA0KN157aysxfS0xPTdcbGVmdCg3XmstMSkrNiQNCg0KJFxcXFwkU2luY2UgJDdeay0xIFxlcXVpdiAwKFxtb2R7Nn0pJCBpcyB0cnVlIGJ5IHRoZSBpbmR1Y3Rpb24gaHlwb3RoZXNpcyBhbmQgNiBpcyBkaXZpc2libGUgYnkgNiwgdGhlIHdob2xlIHN0YXRlbWVudCAkN15uLTEgXGVxdWl2IDAoXG1vZHs2fSkkIGlzIHRydWUu)
Quote from: Sidoh on October 24, 2006, 12:24:43 AM
(http://latex.sidoh.org/?render64=JCQgJFxcKg0KXGxlZnQoN15uLTFccmlnaHQpXG1vZHs2fT0wIFxcDQo3Xm5cZXF1aXYgMVxsZWZ0KFxtb2R7Nn1ccmlnaHQpIFxcDQooN14xLTE9NilcZXF1aXYgMChcbW9kezZ9KSQNCg0KJFxcJEFzc3VtZTogJDdeay0xXGVxdWl2IDAoXG1vZHs2fSlcXA0KN157aysxfS0xPTdcbGVmdCg3XmstMSkrNiQNCg0KJFxcXFwkU2luY2UgJDdeay0xIFxlcXVpdiAwKFxtb2R7Nn0pJCBpcyB0cnVlIGJ5IHRoZSBpbmR1Y3Rpb24gaHlwb3RoZXNpcyBhbmQgNiBpcyBkaXZpc2libGUgYnkgNiwgdGhlIHdob2xlIHN0YXRlbWVudCAkN15uLTEgXGVxdWl2IDAoXG1vZHs2fSkkIGlzIHRydWUu)
You need to specify that you are assuming 7
k-1 is divisible by 6 for some k>=1, then you have shown the formula holds for all n>=1.
A picky detail, but one that is notable in rigorous math courses.
(http://latex.sidoh.org/?render=6%20%7C%20%287%5Ek%20-1%29)
(http://latex.sidoh.org/?render=%286%20%2B%201%29%20%7C%207%5Ek)
(http://latex.sidoh.org/?render=7%20%7C%207%5Ek)
Done.
Quote from: Rule on October 24, 2006, 04:58:55 PM
You need to specify that you are assuming 7k-1 is divisible by 6 for some k>=1, then you have shown the formula holds for all n>=1.
A picky detail, but one that is notable in rigorous math courses.
Oops, yeah. I should have said "Assume 7
k-1 is divisible by 6 for some arbitrary integer greater than one,
k," right?
Another way to do this, which does not require knowledge of modular arithmetic or number theory, is using the polynomial combinatorics whatever theorem:
(http://latex.sidoh.org/?render=7%5E%7Bn%7D%20%3D%20%286%20%2B%201%29%5E%7Bn%7D%20-%201%0D%0A%0D%0A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7Dn%20%5C%5C%200%5Cend%7Barray%7D%5Cright%29%206%5E%7Bn%7D1%5E%7B0%7D%20%5Ctimes%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7Dn%20%5C%5C%201%5Cend%7Barray%7D%5Cright%29%206%5E%7Bn-1%7D1%5E%7B1%7D%20%5Ctimes%20...%20%5Ctimes%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7Dn%20%5C%5C%20n%5Cend%7Barray%7D%5Cright%29%206%5E%7B0%7D1%5E%7Bn%7D%20-%201%0D%0A%0D%0A%0D%0A%0D%0A)
While that may not require knowledge of modular arithmetic and number theory, it does require either (A) set theory (I think) or (B) basic discrete concepts.
2 + 2 requires knowledge of set theory =p
but yeah it requires knowledge of basic combinatorics
Which is Discrete, vis a vis same level as NT :P
Discrete math is much more common to high school math curricula* than number theory is, but yeah, there's no restriction to what math you can use in these problems.
* Grammar Policed on behalf of Sidoh's (pedantic) request
Quote from: Ender on November 01, 2006, 07:36:26 PM
Discrete math is much more common to high school math curricula* than number theory is, but yeah, there's no restriction to what math you can use in these problems.
* Grammar Policed on behalf of Sidoh's (pedantic) request
Hmmm....that means I can use my made up math!
Quote from: Ender on November 01, 2006, 07:36:26 PM
* Grammar Policed on behalf of Sidoh's (pedantic) request
If it's truely pedantic, why did you go through the effort of fixing it?
I just wanted to use the word pedantic because I rarely get to use it and it's really fun to say.
Quote from: Ender on November 01, 2006, 09:40:42 PM
I just wanted to use the word pedantic because I rarely get to use it and it's really fun to say.
That was one of the predictions I'd made to my own question. :P