Clan x86
General Forums => Academic / School => Math and Other Problems => Topic started by: Ender on October 23, 2006, 08:18:47 pm
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Show that 7^n - 1 is divisible by 6 for any positive integer n.
I'll post the solution on Wednesday.
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Simple induction... is this a homework assignment question?
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(http://latex.sidoh.org/?render64=JCQgJFxcKg0KXGxlZnQoN15uLTFccmlnaHQpXG1vZHs2fT0wIFxcDQo3Xm5cZXF1aXYgMVxsZWZ0KFxtb2R7Nn1ccmlnaHQpIFxcDQooN14xLTE9NilcZXF1aXYgMChcbW9kezZ9KSQNCg0KJFxcJEFzc3VtZTogJDdeay0xXGVxdWl2IDAoXG1vZHs2fSlcXA0KN157aysxfS0xPTdcbGVmdCg3XmstMSkrNiQNCg0KJFxcXFwkU2luY2UgJDdeay0xIFxlcXVpdiAwKFxtb2R7Nn0pJCBpcyB0cnVlIGJ5IHRoZSBpbmR1Y3Rpb24gaHlwb3RoZXNpcyBhbmQgNiBpcyBkaXZpc2libGUgYnkgNiwgdGhlIHdob2xlIHN0YXRlbWVudCAkN15uLTEgXGVxdWl2IDAoXG1vZHs2fSkkIGlzIHRydWUu)
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(http://latex.sidoh.org/?render64=JCQgJFxcKg0KXGxlZnQoN15uLTFccmlnaHQpXG1vZHs2fT0wIFxcDQo3Xm5cZXF1aXYgMVxsZWZ0KFxtb2R7Nn1ccmlnaHQpIFxcDQooN14xLTE9NilcZXF1aXYgMChcbW9kezZ9KSQNCg0KJFxcJEFzc3VtZTogJDdeay0xXGVxdWl2IDAoXG1vZHs2fSlcXA0KN157aysxfS0xPTdcbGVmdCg3XmstMSkrNiQNCg0KJFxcXFwkU2luY2UgJDdeay0xIFxlcXVpdiAwKFxtb2R7Nn0pJCBpcyB0cnVlIGJ5IHRoZSBpbmR1Y3Rpb24gaHlwb3RoZXNpcyBhbmQgNiBpcyBkaXZpc2libGUgYnkgNiwgdGhlIHdob2xlIHN0YXRlbWVudCAkN15uLTEgXGVxdWl2IDAoXG1vZHs2fSkkIGlzIHRydWUu)
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.
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(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.
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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 7k-1 is divisible by 6 for some arbitrary integer greater than one, k," right?
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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)
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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.
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2 + 2 requires knowledge of set theory =p
but yeah it requires knowledge of basic combinatorics
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Which is Discrete, vis a vis same level as NT :P
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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
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curriculae
You made that word up. :P
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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!
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* Grammar Policed on behalf of Sidoh's (pedantic) request
If it's truely pedantic, why did you go through the effort of fixing it?
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I just wanted to use the word pedantic because I rarely get to use it and it's really fun to say.
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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