Rules are as per the other marathons. Difficulty should be reasonable, and hints should be provided as deemed necessary. Use any "elementary" techniques, provided you state them, and if it's fairly advanced, outline a proof.
I'll start off simple.
Other things:
For the purposes of this thread, the set of all natural numbers excludes zero.
Vectors and elementary functions of any kind are allowed, but little to no calculus. (this one due to leehuan)
Define terms that the average person following this thread probably wouldn't know.
State any theorems/techniques that may help in solving the problem.
Last edited by Paradoxica; 22 Jan 2016 at 9:03 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Hint: Use Fermat's Little Theorem.
Last edited by Paradoxica; 21 Jan 2016 at 10:24 AM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Isn't this just the Advanced X2 Marathon ?
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Last edited by leehuan; 21 Jan 2016 at 10:40 AM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Solve the equation x^2+y^2+1=xyz over the positive integers.
Hint: First concentrate on determining what possibilities there are for z.
Last edited by seanieg89; 22 Jan 2016 at 8:11 AM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Trivial for n=1.
Suppose the property holds for 1..n-1.
For general n, if their sum is divisible by n, then their sum mod n is 0. Take all elements mod n. If all elements mod n are 0, pick any subset. Otherwise, pick out an element x mod N which is nonzero mod n. 0 <= x < n. Of the remaining elements, take any n-x > 0 elements. These n-x elements have a subset divisible by n-x, so if we then add x to the subset, we have a subset with sum (n-x) + x = 0 mod N, so the subset is divisible by n.
Nice . I actually forgot a term on the LHS from the diophantine equation I was trying to remember though lol, try the edited problem too.
It is slightly harder, but not greatly so. The original hint still stands.
Repost for visibility:
Solve the equation:
over the positive integers.
Hint: Try to determine what z can be first.
Last edited by seanieg89; 22 Jan 2016 at 8:18 AM.
I don't think this is quite right? Just because the sum of a subset is divisible by n-x, it doesn't necessarily mean that the sum is n-x mod N? Or have I misinterpreted what you're saying?
Here's what I had. Suppose that a_i are the elements of the set. Now, consider the sums b_i=a_1+a_2+....+a_i. If one of b_1,b_2.. b_n is 0 mod n, that set has sum divisible by n. If not, then by the pigeon hole principle, two of these are of the same class mod n. Suppose that these two are b_m and b_n, (n>m). Then, b_n-b_m has 0 mod n.
I think we need a bit of clarification here, \pi (3)= 2, but we can definitely have subintervals of length 3 which contain no prime numbers. For instance 20,21,22 is such an interval, and none of the three are prime numbers.
Last edited by Blast1; 26 Jan 2016 at 2:26 PM.
?
In my previous post, I was addressing the second part of your question, not the first (I suppose I should've mentioned this). Basically, in your question you claim that if we have a subinterval of length n (has n elements), then there must exist a subinterval containing k primes for all k which exceeds 0 but is bounded by pi(x). In my previous post, I provided a counter example to this claim, showing how there exists a subinterval of length 3, but within the subinterval which I specifically mentioned (20,21,22) there is no subinterval which contains any primes. According to your question, within the subinterval (20,21,22), there should exist a subinterval which contains 1 prime and also another subinterval which contains 2 primes. Obviously, that's not true in this example.
Also, the first part of your question doesn't quite make sense either. For instance, the elements 2 and 3 form a subinterval of length 2, and clearly any subinterval within this subinterval must contain a prime. Thus your claim is false.
So this is why I'm asking for a clarification, as the question in its current wording isn't true.
Last edited by Blast1; 27 Jan 2016 at 1:08 AM.
zz
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
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