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 June 20th, 2014, 11:21 AM #11 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory I was looking forward for your question. This can be done by counting endpoints of diagonals, I think, which avoids counting diagonals and their meetpoints altogether. The number of triangles in the interior of diagonals with three endpoints is $\binom{6}{3}$. The number of triangles in the interior of diagonals with four endpoints is $4\binom{6}{4}$. Similarly, the number of triangles in the interior of diagonals with five endpoints is $5 \binom{6}{5}$ (note that the diagonals with 5 endpoints must meet themselves in a hexagon) Thus, the number of triangles is $$\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} = 110$$ Thanks from eddybob123
 June 20th, 2014, 02:33 PM #12 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 This is also A006600 in the OEIS.
June 20th, 2014, 03:38 PM   #13
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Quote:
 Originally Posted by CRGreathouse For those interested in the nonmodular answer in its 5782-digit glory, here's a GP script: Code: val(n,p)=my(t);while(n\=p,t+=n);t t=1;forprime(p=2,2014,t*=p^val(2014,p)+1);t
Here is another, hope you'll like it:

Code:
prod(i=1,primepi(2014),g=2014;p=prime(i);(p^sum(j=0,log(g)\log(p),g\=p)+1))
The inaccuracy in log doesn't matter here. Since we start from j = 0, most of the time, 0 is added once too much.

 June 20th, 2014, 04:16 PM #14 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 932 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms I'm amused by how similar our programs are and yet how different they appear. The prod syntax is nicer (no need for a temporary variable to hold the sum) but generating the n-th prime at each step is inefficient. Not that it matters when you're only looping up to 2014, of course! I think we're ready for a new question? Thanks from eddybob123
 June 21st, 2014, 03:31 AM #15 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Here are the question, the one who gets one of the hard questions below gets the control. Easy : Let $f$ be a univariate complex function satisfying $f(x) = f(x + \omega_1) = f(x+\omega_2) = f(x + \omega_3)$ with $\omega_1, \omega_2, \omega_3$ being $\Bbb R$-linearly independent. Can a function like that exist? If so, give example otherwise show proof. [Bonus point 2] Hard : Let $f$ be defined as above except $\omega_1, \omega_2, \omega_3$ being $\Bbb Q$-linearly independent. Can such a function exist? If so, give example otherwise show proof. Hard (this one is for real analysis lovers) : Let $f : \Bbb R \to \Bbb R$ be a function such that for all $x, y \in \Bbb R$ one has $f(x + y^2) - f(x) \geq y$. Can such a function exist? If so, give example otherwise show proof. Balarka . Thanks from eddybob123 Last edited by mathbalarka; June 21st, 2014 at 03:36 AM.
 June 21st, 2014, 04:39 AM #16 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,778 Thanks: 2195 Math Focus: Mainly analysis and algebra What does it mean to say that the $\omega_i$ are $\mathbb{R}$ linearly independant? Aren't they constants? Thanks from eddybob123 Last edited by v8archie; June 21st, 2014 at 04:41 AM.
 June 21st, 2014, 06:22 AM #17 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,778 Thanks: 2195 Math Focus: Mainly analysis and algebra \begin{align*} f(x + y^2) - f(x) &\ge y \implies f(x + y) - f(x) \ge \sqrt{y} \qquad \text{(the positive root since $\sqrt{y} \ge -\sqrt{y}$)} \\[12pt] \text{For } a \in \mathbb{R}^+, n \in \mathbb{Z}^+ \qquad f(x + \frac{a}{n}) - f(x) &\ge \sqrt{\frac{a}{n}} \\ f((x + \frac{a}{n}) + \frac{a}{n}) - f(x + \frac{a}{n}) &\ge \sqrt{\frac{a}{n}} \\ f((x + \frac{2a}{n}) + \frac{a}{n}) - f(x + \frac{2a}{n}) &\ge \sqrt{\frac{a}{n}} \\ &\cdots \\ f((x + \frac{(n-1)a}{n}) + \frac{a}{n}) - f(x + \frac{(n-1)a}{n}) &\ge \sqrt{\frac{a}{n}} \\[6pt] \text{Summing the above equations gives} \qquad f(x+a) - f(x) &\ge \sqrt{an} \\ \end{align*} Then, letting $n \to \infty$ we see that $f(x+a) - f(x) \to \infty$. This is true for all $x$ and all $a \gt 0$ so no such function exists. Thanks from eddybob123 Last edited by v8archie; June 21st, 2014 at 06:25 AM.
June 21st, 2014, 06:47 AM   #18
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Quote:
 Originally Posted by v8archie What does it mean to say that the $\omega_i$ are $\mathbb{R}$ linearly independant? Aren't they constants?
Yes. For example, $1$ and $\pi$ are $\Bbb Q$-linearly independent.

June 21st, 2014, 06:51 AM   #19
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Quote:
 Originally Posted by v8archie Then, letting $n \to \infty$ we see that $f(x+a) - f(x) \to \infty$. This is true for all $x$ and all $a \gt 0$ so no such function exists.
Nice work. You get to post the next question.

PS : There is actually a swifter argument, IMHO, using the Basel problem.

 June 21st, 2014, 07:31 AM #20 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,778 Thanks: 2195 Math Focus: Mainly analysis and algebra I'm not familiar with the Basel problem. I first proved it to myself using l'Hopital's Rule on$$\lim_{y \to 0} \frac{f(x + y^2) - f(x)}{y} \ge 1$$ but I wasn't completely happy with the validity of the argument. Thanks from eddybob123

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