Problem 1 Let $ \sigma(\cdot): \mathbb{N}_0 \to \mathbb{N}_0$ be the function from every positive integer $ n$ to the sum of divisors $ \sum_{d \mid n}{d}$ (i.e. $ \sigma(6) = 6 + 3 + 2 + 1$ and $ \sigma(8) = 8 + 4 + 2 + 1$). Find all primes $ p$ such that $ p \mid \sigma(p - 1)$. (Salvatore Tringali) Problem 2 Define $ \phi$ the positive real root of $ x^2 - x - 1$ and let $ a,b,c,d$ be positive real numbers such that $ (a + 2b)^2 = 4c^2 + 1$. Show that $ \displaystyle 2d^2 + a^2\left(\phi - \frac {1}{2}\right) + b^2\left(\frac {1}{\phi - 1} + 2\right) + 2 \ge 4(c - d) + 2\sqrt {d^2 + 2d}$ and find all cases of equality. (A.Naskov) Problem 3 Let a cyclic quadrilateral $ ABCD$, $ AC \cap BD = E$ and let a circle $ \Gamma$ internally tangent to the arch $ BC$ (that not contain $ D$) in $ T$ and tangent to $ BE$ and $ CE$. Call $ R$ the point where the angle bisector of $ \angle ABC$ meet the angle bisector of $ \angle BCD$ and $ S$ the incenter of $ BCE$. Prove that $ R$, $ S$ and $ T$ are collinear. (Gabriel Giorgieri) Problem 4 Let $ m$ a positive integer and $ p$ a prime number, both fixed. Define $ S$ the set of all $ m$-uple of positive integers $ \vec{v} = (v_1,v_2,\ldots,v_m)$ such that $ 1 \le v_i \le p$ for all $ 1 \le i \le m$. Define also the function $ f(\cdot): \mathbb{N}^m \to \mathbb{N}$, that associates every $ m$-upla of non negative integers $ (a_1,a_2,\ldots,a_m)$ to the integer $ \displaystyle f(a_1,a_2,\ldots,a_m) = \sum_{\vec{v} \in S} \left(\prod_{1 \le i \le m}{v_i^{a_i}} \right)$. Find all $ m$-uple of non negative integers $ (a_1,a_2,\ldots,a_m)$ such that $ p \mid f(a_1,a_2,\ldots,a_m)$. (Pierfrancesco Carlucci) Problem 5 Let $ X: = \{x_1,x_2,\ldots,x_{29}\}$ be a set of $ 29$ boys: they play with each other in a tournament of Pro Evolution Soccer 2009, in respect of the following rules: i) every boy play one and only one time against each other boy (so we can assume that every match has the form $ (x_i \text{ Vs } x_j)$ for some $ i \neq j$); ii) if the match $ (x_i \text{ Vs } x_j)$, with $ i \neq j$, ends with the win of the boy $ x_i$, then $ x_i$ gains $ 1$ point, and $ x_j$ doesn’t gain any point; iii) if the match $ (x_i \text{ Vs } x_j)$, with $ i \neq j$, ends with the parity of the two boys, then $ \frac {1}{2}$ point is assigned to both boys. (We assume for simplicity that in the imaginary match $ (x_i \text{ Vs } x_i)$ the boy $ x_i$ doesn’t gain any point). Show that for some positive integer $ k \le 29$ there exist a set of boys $ \{x_{t_1},x_{t_2},\ldots,x_{t_k}\} \subseteq X$ such that, for all choice of the positive integer $ i \le 29$, the boy $ x_i$ gains always a integer number of points in the total of the matches $ \{(x_i \text{ Vs } x_{t_1}),(x_i \text{ Vs } x_{t_2}),\ldots, (x_i \text{ Vs } x_{t_k})\}$. (Paolo Leonetti) Good luck! For all doubts ask me in mp.

Name of users who send me solutions (if you are not here contact me): Ahwingsecretagent, NickNafplio, Zhero, Bugi, azjps, Palina.41, mavropnevma, geda, mod_2, TBPL, dario2994, exodd, Giuseppe_R, kn, Maioc92, String. Here there are links where you can post your solution: Problem1 Problem2 Problem3 Problem4 Problem5.

When will the results be available?

Hi Bugi, I hope to publish result up to some days, together with official solutions.. as I hope to get the second round on next monday at same hours Best regards Paolo

Tuesday is past... Just a reminder

Thanks for remembering me Here are the results for the first round: mavropnevma- 7/7/0/7/7 TBPL- 7/7/0/3/7 exodd- 7/7/0/5/0 azjps- 7/5/0/7/0 Zhero- 7/3/0/7/0 Maioc92- 7/7/0/0/0 kn- 7/0/0/7/0 NickNapflio- 7/7/0/0/0 dario2994- 2/4/0/7/0 mod_2- 7/1/0/5/0 Bugi- 6/6/0/0/0 String- 7/0/0/0/0 Palina.41- 7/0/0/0/0 geda- 7/0/0/0/0 Ahwingsecretagent- 7/0/0/0/0 Giuseppe_R-1/0/0/0/0 And my personal congrats to mavropnevma-pdf-solutions! Second round will be taken next week (I'll open a new topic for that one ) Thanks again for partecipating!

Total sum on no. 3: 0 points! 12 points enough, although I over-estimated that no. 4...

Bugi wrote: Total sum on no. 3: 0 points! Yes, I know.. although on ML that problem received a lot of solutions, I think that for the second round at least geometry problem will be simpler.. Bugi wrote: .. although I over-estimated that no. 4... Sure Cheers Bugi and all other! Ps. For all doubt about points contact me with pm Ps1. I cannot edit the previous message, however this is a edit about Nicknapflio points: 7/7/0/2/0.

Will there be a second round? If so, when will it start?

It'll start this Friday Look here