120 is atresvido, so then minimal number of divisors must be at most 16. It is easy to prove that if $ n$ is a number of the form $ p^{k}$, $ pq$, $ p^{2}q$ and $ pqr$ ($ p,q,r$ distinct primes), then $ n$ can not be atresvido. My conjecture is for satisfies this condition, $ n$ must be at least of the form $ p^{3}qr$ ($ p,q,r$ distinct primes). Really I'm trying to make a relation between the atresvido condition and the descomposition of a group into Sylow p-subgroups product.

i have done this by brute force.checking all the cases and making some inequalities with the sum of divisors formula.

It is obvious that the sum of divisors of $ N$ is at least $ 3N$ if $ N$ is atresvido, since one of the subsets must contain $ N$. The sum of divisors of any positive integer $ N$ with $ 2n$ divisors has as an upper bound $ \sum_{k = 1}^{n}\left(k+\frac{N}{k}\right)$, and if it has $ 2n+1$ divisors (i.e., if $ N$ is a perfect square), then the upper bound is $ \sqrt{N}+\sum_{k = 1}^{n}\left(k+\frac{N}{k}\right)$. This comes from the fact that, for any two divisors $ 1\leq d < D\leq\sqrt{N}$, then $ d+\frac{N}{d}\geq D+\frac{N}{D}$. Then, 1) any $ N$ atresvido must be the product of at least 4 (not necessarily different) primes. Otherwise, it has at most 8 divisors (includind 1 and itself), and it must hold that $ 3N\leq10+\frac{25N}{12}$, or $ N\leq\frac{120}{11}< 11$. It is a trivial exercise to show that no $ N\leq10$ can be atresvido. 2) any $ N$ atresvido with at most 15 divisors must be less than 100, because if it has exactly 15 divisors, then it must hold that $ 3N\leq\sqrt{N}+\frac{1089}{420}+28$, or equivalently, $ \frac{171N}{420}\leq28+\sqrt{N}$. If $ N\geq100$, then $ \frac{42N}{420}\geq\sqrt{N}$, and $ \frac{129N}{420}>\frac{126N}{420}\geq30 > 28$. Checking the squares below 100, we find no atresvido number, so if any atresvido number has less than 16 divisors, it has at most 14. 3) any $ N$ atresvido with at most 14 divisors must be less than 70, because it must hold that $ 3N\leq\frac{1089N}{420}+28$, or equivalently, $ N\leq\frac{28\cdot420}{171}<\frac{28\cdot420}{168}= 70$. So, since no perfect square below 100 may be atresvido, the only possible candidates for atresvido numbers with less than 16 divisors are numbers less than 70, which are the product of at least 4 (not necessarily different) primes, and which are not perfect squares. This leaves 24, 32, 40, 48, 54, 56 and 60, which can be shown not to be atresvidos. So no number may be atresvido with less than 16 divisors, or 16 is the minimum. PS: I am waiting to see the official solution, because I cannot come up with anything simpler than this...

yes your solution is more elegant then mine. but checking is easy i think. if you have 30 minutes you can do it in some way.

anonymous1173 wrote: but checking is easy i think The way that I used to check whether it could be atresvido or not is just finding all their divisors, adding them up, and seeing that they did not amount to thrice the number... If anybody cares to suggest a simpler way...

daniel73 wrote: PS: I am waiting to see the official solution, because I cannot come up with anything simpler than this... I propose this problem , and the solution i have is not short ( i analize the number of primer factors of an atresvido number) maybe the jury found a shorter solution. Pd: ¿ do you know what means "atresvido" ? , i propose the problem with the name "tripartite". $ Tipe$

it is just a play in words between "tres" (three) and "atrevido" (daring) - I guess. Translations are provided free of charge for our non-spanish-speaking friends all over the world I actually proposed a loose translation for the problem for this forum where instead of calling the number "atresvido", it could be called "threetening" in english. Kind of lame, actually I just wanted to keep the part of setting a pseudo-fun name based on a variation of the name of 3 in the language in which the problem was stated...

tipe wrote: I propose this problem , and the solution i have is not short ( i analize the number of primer factors of an atresvido number) maybe the jury found a shorter solution. So Tipe, you proposed this problem!! Very nice! Congratulations, I really liked it!

Jutaro wrote: tipe wrote: I propose this problem , and the solution i have is not short ( i analize the number of primer factors of an atresvido number) maybe the jury found a shorter solution. So Tipe, you proposed this problem!! Very nice! Congratulations, I really liked it! I concur, very nice problem. Congratulations on getting it selected, Tipe!!!

I'm very happy that the jury selected my problem . The last year (IbMO-2006) problem 5 was also from Perú (the problem of geometry). $ Tipe$

I didn't like it

scorpius119 wrote: Let $ n$ be atresvido. The divisor $ n$appears in one of the three subsets, so we have $ \sigma (n)\geq 3n$. ($ \sigma$ is sum of positive divisors). We have the bound \[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p - 1} \] where the product runs of primes. As a result, $ n$ must have at least three distinct prime divisors (if it had at most two, the upper bound would be at most $ \frac {2}{1}\cdot\frac {3}{2} = 3$, wihch is not possible. Now in the case $ n = pqr$ or $ n = p^{2}qr$ where $ p,q$ distinct primes, we have \[ \frac {\sigma (n)}{n}\leq\frac {p^{2} + p + 1}{p^{2}}\cdot\frac {q + 1}{q}\cdot\frac {r + 1}{r} \] \[ \leq\frac {7}{4}\cdot\frac {4}{3}\cdot\frac {6}{5} = \frac {14}{5} < 3 \] which is also not possible. So either $ n$ has at least four distinct prime divisors, or $ n$ is a product of three distinct prime divisors, in which case either one prime is raised to the third power or greater or at least two primes are raised to the second power or greater. Because of this, $ n$ has at least \[ \min\{ 2^{4},4\cdot 2^{2},3^{2}\cdot 2\} = 16 \] positive divisors. But we have $ n = 120 = 2^{3}\cdot 3\cdot 5$ is atresvido, since \[ n = 120 = 60 + 40 + 20 = \sigma (120) - 60 - 40 - 20 \] and $ \sigma (120) = 360$. So the minimum is $ \boxed{16}$. Sorry why \[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p - 1} \]

csbonny wrote: Sorry why \[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p - 1} \] It's not too hard to see. You should spend more time with it. It's a basic & trivial problem if you are being in training for olympiad. เอาเล้ยยยยยยยยยยย บิ๊กเชียร์อยู่

csbonny wrote: scorpius119 wrote: Let $ n$ be atresvido. The divisor $ n$appears in one of the three subsets, so we have $ \sigma (n)\geq 3n$. ($ \sigma$ is sum of positive divisors). We have the bound \[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p - 1} \] where the product runs of primes. As a result, $ n$ must have at least three distinct prime divisors (if it had at most two, the upper bound would be at most $ \frac {2}{1}\cdot\frac {3}{2} = 3$, wihch is not possible. Now in the case $ n = pqr$ or $ n = p^{2}qr$ where $ p,q$ distinct primes, we have \[ \frac {\sigma (n)}{n}\leq\frac {p^{2} + p + 1}{p^{2}}\cdot\frac {q + 1}{q}\cdot\frac {r + 1}{r} \] \[ \leq\frac {7}{4}\cdot\frac {4}{3}\cdot\frac {6}{5} = \frac {14}{5} < 3 \] which is also not possible. So either $ n$ has at least four distinct prime divisors, or $ n$ is a product of three distinct prime divisors, in which case either one prime is raised to the third power or greater or at least two primes are raised to the second power or greater. Because of this, $ n$ has at least \[ \min\{ 2^{4},4\cdot 2^{2},3^{2}\cdot 2\} = 16 \] positive divisors. But we have $ n = 120 = 2^{3}\cdot 3\cdot 5$ is atresvido, since \[ n = 120 = 60 + 40 + 20 = \sigma (120) - 60 - 40 - 20 \] and $ \sigma (120) = 360$. So the minimum is $ \boxed{16}$. Sorry why \[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p - 1} \] Thats a good bound.. I wonder how many people solved the problem since deriving that inequality is hard and this is a problem 5. Anyways, csbony, it is well known that if $n= p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}$ then ${\sigma (n)}= (\frac {p_1^{\alpha_1+1}-1}{p_1-1})(\frac {p_2^{\alpha_2+1}-1}{p_2-1})...(\frac {p_n^{\alpha_n+1}-1}{p_n-1})$ So $[ \frac{\sigma (n)}{n}<\prod_{p|n}\frac{p}{p-1} ]$ is equivalent to proving that $({p_1^{\alpha_1+1}-1})({p_2^{\alpha_2+1}-1})....({p_n^{\alpha_n+1}-1}) \le (p_1^{\alpha_1+1})(p_2^{\alpha_2+1})....(p_n^{\alpha_n+1})$ which is true. Sorry for reviving this old thread, but I will like to see new solutions to the problem.