nicetry007 wrote: Peter wrote: nicetry007 wrote: The numbers are 2^5. 3^2. 5^2. 7^1 , 2^6. 3^3. 5^1. 7^1 , 2^2. 3^2. 5^2. 7^1. 11^1 , 2^4. 3^2. 5^1. 7^1. 11^1. I'm not posting my solution here for it is not a solution but a mere enumeration of all possible cases. If there is a nice way of doing the problem, please post your ideas. Otherwise, I feel this isn't a problem that deserves to be on the IMO shortlist. I don't see the problem? You gave a correct example, that is all that the question asks. How can you expect a shorter solution than just 4 numbers? I only provided the answers and not the solution. In other words, I didn't outline a method for identifying the numbers. The reason for not doing so is that it is very tedious.

I will outline here an approach for the general problem: Find the positive integers $ n\in\overline{2,m}$ for which $ \tau(n)\ge t$, where $ \tau(n)$ represents the number of positive divisors of $ n$. Assume that $ n$ is a number satisfying the conditions. Write $ n = p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{r}^{\alpha_{r}}$, where $ p_{1}< p_{2}< ... < p_{r}$ are primes and $ \alpha_{1},\alpha_{2}, ...,\alpha_{r}$ are positive integers. So $ \tau(n) = (\alpha_{1}+1)(\alpha_{2}+1)\cdots (\alpha_{r}+1)\ge t$. Now, notice that $ m\ge n\ge p_{1}p_{2}...p_{r}\ge\mathcal{P}(r)$, where $ \mathcal{P}(r)$ is the product of the first $ r$ prime numbers. This gives a first upper bound for $ r$: denote it by $ r_{1}\in\mathbb{Z}$. Next, by AM-GM, $ \alpha_{1}+\alpha_{2}+...+\alpha_{r}\ge r(\sqrt [r]{t}-1)$, so $ m\ge n\ge 2^{\alpha_{1}+...+\alpha_{r}}\ge 2^{r(\sqrt [r]{t}-1)}$, which shows that $ r(\sqrt [r]{t}-1)\le\log_{2}(m)$. We infer some lower bound for $ r$, call it $ r_{0}\ge 1$. Now, we take each $ r\in [r_{0}, r_{1}]$ and try to find the corresponding set of solutions. One idea is to use the fact that $ m\ge n\ge p_{1}p_{2}...p_{r}\cdot 2^{\alpha_{1}+...+\alpha_{r}-r}$. Now we get that $ \alpha_{1}+\alpha_{2}+...+\alpha_{r}\le k$ for some positive integer $ k$ (for example, $ m = 70000$ and $ r = 4$ implies $ \alpha_{1}+...+\alpha_{r}\le 12$). Thus, if $ \beta_{1}, ...,\beta_{r}$ are the $ \alpha$s in increasing order, by the condition $ \tau(n)\ge t$ we see that $ (\beta_{1}, ...,\beta_{r})$ is in some set $ S$, and by rearrangement inequality $ m\ge n\ge q_{1}^{\beta_{r}}...q_{r}^{\beta_{1}}$, where $ q_{1}, ..., q_{r}$ are the first $ r$ primes, $ q_{1}< q_{2}< ... < q_{r}$. Eventually we will get a few $ r$-uples to deal with. In our case, $ m = 70000$, $ t = 100$, we get $ n\in\{ 45360, 50400, 55440, 60480, 65520, 69300\}$. The problem would be more suitable with lesser and more adequate bounds, still for the students not to try "computational" methods.