My solution during the contest: Take $n=13p^3$ for an arbitarily prime $p\neq13$. Then $n$ has $8$ divisors and we have $\lfloor\sqrt{3}d(n)\rfloor=\lfloor8\sqrt{3}\rfloor=13$.

Tintarn wrote: Denote by $d(n)$ the number of positive divisors of a positive integer $n$. Prove that there are infinitely many positive integers $n$ such that $\left\lfloor\sqrt{3}\cdot d(n)\right\rfloor$ divides $n$. Let $p$ be a prime sufficiently large enough. Let us say that $n$ is of form $43^4p^4$. Then $\left \lfloor \sqrt{3}d(n) \right \rfloor = 43$ and this divides $43^4 p^4$, and basically there exists infinitely many primes, hence we're done.

DapperPeppermint wrote: Let us say that $n$ is of form $3p$. Then $\left \lfloor \sqrt{3}d(n) \right \rfloor = 3$ Ehm, no? Then $d(n)=4$ and I don't think that $\lfloor 4\sqrt{3}\rfloor=3$...

Tintarn wrote: DapperPeppermint wrote: Let us say that $n$ is of form $3p$. Then $\left \lfloor \sqrt{3}d(n) \right \rfloor = 3$ Ehm, no? Then $d(n)=4$ and I don't think that $\lfloor 4\sqrt{3}\rfloor=3$... Ugh my bad I'll edit my solution

Slightly more interesting problem: Show that this remains true if we replace $\sqrt{3}$ by an arbitrary irrational number $0<\alpha<2$. (Two of the three complete solutions submitted at the competition immediately generalize to this case, but of course the shortest one from #2 does not.) Still slightly more interesting problem: Show that this remains true if we replace $\sqrt{3}$ by an arbitrary number $0<\alpha<2$. (Only one of the three solutions still works.)

I find a solution using dirichlet and equidistribution theorem, proving existence of such $n$ , without giving any existence example, Plz let me know whether it is allowed to use it in the contest, so I can post it.

Well, it doesn't matter whether it's allowed to use in the contest, right? If it's an interesting solution, just go ahead and post it!

So here's my solution , I am new to Analytic NT , So plz let me know if there is any mistake - So here we want $\left\lfloor\sqrt{3}\cdot d(n)\right\rfloor$ to divides $n$ , so select $n = p^k$ , so now we want $$\left\lfloor\sqrt{3}\cdot (k+1)\right\rfloor \mid p^k$$, so if we can make it divides $p$ only , then also we will be done. so now we want existence of $k$ and a prime $p$ such that $$p < \sqrt{3}\cdot(k+1) < p+1$$, so by doing some easy computations we want to prove the existence of $k$ and a prime $p$ such that $$\frac{p - \sqrt{3}}{\sqrt{3}} < k < \frac{p - \sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}}$$or equivalently such a natural $k$ will exist if there exist a prime $p$ such that $$ 1 - \frac{1}{\sqrt{3}} < \left \{ \frac{p - \sqrt{3}}{\sqrt{3}} \right \} < 1$$, or equivalently $$ \frac{\sqrt{3} - 1}{\sqrt{3}} < \left \{ \frac{p}{\sqrt{3}} \right \} <1 $$, \ this will have density $\frac{1}{\sqrt{3}} > \frac{1}{2}$ , by Vinogradov's Equidistribution results as $\frac{1}{\sqrt{3}}$ is an irrational number , we can say that such primes will exist. $\square$

MatBoy-123 wrote: So by density of prime numbers in $\mathbb{N}$ , there must exist such a prime $p$ , and we are done. Unfortunately, the primes have density $0$, so this part of the argument does not work. Indeed, there is an equidistribution result for primes saying that for any fixed irrational $\alpha$, the sequence $\{\alpha p\}$ is equidistributed modulo $1$ where $p$ runs through the sequence of prime numbers. This is classically due to Vinogradov but it is much harder than the corresponding result for $\{\alpha n\}, n \in \mathbb{N}$.

Tintarn wrote: MatBoy-123 wrote: So by density of prime numbers in $\mathbb{N}$ , there must exist such a prime $p$ , and we are done. Unfortunately, the primes have density $0$, so this part of the argument does not work. Indeed, there is an equidistribution result for primes saying that for any fixed irrational $\alpha$, the sequence $\{\alpha p\}$ is equidistributed modulo $1$ where $p$ runs through the sequence of prime numbers. This is classically due to Vinogradov but it is much harder than the corresponding result for $\{\alpha n\}, n \in \mathbb{N}$. So can we say from this adding to my argument that such primes exist?

MatBoy-123 wrote: So can we say from this adding to my argument that such primes exist? You mean from Vinogradov's Equidistribution result? Yes, this would imply the result.

Construction: Take $n=2^4\cdot 17\cdot p$ with $p$ prime. then $d(n)=(4+1)(1+1)(1+1)=20$ and $\lfloor \sqrt{3}\cdot d(n) \rfloor=\lfloor \sqrt{3}\cdot 20 \rfloor=34$ and $34\mid 2^4\cdot 17\cdot p$