Divisors problem Prove that there exists such $a \in \mathbb{N}$ which has at least 2018 divisors $b$ such that $$1 \leq \frac{b}{\sqrt{a}}<1.01$$
 A: Since the deadline passed, I post a solution. 
Let $\{p_n\}$ be the sequence of all primes in increasing order. By prime number theorem, we can find $x>0$ such that $\pi(1.01 x) - \pi (x) > 2019$. Let $p_n, p_{n+1}, \ldots, p_{n+2018}$ be the primes in the interval $(x, 1.01x)$. Then
$$
1<\frac{p_{n+k}}{p_n}< 1.01, \ \textrm{ for all }k=1, \ldots, 2018.$$
Then take $a=(p_n p_{n+1} \cdots p_{n+2018})^2$. We have
$\sqrt a = p_n p_{n+1} \cdots p_{n+2018}$. Also, 
$$
b_1=\sqrt a \frac{p_{n+1}}{p_n}=p_{n+1}^2 p_{n+2} \cdots p_{n+2018}, 
$$
$$
b_2=\sqrt a \frac{p_{n+2}}{p_n}= p_{n+1}  p_{n+2}^2 p_{n+3} \cdots p_{n+2018}, 
$$
$$\cdots $$
$$
b_{2018}=\sqrt a \frac{p_{n+2018}}{p_n}=p_{n+1}  p_{n+2} \cdots p_{n+2017}p_{n+2018}^2, 
$$
These $2018$ divisors $b_1, \ldots, b_{2018}$ satisfy $\sqrt a \leq b_i< 1.01\sqrt a$. 
Note that this method, in fact, shows that there are infinitely many such $a$, and the number $2018$ can be given arbitrarily large, and $1.01$ can be given arbitrary $\lambda>1$.   
