Consider $A$, the set of natural numbers with exactly $2019$ natural divisors , and for each $n \in A$, denote $$S_n=\frac{1}{d_1+\sqrt{n}}+\frac{1}{d_2+\sqrt{n}}+...+\frac{1}{d_{2019}+\sqrt{n}}$$where $d_1,d_2, .., d_{2019}$ are the natural divisors of $n$. Determine the maximum value of $S_n$ when $n$ goes through the set $ A$.
Problem
Source: 2019 Romanian NMO grade VIII P1
Tags: inequalities, number theory, Divisors
rms
10.09.2024 23:51
parmenides51 wrote: Consider $A$, the set of natural numbers with exactly $2019$ natural divisors , and for each $n \in A$, denote $$S_n=\frac{1}{d_1+\sqrt{n}}+\frac{1}{d_2+\sqrt{n}}+...+\frac{1}{d_{2019}+\sqrt{n}}$$where $d_1,d_2, .., d_{2019}$ are the natural divisors of $n$. Determine the maximum value of $S_n$ when $n$ goes through the set $ A$. Translated from here:
**Solution.** Since \( 2019 = 3 \cdot 673 \) and 673 is prime, it follows that any number in \( A \) has one of the forms: \( p^{2018} \), where \( p \) is a prime number, or \( p^2 \cdot q^{672} \), where \( p, q \) are distinct primes. Thus, the smallest number in the set \( A \) is \( 3^2 \cdot 2672 \)
Since \( d_i \) divides \( n \) if and only if \( \frac{n}{d_i} \) divides \( n \), we obtain that:
\[
2 \cdot S_n = \sum_{i=1}^{2019} \left( \frac{1}{d_i + \sqrt{n}} + \frac{1}{\frac{n}{d_i} + \sqrt{n}} \right) = \sum_{i=1}^{2019} \left( \frac{1}{d_i + \sqrt{n}} + \frac{1}{\sqrt{n}} \cdot \frac{d_i}{d_i + \sqrt{n}} \right) = \frac{2019}{\sqrt{n}}
\]It results that
\[
S_n = \frac{2019}{2} \cdot \frac{1}{\sqrt{n}} \leq \frac{3 \cdot 673}{2 \cdot \sqrt{3^2 \cdot 2672}} = \frac{673}{2337},
\]the maximum value of \( S_n \) when \( n \) belongs to \( A \). Equality is reached for \( n = 3^2 \cdot 2672 \).