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Number theory Difficulty 6.4 National Olympiad Prove it United States

Problem:
Kelvin the Frog was bored in math class one day, so he wrote all ordered triples $(a, b, c)$ of positive integers such that $a b c = 2310$ on a sheet of paper. Find the sum of all the integers he wrote down. In other words, compute
$\sum_{\substack{a b c = 2310 \\ a, b, c \in \mathbb{N}}} (a + b + c)$
where $\mathbb{N}$ denotes the positive integers.

Solution

Solution:
Note that $2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ . The given sum clearly equals $3 \sum_{a b c = 2310} a$ by symmetry. The inner sum can be rewritten as
$\sum_{a \mid 2310} a \cdot \tau\left(\frac{2310}{a}\right)$
as for any fixed $a$ , there are $\tau\left(\frac{2310}{a}\right)$ choices for the integers $b, c$ .
Now consider the function $f(n) = \sum_{a \mid n} a \cdot \tau\left(\frac{n}{a}\right)$ . Therefore, $f = n * \tau$ , where $n$ denotes the function $g(n) = n$ and $*$ denotes Dirichlet convolution. As both $n$ and $\tau$ are multiplicative, $f$ is also multiplicative.
It is easy to compute that $f(p) = p + 2$ for primes $p$ . Therefore, our final answer is $3(2 + 2)(3 + 2)(5 + 2)(7 + 2)(11 + 2) = 49140$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.