On the Sum of Some of the Factors of $2310$

Question

Given 3 natural numbers a,b,c such that $a\times b\times c=2310$. Find the sum of $\sum a+b+c$

My attempt:

Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$

So now since $a,b,c$ are symmetric , their sum should be the same , and I should get , $\sum a+b+c= 3 \sum a$

But computing this is becoming tedious. As when $a$ is $1$ ,then $b,c$ will have $2^5 $combinations. When $a = 2$ or $3$ or $5$ or $7$ or $11$ ,then $b,c$ will have $2^4$ combination. So, the sum. Will become $$3 \sum a = 3 \{2^5 + (2+3+5+7+11) 2 ^4 + (2×3 +2×5... +3×5 +....) 2^3 + (2×3×5 +... ) 2^2 ..... +( 2×3×5×7×11)2^0 \} $$

What next ?? , Because this sum really starts to become too big. I'm sure there must be a shorter way as it is an exam type problem supposed to be solved in a limited time. Or is there any better approach ?? Please help :)

Answer 1

Consider $f(x)=2^5(x+\dfrac{2}{2})(x+\dfrac{3}{2})(x+\dfrac{5}{2})(x+\dfrac{7}{2})(x+\dfrac{11}{2})$ Your sum equals $3f(1)=3[2^5(1+\dfrac{2}{2})(1+\dfrac{3}{2})(1+\dfrac{5}{2})(1+\dfrac{7}{2})(1+\dfrac{11}{2})]=3(2+2)(2+3)(2+5)(2+7)(2+11)$

Answer 2

Silly overkill method:

For any (positive) natural number $N$, set $$d_N=\sum_{abc=N}(a+b+c)\text{.}$$ Consider the Dirichlet series $$f(s)=\sum_{N}\frac{d_N}{N^s}\text{.}$$

Then $$\begin{split}f(s)&=\sum_{a,b,c}\frac{a+b+c}{a^sb^sc^s}\\ &=3\zeta(s-1)\zeta(s)^2\end{split}$$ so that $f(s)$ admits the Euler product $$\begin{align} \frac{f(s)}{3}&=\prod_{p\text{ prime}}f_p(p^{-s}) \\ f_p(T)&=\frac{1}{(1-pT)(1-T)^2}=1+(p+2)T+\ldots \end{align}$$ Now suppose that $N$ has no repeated prime factors. Then reading off the coefficient of $N^{-s}$ gives $$d_N=3\prod_{p|N}(p+2)\text{.}$$