Determine with a proof the largest number which can be written as a product of natural numbers which have sum 2012

Question

I'm having trouble finding the largest number and proving it.

Answer 1

Use as many threes as you can, make up the remainder with twos, as necessary. Proof left to reader.

Answer 2

Let the maximum product be $p$. From AM-GM, we have $$\dfrac{\sum_{k=1}^n a_k}n \geq \left(\prod_{k=1}^n a_k \right)^{1/n}$$ Hence, we get that $$\dfrac{2012}n \geq p^{1/n}$$ Hence, we have that $$p \leq \left(\dfrac{2012}n \right)^n$$ Now study the behavior of the function $f(x) = \left(\dfrac{2012}x \right)^x$ or equivalently the behavior of the function $g(x) = \log(f(x)) = x \log (2012) - x \log(x)$. We then have $$g'(x) = \log(2012) - 1 - \log(x)$$ $$g'(x) = 0 \implies x = \dfrac{2012}e$$ Hence, $n \approx \dfrac{2012}e$ and to maximize the product all number must be more or less equal to $e$. Since, all the numbers are natural numbers, they must be either $2$ or $3$. Now make use of the fact that $2 + 2 + 2 = 3 + 3$ and $2^3 < 3^2$ to conclude that we need six hundred and seventy $3's$ and a $2$ to maximize the product.