number theory
what is the highest product of the numbers that sum to 100
for example $100 = 1+1+1+1+1+1+1+\ldots+1$ the product of these is just $1^{100} = 1$
$100 = 99 + 1$ the product of these is $99\times 1$
the numbers have to be positive integers
do the numbers all have to be the same – for example I think it is $2^{50}$
To add pessimism in finding such prime $p$ that cannot be written as sum of $3$ co-divisors of cube, I've drawn images:
Number of ways to write an integer number $p$ as the sum $x+y+z$, where $x\cdot y\cdot z=c^3$.
($\color{red}{\bf{red}}$ dots $-$ composite numbers, $\bf{black}$ dots $-$ prime numbers)
$$p\le 500$$
$$p\le 5\;000$$
$$p\le 50\;000$$
$$p\le 500\;000$$
$$p\le 5\;000\;000$$
Control points:
$p=486$ (composite): $1$ way: $486 = 162+162+162$;
$p=2048$ (composite): $2$ ways: $2048 = 128+720+1200 = 224+256+1568$;
$p=6656$ (composite): $3$ ways: $\small {6656 = 416+2340+3900 = 512+1536+4608 = 728+832+5096}$;
$p=7559$ (prime): $4$ ways: $\scriptsize{7559 = 9+50+7500 = 114+225+7220 = 135+1024+6400 = 722+2809+4028}$;
$p=26624$ (composite): $5$ ways;
$p=58757$ (prime): $10$ ways;
$p=80429$ (prime): $13$ ways;
$p=111611$ (prime): $15$ ways;
$...$
These images were made in style like here.
(Since I've seen images for Goldbach's conjecture, I lost any hope to find contradiction for it).
Method of search:
to build array $a[3N]$ for storing number of ways;
$a[p]$ is the number of ways to write $p$ as such sum; (initially, each $a[j]=0$);
for $c = 1 ... N$
$\quad$ create list of prime divisors of $c^3$;
$\quad$ (there is enough to know prime decomposition of $c$)
$\quad$ for each pair $x,y$ of divisors of $c^3$ to find $z=\dfrac{c^3}{xy}$;
$\quad$ if $z\in\mathbb{Z}$ and if $x\le y\le z$, then increase $a[x+y+z]$.
(if $c>N$, then sum of co-divisors of $c^3$ is greater than $3N$).
$x+y + xy=10x + y\implies 9x=xy$
If $x\ne 0$ then $y=9$
Best Answer
The arithmetic-geometric-mean equation says that $\sum_{i=1}^{n}{\frac{a_i}{n}}\geq\sqrt[n]{\prod_{i=1}^n{a_i}}$.
If you enter your condition on the left side and look at different values for $n$, you might find a solution.
Edit: This method works well to guess a solution, which is that most of your numbers will be $3$, and by trying that we can find $3^{32}\cdot 2^2$. Now we can go about proving that this indeed the maximum:
Assume that any $a_k$ of your numbers is greater than $4$. We could then substitute this number by $\frac{a_k}{2}+\frac{a_k}{2}$, which have a product larger than $a_k$, since $\frac{a_k}{2}^2\geq a_k \Leftrightarrow a_k\geq 4$.
(If $a_k$ is odd, we can do the same thing with two numbers with difference $1$.)
From this we can conclude that there can be no numbers greater than $3$ in your sum. Also, there can be no $1$s, for obvious reasons.
Hence, we have a selection of $a$ $3$s, and $b$ $2$s, where $3a+2b\leq 100$. Formally, we have to look at the cases $3a+2b=100, 3a+2b=99$, since if the sum was smaller than $98$ we could simply add a $2$.
Thus, we can write our product as $3^a2^{\frac{100-3a}{2}}$. If we look at this function and its maximum/monotony properties the only possible values for a maximum are $a=32, a=33$. Comparison leads to the maximum being at $a=32, b=2$.
Another possibility with less calculus would be to look at $3\cdot 3\geq 6$, immediately telling us that there can be no more than $3$ $2$s, else we could, again, substitute them with $2$ $3$s. In hindsight, this way might actually be a lot quicker and requires no differentiation...
In any case, with both variants we are done, and you have your maximum being what quite some people have pointed out so far.