highest product of the numbers that sum to $100$ 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 $2^{50}$
 A: 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.  
A: If we write a continuous version $f(x)=x^{\frac{100}{x}}$ it maximum is when $x=e$, the closest natural number is 3 so the maximum is $3^{32}*4$.
I will add that $f(x)=x^{\frac{b}{x}};\ \forall b\geq 1$ have a maximum on $x=e$.
We need to see now that any composition will be lesser number that just an unique exponential. I write $$f(x's)=x_0^{\frac{b-\sum_{i}c_i}{x_0}}\cdot x_1^{\frac{c_1}{x_1}}\cdot x_2^{\frac{c_2}{x_2}}\cdots x_n^{\frac{c_n}{x_n}};\ c_i\neq c_j,\ \sum c_i<b\\
x_i,b,c_i>1$$
From above we can see that the maximum for any multiplier is when it base is $x_i=e$ so you can see that for any $c_i$ we choose it maximum value will be related to a base $x_i=e$ so doesnt exist any composition with $x_i\neq e$ that make the function to have a greater maximum choosing any $c_i$ decomposition that you want.
This translated to natural numbers put the maximum with $x=3$, i.e. no closest natural number to $e$ that $3$ than lead to a number f(n) closest to f(e):
$$|f(e)-f(3)|<|f(e)-f(n)|\ \forall n\in \Bbb N-\{3\}$$
And we can add that if $0\not\equiv b \mod 3$ then the next number close to $f(e)$ is $f(2)$ so we must compose the number with powers of base 3 and 2. I.e.:
$$r\equiv b\mod 3\ \rightarrow f(b)=3^{\frac{b-c}{3}}2^{\frac{c}{2}}\begin{cases}r\equiv 0, f(b)=3^{\frac{b}{3}}\\r\equiv 1, f(b)=3^{\frac{b-4}{3}}*4\\r\equiv 2, f(b)=3^{\frac{b-2}{3}}*2\end{cases}$$
A: Let's look at a factorisation containing the term $n$. We have $2(n-2)=2n-4$ so if $n\ge 4$ we have $2(n-2)= 2n-4 \ge n$, and replace $n$ with $2, n-2$, which does not reduce the product and may increase it.
If $1$ appears in the sum we can add it to another summand, which clearly increases the product without changing the sum.
So we get a sum consisting of terms which are either $2$ or $3$. Then we note Barak's observation that $3\cdot 3 \gt 2\cdot 2 \cdot 2$
A: Let's say a set $S=\{x_i\}:i\in \{1\to n\}$ for some $n\in\mathbb{N}$ is the optimal set.
Since $\forall x \geqslant 4 : x * 2 \geqslant x + 2$, the set doesn't have any element $x \geqslant 4$ because we can take it out of the set and add a two and a $(x-2)$ instead.
now for elements strictly smaller than 4:
adding a $1$ want help since $x*1 = x$.
However, comparing 2 to 3, we can find that $3*2 = 2*3$ which means adding 3 two times is equal to add 2 three times but $3*3>2*2*2$ which means we should switch as many sets of (2*2*2) to (3*3) as possible.
Here we have three different cases which are the values for sum modulo 3.
The case of 100 is 1 but let's speak generally.

*
for 0 the number can be divided by 3 and we are done.


*
for 2 we can add a 2 at the end with affecting threes so the answer is a 2 and $x/3$ threes.


*
The case of one is a bit tricky that when divided by 3 we have a rest of one so we even ignore it since $x*1 = x$ or we can take a three out if $x > 3 (x\neq1)$ and add two twos so we're comparing 2*2 to 1*3 and 2*2 is bigger so in this case we have 2 twos and a rest of threes.

In conclusion the answer to your question should be 2 twos and $(96\div3=32)$ threes
