$a_1+a_2+…+a_n=2008$ where all $a_i$ are positive integers. If $A_k = a_1 a_2 … a_k$ , what is the largest possible value of $A_1+A_2+…+A_n$? The sum of a set of positive integers = $2008$.
$a_1+a_2+…+a_n=2008$ where all $a_i$ are positive integers.
If $A_k=a_1 a_2 … a_k$ , what is the largest possible value of $A_1+A_2+…+A_n$ ?
 A: All of the $a_i$ must come in weakly decreasing order.
If $a_1+\dots+a_n=2008$ with $a_1$ an even number bigger than $2$, then consider $$(a_1/2)+2+a_2+\dots+a_n+1+\dots+1=2008.$$
If $a_1+\dots+a_n=2008$ with $a_1$ odd and at least $5$, say $a_1=2k+1$, consider 
$$
(a_1-2)+2+a_2\dots+a_n=2008
$$
(Clearly $a_1=1$ cannot be maximal). 
Once you have figured out $a_1$, apply the same argument to $a_2$, $a_3$, etc.
(edit: at each step above, replace $j$ trailing copies of $1$ by the integer $j$; that is, instead of using $2006+1+1$, use $2006+2$. Doing this does not change the result). 
This shows that each $a_i$ must be $3$ or $2$. Simplifying the terms $A_1+\dots+A_n$ is now just adding terms of a 2 geometric series. Finally, if there are $k$ $3s$ then there are $\frac{2008-3k}{2}$ $2s$; thus we want to maximize 
$$
\frac{3(3^k-1)}{2}+\frac{3(3^k-1)}{2}\cdot2\cdot\left(2^\frac{2008-3k}{2}-1\right)
$$
From Wolframalpha this gives $668$ copies of $3$ and $2$ copies of $2$ for an answer of roughly $3\times 10^{319}$
