The expression

$$
\left(\begin{array}{lllll}
1 & 1 & 1 & \cdots & 1
\end{array}\right)
$$

is written on a board, with 2013 ones in between the outer parentheses. Between each pair of consecutive ones you may write either "+" or ")(" (you cannot leave the space blank). What is the maximum possible value of the resulting expression?
The answer is $3^{671}$, formed by dividing the ones into groups of three as the answer suggests.

It should be obvious that a maximum value exists, since there are only finitely many ways to place the pluses and parentheses and one (or more) of them must give the largest value.

If there is a factor greater than or equal to 5 , we can prove that the arrangement is not optimal by splitting off two 1 's, e.g. changing

$$
(1+1+1+1+1) \text { to }(1+1)(1+1+1) \text {. }
$$

This is tantamount to changing a factor of $n+2$ to $2 n$ for $n \geq 3$, which increases the value of the expression.

If we have a factor containing only one 1 , then we can merge it with any other factor $n$; this changes $n$ to $n+1$ and therefore also increases the value of the expression.

Therefore we need only consider expressions in which every factor is 2,3 , or 4 . Because 4 can be broken up into $2 \cdot 2$ without changing the value of the expression, we will restrict our attention to expressions made up of 2's and 3's.

Now if there are at least three factors of 2 , we can change them to two factors of 3 without altering the total number of 1 's in the expression. This replaces a factor of 8 by a factor of 9 , thus increasing the expression.

So the optimal way of forming the expression consists of zero, one, or two 2's and some undetermined number of 3's. Given that the total number of ones is 2013 , it is clear that there must be no twos and 671 threes, implying that $3^{671}$ is the maximum value.