2009 AMC 10B Problem 7

Parentheses around a string of multiplications change nothing, because $2 \times 3 \times 4$ is the same in any order. So the only thing parentheses can really change is when the addition happens.

1. Without any parentheses, the multiplications go first: $2 \times 3 + 4 \times 5 = 6 + 20 = 26$.
2. Make the addition happen before the multiplication on its right. The $3$ and the $4 \times 5$ get added first:
$2 \times (3 + 4 \times 5) = 2 \times 23 = 46$.
3. Make the addition happen before the multiplication on its left. The $2 \times 3$ and the $4$ get added first:
$(2 \times 3 + 4) \times 5 = 10 \times 5 = 50$.
4. Make the addition happen before both multiplications:
$2 \times (3 + 4) \times 5 = 2 \times 7 \times 5 = 70$.
5. Any other way of placing parentheses does one of these four things, so it gives one of these four numbers.

The four values $26, 46, 50, 70$ are all different, so there are $4$ values.

The answer is $\boxed{\textbf{(C)}\ 4}$.

Parentheses only matter where they change the order of different operations. Focus on the one operation that is different (the addition) and ask which multiplications it can come before: the left one or not, the right one or not. That gives $2 \times 2 = 4$ cases. The question asks for different values, so compute each case and check none repeat.

Multiplication is associative, so regrouping a product alone changes nothing. The only thing parentheses can do is decide whether the addition happens before one or both of the neighboring multiplications. There are four possibilities:

• Neither: $2\times3 + 4\times5 = 26$ (the usual value).
• Addition before the right product: $2\times(3 + 4\times5) = 2 \times 23 = 46$.
• Addition before the left product: $(2\times3 + 4)\times5 = 10 \times 5 = 50$.
• Addition first, before both: $2\times(3+4)\times5 = 70$.

Any other placement of parentheses reproduces one of these, so there are $4$ distinct values.

The answer is $\boxed{\textbf{(C)}\ 4}$.