2009 AMC 10B Problem 7

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 $\boxed{\textbf{(C)}\ 4}$.

Parentheses only matter where they change the order of different operations. Focus on the lone addition and ask which multiplications it precedes; that gives an exhaustive list of $2 \times 2$ groupings. Distinct expressions are not the question; distinct values are, so compute each.

• Counting different-looking parenthesizations that evaluate to the same number, such as (2x3+4)x5 and ((2x3)+4)x5, as different values.
• Missing the case $2\times(3+4\times5) = 46$ and answering $3$.