2009 AMC 10B Problem 7

Attempt Problem 7 of the 2009 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

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By inserting parentheses, it is possible to give the expression $2\times3+4\times5$ several values. How many different values can be obtained?

$2$

$3$

$4$

$5$

$6$

There are five full parenthesizations of four numbers joined by three operations. Evaluating all five gives

$((2\times3)+4)\times5=50$ and $(2\times(3+4))\times5=70.$

$(2\times3)+(4\times5)=26.$

$2\times((3+4)\times5)=70$ and $2\times(3+(4\times5))=46.$

Thus the distinct values are $26, 46, 50,$ and $70,$ for a total of $4.$

Thus, the correct answer is C.