2009 AMC 10B Problems/Problem 7

The following problem is from both the 2009 AMC 10B #7 and 2009 AMC 12B #6, so both problems redirect to this page.

Problem

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

$\mathrm{(A)}\ 2\qquad \mathrm{(B)}\ 3\qquad \mathrm{(C)}\ 4\qquad \mathrm{(D)}\ 5\qquad \mathrm{(E)}\ 6$

Solution

The three operations can be performed on any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions $\begin{align*} (2\times3) + (4\times5) &= 26,\\ (2\times3 + 4)\times5 &= 50,\\ 2\times(3 + (4\times5)) &= 46,\\ 2\times(3 + 4)\times5 &= 70 \end{align*}$ are in fact all distinct. So the answer is $\boxed{4}$ , which is choice $\mathrm{(C)}$ .

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2009 AMC 10B Problems/Problem 7

Problem

Solution

See also