Difference between revisions of "2009 AMC 10B Problems/Problem 7"

(Solution)
Line 17: Line 17:
 
(2\times3) + (4\times5) &= 26,\\
 
(2\times3) + (4\times5) &= 26,\\
 
(2\times3 + 4)\times5 &= 50,\\
 
(2\times3 + 4)\times5 &= 50,\\
2\times(3 + (4\times5)) &= 46,\\
+
2\times(3 + 4\times5) &= 46,\\
 
2\times(3 + 4)\times5 &= 70
 
2\times(3 + 4)\times5 &= 70
 
\end{align*}</cmath>
 
\end{align*}</cmath>

Revision as of 14:38, 9 August 2020

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 $\mathrm{(C)}$.

See also

2009 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png