Difference between revisions of "2007 AMC 10B Problems/Problem 2"
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<math>\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 16</math> | <math>\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 16</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Substitute and simplify. | Substitute and simplify. | ||
| − | <cmath>(3+5)5 - (5+3)3 = (3+5)2 = ( | + | <cmath>(3+5)5 - (5+3)3 = (3+5)2 = 8\cdot2 = \boxed{\textbf{(E) }16}</cmath> |
| + | |||
| + | ==Solution 2== | ||
| + | Note that | ||
| + | <math>(a \star b) - (b \star a) = (a+b)b - (b+a)a= (a+b)(b-a)= b^2 - a^2</math>. We can substitute <math>a=3</math> and <math>b=5</math> to get <math>5^2 - 3^2 = \boxed{\textbf{(E) }16}</math>. | ||
| + | |||
| + | ~MathFun1000 (Minor Edits) | ||
==See Also== | ==See Also== | ||
| − | {{AMC10 box|year=2007|ab=B|num-b=1|num-a= | + | {{AMC10 box|year=2007|ab=B|num-b=1|num-a=3}} |
| + | {{MAA Notice}} | ||
Latest revision as of 09:33, 7 March 2022
Contents
Problem
Define the operation 
by 
What is 
Solution 1
Substitute and simplify.
![\[(3+5)5 - (5+3)3 = (3+5)2 = 8\cdot2 = \boxed{\textbf{(E) }16}\]](http://latex.artofproblemsolving.com/1/7/a/17a3171fd336e8778f58d182a02e1600a87f8bcb.png)
Solution 2
Note that
. We can substitute 
and 
to get 
.
~MathFun1000 (Minor Edits)
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
