Difference between revisions of "2007 AMC 10B Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
Note that | Note that | ||
| − | <math>(a \star b) - (b \star a) = (a+b)b - (b+a)a | + | <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>. |
| − | = (a+b)(b-a) | + | |
| − | = b^2 - a^2 | + | ~MathFun1000 (Minor Edits) |
| − | = 5^2 - 3^2 = \boxed{\textbf{(E) }16}</math> | ||
| − | |||
==See Also== | ==See Also== | ||
Revision as of 10:30, 7 March 2022
Contents
Problem
Define the operation 
by 
What is 
Solution 1
Substitute and simplify.
![\[(3+5)5 - (5+3)3 = (3+5)2 = (8)2 = \boxed{\textbf{(E) }16}\]](http://latex.artofproblemsolving.com/1/2/6/126b27b43d8364fd141008df3e65dc050250f3fe.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. 
