Difference between revisions of "2005 AMC 10A Problems/Problem 2"
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==Problem== | ==Problem== | ||
| − | For each pair of real numbers <math>a | + | For each pair of real numbers <math>a \neq b</math>, define the [[operation]] <math>\star</math> as |
<math> (a \star b) = \frac{a+b}{a-b} </math>. | <math> (a \star b) = \frac{a+b}{a-b} </math>. | ||
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<math> ((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \mathrm{(C)}</math> | <math> ((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \mathrm{(C)}</math> | ||
| − | == | + | ==Video Solution== |
| − | + | CHECK OUT Video Solution: https://youtu.be/5g_m3_nck8E | |
| − | + | ==See also== | |
| + | {{AMC10 box|year=2005|ab=A|num-b=1|num-a=3}} | ||
| − | + | [[Category:Introductory Number Theory Problems]] | |
| + | {{MAA Notice}} | ||
Revision as of 21:00, 30 October 2020
Contents
Problem
For each pair of real numbers 
, define the operation 
as
.
What is the value of 
?
Solution
Video Solution
CHECK OUT Video Solution: https://youtu.be/5g_m3_nck8E
See also
| 2005 AMC 10A (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. 
