Difference between revisions of "2005 AMC 12B Problems/Problem 23"
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<cmath>= 10^z\left(10^{z + 1} - \frac{10^{2z} - 10^{z + 1}}{2}\right)</cmath> | <cmath>= 10^z\left(10^{z + 1} - \frac{10^{2z} - 10^{z + 1}}{2}\right)</cmath> | ||
<cmath>= 10^{2z + 1} - \frac{10^{3z} - 10^{2z + 1}}{2}</cmath> | <cmath>= 10^{2z + 1} - \frac{10^{3z} - 10^{2z + 1}}{2}</cmath> | ||
− | <cmath>= -\frac{1}{2} \cdot 10^{3z} + \frac{3}{2} \cdot 10^{2z}</cmath> | + | <cmath>= -\frac{1}{2} \cdot 10^{3z} + \frac{3}{2} \cdot 10^{2z + 1}</cmath> |
<cmath>= -\frac{1}{2} \cdot 10^{3z} + 15 \cdot 10^{2z}.</cmath> | <cmath>= -\frac{1}{2} \cdot 10^{3z} + 15 \cdot 10^{2z}.</cmath> | ||
Revision as of 14:04, 30 July 2015
Problem
Let be the set of ordered triples of real numbers for which
There are real numbers and such that for all ordered triples in we have What is the value of
Solution
Let and . Then, implies ,so . Therefore, . Since , we find that . Thus, .
Alternate Solution
First, remember that factors to . By the givens, and . These can be used to find :
Therefore,
It follows that and , thus
See Also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.