Difference between revisions of "2005 AMC 12B Problems/Problem 23"
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Then, <math>\log(s)=z</math> implies <math>\log(10s) = z+1= \log(t)</math>,so <math>t=10s</math>. | Then, <math>\log(s)=z</math> implies <math>\log(10s) = z+1= \log(t)</math>,so <math>t=10s</math>. | ||
Therefore, <math>x^3 + y^3 = s*\frac{3t-s^2}{2} = s(15s-\frac{s^2}{2})</math>. | Therefore, <math>x^3 + y^3 = s*\frac{3t-s^2}{2} = s(15s-\frac{s^2}{2})</math>. | ||
− | Since <math>s = 10^z</math>, we find that <math>x^3 + y^3 = 15\times10^2z - (1/2)\times10^3z</math>. | + | Since <math>s = 10^z</math>, we find that <math>x^3 + y^3 = 15\times10^(2z) - (1/2)\times10^(3z)</math>. |
Thus, <math>a+b = \boxed{\frac{29}{2}}</math>. | Thus, <math>a+b = \boxed{\frac{29}{2}}</math>. | ||
Revision as of 16:33, 22 April 2016
Contents
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 1
Let and
.
Then,
implies
,so
.
Therefore,
.
Since
, we find that
.
Thus,
.
Solution 2
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.