Difference between revisions of "2006 AIME I Problems/Problem 5"
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Since <math>a</math>, <math>b</math>, and <math>c</math> are integers, we can match coefficients: | Since <math>a</math>, <math>b</math>, and <math>c</math> are integers, we can match coefficients: | ||
− | <cmath> 2ab\sqrt{6} &= | + | <cmath> |
− | 2ac\sqrt{10} &= | + | \begin{align*} |
− | 2bc\sqrt{15} &= | + | 2ab\sqrt{6} &= 104\sqrt{6} \\ |
− | 2a^2 + 3b^2 + 5c^2 &= | + | 2ac\sqrt{10} &=468\sqrt{10} \\ |
+ | 2bc\sqrt{15} &=144\sqrt{15}\\ | ||
+ | 2a^2 + 3b^2 + 5c^2 &=2006 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
Solving the first three equations gives: | Solving the first three equations gives: |
Latest revision as of 18:24, 10 March 2015
Contents
Problem
The number can be written as where and are positive integers. Find .
Solution 1
We begin by equating the two expressions:
Squaring both sides yields:
Since , , and are integers, we can match coefficients:
Solving the first three equations gives:
Multiplying these equations gives .
Solution 2
We realize that the quantity under the largest radical is a perfect square and attempt to rewrite the radicand as a square. Start by setting , , and . Since
we attempt to rewrite the radicand in this form:
Factoring, we see that , , and . Setting , , and , we see that
so our numbers check. Thus . Square rooting gives us and our answer is
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.