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: | ||
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<cmath>2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5</cmath> | <cmath>2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5</cmath> | ||
| − | so our numbers check. Thus <math>104\sqrt{2}+468\sqrt{3}+144\sqrt{5}=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2</math>. Square rooting gives us <math>13\sqrt{2}+4\sqrt{3}+18\sqrt{5}</math> and our answer is <math>13\cdot4\cdot18=\boxed{936}</math> | + | so our numbers check. Thus <math>104\sqrt{2}+468\sqrt{3}+144\sqrt{5}+2006=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2</math>. Square rooting gives us <math>13\sqrt{2}+4\sqrt{3}+18\sqrt{5}</math> and our answer is <math>13\cdot4\cdot18=\boxed{936}</math> |
== See also == | == See also == | ||
| Line 92: | Line 96: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
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:
![\[a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}\]](http://latex.artofproblemsolving.com/5/6/7/56755e40101b654cd62aaaf4f16dc4760f70b9c1.png)
Squaring both sides yields:
![\[2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006\]](http://latex.artofproblemsolving.com/3/b/a/3ba9d82307df69713fdd1fffeaf5e555a12725c5.png)
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
![\[(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)\]](http://latex.artofproblemsolving.com/2/e/f/2ef72c62a6e6b8b5597eddd75764f62195081bdc.png)
we attempt to rewrite the radicand in this form:
![\[2006+2(52xy+234xz+72yz)\]](http://latex.artofproblemsolving.com/a/7/3/a7302c22b2e8ea1e719616b856e4bb1f31a2723a.png)
Factoring, we see that 
, 
, and 
. Setting 
, 
, and 
, we see that
![\[2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5\]](http://latex.artofproblemsolving.com/2/c/6/2c66274bf160f62ac98ceb67bcabbdb7dcd38774.png)
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. 
