Difference between revisions of "2006 AIME A Problems/Problem 5"

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== Problem ==
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#REDIRECT [[2006 AIME I Problems/Problem 5]]
The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </math> where <math> a, b, </math> and <math> c </math> are [[positive]] [[integer]]s. Find <math>abc</math>.
 
 
 
== Solution ==
 
We begin by [[equate | equating]] the two expressions:
 
 
 
<math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math>
 
 
 
Squaring both sides yeilds:
 
 
 
<math> 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 </math>
 
 
 
Since <math>a</math>, <math>b</math>, and <math>c</math> are integers:
 
 
 
1: <math> 2ab\sqrt{6} = 104\sqrt{6} </math>
 
 
 
2: <math> 2ac\sqrt{10} = 468\sqrt{10} </math>
 
 
 
3: <math> 2bc\sqrt{15} = 144\sqrt{15} </math>
 
 
 
4: <math> 2a^2 + 3b^2 + 5c^2 = 2006 </math>
 
 
 
Solving the first three equations gives:
 
 
 
<math> ab = 52 </math>
 
 
 
<math> ac = 234 </math>
 
 
 
<math> bc = 72 </math>
 
 
 
Multiplying these equations gives:
 
 
 
<math> (abc)^2 = 52 \cdot 234 \cdot 72</math>
 
 
 
<math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
 
 
 
If it was required to solve for each variable, dividing the product of the three variables by the product of any two variables would yield the third variable. Doing so yields:
 
 
 
<math>a=13</math>
 
 
 
<math>b=4</math>
 
 
 
<math>c=18</math>
 
 
 
Which clearly fits the fourth equation:
 
<math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math>
 
 
 
== See also ==
 
{{AIME box|year=2006|n=II|num-b=4|num-a=6}}
 
 
 
[[Category:Intermediate Combinatorics Problems]]
 

Latest revision as of 17:41, 26 June 2009