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

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== Problem ==
 
== Problem ==
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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 integers. Find <math> a\cdot b\cdot c.  </math>
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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>.
  
  
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<math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
 
<math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
  
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If it was required to solve for each variable, dividing the product of the three variables by the product of any two variables would yeild the third variable. Doing so yeilds:  
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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>a=13</math>
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== See also ==
 
== See also ==
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* [[2006 AIME I Problems/Problem 4 | Previous problem]]
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{{AIME box|year=2006|n=I|num-b=4|num-a=6}}
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* [[2006 AIME I Problems/Problem 6 | Next problem]]
 
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* [[2006 AIME I Problems]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 20:17, 11 March 2007

Problem

The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers. Find $abc$.



Solution

We begin by equating the two expressions:

$a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$

Squaring both sides yeilds:

$2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006$

Since $a$, $b$, and $c$ are integers:

1: $2ab\sqrt{6} = 104\sqrt{6}$

2: $2ac\sqrt{10} = 468\sqrt{10}$

3: $2bc\sqrt{15} = 144\sqrt{15}$

4: $2a^2 + 3b^2 + 5c^2 = 2006$

Solving the first three equations gives:

$ab = 52$

$ac = 234$

$bc = 72$

Multiplying these equations gives:

$(abc)^2 = 52 \cdot 234 \cdot 72$

$abc = \sqrt{52 \cdot 234 \cdot 72} = 936$

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:

$a=13$

$b=4$

$c=18$

Which clearly fits the fourth equation: $2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006$

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
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