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

(the second half really isn't necessary, and taking sqrts aren't hard anyway..)
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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>.
 
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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== Solution ==
+
== Solution 1 ==
 
We begin by [[equate | equating]] the two expressions:
 
We begin by [[equate | equating]] the two expressions:
  
Line 72: Line 72:
 
<math>abc=\boxed{936}</math>
 
<math>abc=\boxed{936}</math>
 
-->
 
-->
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 +
== 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 <math>x=\sqrt{2}</math>, <math>y=\sqrt{3}</math>, and <math>z=\sqrt{5}</math>.  Since
 +
 +
<cmath>(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)</cmath>
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 +
we attempt to rewrite the radicand in this form:
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<cmath>2006+2(52xy+234xz+72yz)</cmath>
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Factoring, we see that <math>52=13\cdot4</math>, <math>234=13\cdot18</math>, and <math>72=4\cdot18</math>. Setting <math>p=13</math>, <math>q=4</math>, and <math>r=18</math>, we see that
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<cmath>2006=13^2\cdotx^2+4^2\cdoty^2+18^2\cdotz^2=169\cdot2+16\cdot3+324\cdot5</cmath>
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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>
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== See also ==
 
== See also ==
 
{{AIME box|year=2006|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2006|n=I|num-b=4|num-a=6}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 00:35, 13 July 2012

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 1

We begin by equating the two expressions:

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

Squaring both sides yields:

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

Since $a$, $b$, and $c$ are integers, we can match coefficients: 

\[2ab\sqrt{6} &=& 104\sqrt{6} \\
 2ac\sqrt{10} &=& 468\sqrt{10} \\
 2bc\sqrt{15} &=& 144\sqrt{15}\\
 2a^2 + 3b^2 + 5c^2 &=& 2006\] (Error compiling LaTeX. Unknown error_msg)

Solving the first three equations gives: \begin{eqnarray*}ab &=& 52\\ ac &=& 234\\ bc &=& 72 \end{eqnarray*}

Multiplying these equations gives $(abc)^2 = 52 \cdot 234 \cdot 72 = 2^63^413^2 \Longrightarrow abc = \boxed{936}$.


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 $x=\sqrt{2}$, $y=\sqrt{3}$, and $z=\sqrt{5}$. Since

\[(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)\]

we attempt to rewrite the radicand in this form:

\[2006+2(52xy+234xz+72yz)\]

Factoring, we see that $52=13\cdot4$, $234=13\cdot18$, and $72=4\cdot18$. Setting $p=13$, $q=4$, and $r=18$, we see that 

\[2006=13^2\cdotx^2+4^2\cdoty^2+18^2\cdotz^2=169\cdot2+16\cdot3+324\cdot5\] (Error compiling LaTeX. Unknown error_msg)

so our numbers check. Thus $104\sqrt{2}+468\sqrt{3}+144\sqrt{5}=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2$. Square rooting gives us $13\sqrt{2}+4\sqrt{3}+18\sqrt{5}$ and our answer is $13\cdot4\cdot18=\boxed{936}$

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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