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2006 AIME I Problems/Problem 5: Difference between revisions

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== Problem ==
== Problem ==
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>
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 1 ==
We begin by [[equate | equating]] the two expressions:


<cmath> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</cmath>


Squaring both sides yields:


<cmath> 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 </cmath>


== Solution ==
Since <math>a</math>, <math>b</math>, and <math>c</math> are integers, we can match coefficients:  
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>
<cmath>  
\begin{align*}
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
\end{align*}
</cmath>  


Squaring both sides yeilds:  
Solving the first three equations gives:  
<cmath>\begin{eqnarray*}ab &=& 52\\
ac &=& 234\\
bc &=& 72 \end{eqnarray*}</cmath>


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


Since <math>a</math>, <math>b</math>, and <math>c</math> are integers:
<!--
Since this is the AIME and you do not have a calculator solving <math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math> might prove difficult.
So instead use the three equations given above.


1: <math> 2ab\sqrt{6} = 104\sqrt{6} </math>  
<math> ab = 52 </math>  


2: <math> 2ac\sqrt{10} = 468\sqrt{10} </math>
<math> ac = 234 </math>  


3: <math> 2bc\sqrt{15} = 144\sqrt{15} </math>  
<math> bc = 72 </math>  


4: <math> 2a^2 + 3b^2 + 5c^2 = 2006 </math>  
Thus <math> a = 52/b = 234/c </math>


Solving the first three equations gives:  
<math> 52c = 234b </math>
 
<math> c = 234/52b </math>
 
<math> c = 9/2b </math>
 
Plugging into last equation leads to:
 
<math> 9/2b^2 = 72 </math>


<math> ab = 52 </math>  
<math> b = 4 </math>


<math> ac = 234 </math>
Plugging into others you get


<math> bc = 72 </math>  
<math>a=13</math>


Multiplying these equations gives:
<math>b=4</math>


<math> (abc)^2 = 52 \cdot 234 \cdot 72</math>
<math>c=18</math>


<math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
Much easier than taking crazy square roots without a calculator!


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:  
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>
Line 49: Line 73:
Which clearly fits the fourth equation:
Which clearly fits the fourth equation:
<math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math>
<math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math>
<math>abc=\boxed{936}</math>
-->
== 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>
we attempt to rewrite the radicand in this form:
<cmath>2006+2(52xy+234xz+72yz)</cmath>
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
<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}+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 ==
* [[2006 AIME I Problems/Problem 4 | Previous problem]]
{{AIME box|year=2006|n=I|num-b=4|num-a=6}}
* [[2006 AIME I Problems/Problem 6 | Next problem]]
* [[2006 AIME I Problems]]


[[Category:Intermediate Algebra Problems]]
[[Category:Intermediate Algebra Problems]]
{{MAA Notice}}

Latest revision as of 17:24, 10 March 2015

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:

\begin{align*} 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 \end{align*}

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

\[2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5\]

so our numbers check. Thus $104\sqrt{2}+468\sqrt{3}+144\sqrt{5}+2006=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2$. 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
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