2006 AIME I Problems/Problem 5


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


we attempt to rewrite the radicand in this form:


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


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 (ProblemsAnswer KeyResources)
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. AMC logo.png