2006 AIME I Problems/Problem 5

Revision as of 17:42, 28 November 2007 by Azjps (talk | contribs) (the second half really isn't necessary, and taking sqrts aren't hard anyway..)

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

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