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2006 AIME I Problems/Problem 5

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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:

$2ab\sqrt{6} &=& 104\sqrt{6} \\ 2ac\sqrt{10} &=& 468\sqrt{10} \\ 2bc\sqrt{15} &=& 144\sqrt{15}\\ 2a^2 ...$

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 (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Category: Intermediate Algebra Problems

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