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2006 AIME I Problems/Problem 5

Revision as of 02:26, 12 July 2006 by MCrawford (talk | contribs) (adding a note as to the method used at the beginning of the problem)

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 $a\cdot b\cdot c.$



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 yeilds:

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

Since $a$, $b$, and $c$ are integers:

1: $2ab\sqrt{6} = 104\sqrt{6}$

2: $2ac\sqrt{10} = 468\sqrt{10}$

3: $2bc\sqrt{15} = 144\sqrt{15}$

4: $2a^2 + 3b^2 + 5c^2 = 2006$

Solving the first three equations gives:

$ab = 52$

$ac = 234$

$bc = 72$

Multiplying these equations gives:

$(abc)^2 = 52 \cdot 234 \cdot 72$

$abc = \sqrt{52 \cdot 234 \cdot 72} = 936$

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:

$a=13$

$b=4$

$c=18$

Which clearly fits the fourth equation: $2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006$

See also

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