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− | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 5]] |
− | 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>.
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− | == Solution ==
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− | We begin by [[equate | equating]] the two expressions:
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− | <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math>
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− | Squaring both sides yeilds:
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− | <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>
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− | Since <math>a</math>, <math>b</math>, and <math>c</math> are integers:
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− | 1: <math> 2ab\sqrt{6} = 104\sqrt{6} </math>
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− | 2: <math> 2ac\sqrt{10} = 468\sqrt{10} </math>
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− | 3: <math> 2bc\sqrt{15} = 144\sqrt{15} </math>
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− | 4: <math> 2a^2 + 3b^2 + 5c^2 = 2006 </math>
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− | Solving the first three equations gives:
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− | <math> ab = 52 </math>
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− | <math> ac = 234 </math>
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− | <math> bc = 72 </math>
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− | Multiplying these equations gives:
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− | <math> (abc)^2 = 52 \cdot 234 \cdot 72</math>
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− | <math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
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− | 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:
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− | <math>a=13</math>
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− | <math>b=4</math>
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− | <math>c=18</math>
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− | Which clearly fits the fourth equation:
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− | <math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math>
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− | == See also ==
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− | {{AIME box|year=2006|n=II|num-b=4|num-a=6}}
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− | [[Category:Intermediate Combinatorics Problems]]
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