Difference between revisions of "2006 AIME I Problems/Problem 5"
The Anomaly (talk | contribs) (→Solution) |
(the second half really isn't necessary, and taking sqrts aren't hard anyway..) |
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We begin by [[equate | equating]] the two expressions: | We begin by [[equate | equating]] the two expressions: | ||
| − | < | + | <cmath> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</cmath> |
| − | Squaring both sides | + | Squaring both sides yields: |
| − | < | + | <cmath> 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 </cmath> |
| − | Since <math>a</math>, <math>b</math>, and <math>c</math> are integers: | + | Since <math>a</math>, <math>b</math>, and <math>c</math> are integers, we can match coefficients: |
| − | + | <cmath> 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 </cmath> | |
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Solving the first three equations gives: | Solving the first three equations gives: | ||
| + | <cmath>\begin{eqnarray*}ab &=& 52\\ | ||
| + | ac &=& 234\\ | ||
| + | bc &=& 72 \end{eqnarray*}</cmath> | ||
| − | + | Multiplying these equations gives <math> (abc)^2 = 52 \cdot 234 \cdot 72 = 2^63^413^2 \Longrightarrow abc = \boxed{936}</math>. | |
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| − | Multiplying these equations gives | ||
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| − | <math> (abc)^2 = 52 \cdot 234 \cdot 72 | ||
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| + | <!-- | ||
Since this is the AIME and you do not have a calculator solving <math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math> might prove difficult. | Since this is the AIME and you do not have a calculator solving <math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math> might prove difficult. | ||
So instead use the three equations given above. | So instead use the three equations given above. | ||
| Line 82: | Line 71: | ||
<math>abc=\boxed{936}</math> | <math>abc=\boxed{936}</math> | ||
| − | + | --> | |
== See also == | == See also == | ||
{{AIME box|year=2006|n=I|num-b=4|num-a=6}} | {{AIME box|year=2006|n=I|num-b=4|num-a=6}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 17:42, 28 November 2007
Problem
The number 
can be written as 
where 
and 
are positive integers. Find 
.
Solution
We begin by equating the two expressions:
![\[a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}\]](http://latex.artofproblemsolving.com/5/6/7/56755e40101b654cd62aaaf4f16dc4760f70b9c1.png)
Squaring both sides yields:
![\[2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006\]](http://latex.artofproblemsolving.com/3/b/a/3ba9d82307df69713fdd1fffeaf5e555a12725c5.png)
Since 
, 
, and 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:
Multiplying these equations gives 
.
See also
| 2006 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
