Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AIME I Problems/Problem 5"

m
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
The number <center><math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math></center> 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 integers. Find <math> a\cdot b\cdot c.  </math>
+
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 integers. Find <math> a\cdot b\cdot c.  </math>
  
  
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
 +
<math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math>
  
 +
Squaring both sides yeilds:
  
 +
<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>
  
 +
Since <math>a</math>, <math>b</math>, and <math>c</math> are integers:
 +
 +
1: <math> 2ab\sqrt{6} = 104\sqrt{6} </math>
 +
 +
2: <math> 2ac\sqrt{10} = 468\sqrt{10} </math>
 +
 +
3: <math> 2bc\sqrt{15} = 144\sqrt{15} </math>
 +
 +
4: <math> 2a^2 + 3b^2 + 5c^2 = 2006 </math>
 +
 +
Solving the first three equations gives:
 +
 +
<math> ab = 52 </math>
 +
 +
<math> ac = 234 </math>
 +
 +
<math> bc = 72 </math>
 +
 +
Multiplying these equations gives:
 +
 +
<math> (abc)^2 = 52 \cdot 234 \cdot 72</math>
 +
 +
<math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
 +
 +
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:
 +
 +
<math>a=13</math>
 +
 +
<math>b=4</math>
 +
 +
<math>c=18</math>
 +
 +
Which clearly fits the fourth equation:
 +
<math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math>
 +
 +
--[[User:Xantos C. Guin|Xantos C. Guin]] 17:15, 30 June 2006 (EDT)
  
 
== See also ==
 
== See also ==
 
* [[2006 AIME I Problems]]
 
* [[2006 AIME I Problems]]

Revision as of 16:15, 30 June 2006

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

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

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: $2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006$

--Xantos C. Guin 17:15, 30 June 2006 (EDT)

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_5&oldid=5618"