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"

(→‎Solution)
(→‎Solution)
Line 34: Line 34:
  
 
<math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
 
<math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
 +
 +
ALTERNATIVELY:
 +
 +
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.
 +
 +
<math> ab = 52 </math>
 +
 +
<math> ac = 234 </math>
 +
 +
<math> bc = 72 </math>
 +
 +
Thus <math> a = 52/b = 234/c </math>
 +
 +
<math> 52c = 234b </math>
 +
 +
<math> c = 234/52b </math>
 +
 +
<math> c = 9/2b </math>
 +
 +
Plugging into last equation leads to:
 +
 +
<math> 9/2b^2 = 72 </math>
 +
 +
<math> b = 4 </math>
 +
 +
Plugging into others you get
 +
 +
<math>a=13</math>
 +
 +
<math>b=4</math>
 +
 +
<math>c=18</math>
 +
 +
Much easier than taking crazy square roots without a calculator!
  
 
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:  
 
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:  

Revision as of 23:43, 22 November 2007

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 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:

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

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

ALTERNATIVELY:

Since this is the AIME and you do not have a calculator solving $abc = \sqrt{52 \cdot 234 \cdot 72} = 936$ might prove difficult. So instead use the three equations given above.

$ab = 52$

$ac = 234$

$bc = 72$

Thus $a = 52/b = 234/c$

$52c = 234b$

$c = 234/52b$

$c = 9/2b$

Plugging into last equation leads to:

$9/2b^2 = 72$

$b = 4$

Plugging into others you get

$a=13$

$b=4$

$c=18$

Much easier than taking crazy square roots without a calculator!

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:

$a=13$

$b=4$

$c=18$

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

$abc=\boxed{936}$

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_5&oldid=19845"