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"

(→‎Problem)
m
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
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>.
  
−
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face <math> F </math> is greater than 1/6, the probability of obtaining the face opposite is less than 1/6, the probability of obtaining any one of the other four faces is 1/6, and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is <math>47/288</math>. Given that the [[probability]] of obtaining face <math> F </math>  is <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] positive integers, find <math> m+n. </math>
+
== Solution ==
 +
We begin by [[equate | equating]] the two expressions:
  
−
== 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:
  
−
For now, assume that face <math>F</math> has a 6, so the opposite face has a 1.  Let <math>A(n)</math> be the probability of rolling a number <math>n</math> on one die and let <math>B(n)</math> be the probability of rolling a number <math>n</math> on the other die. 7 can be obtained by rolling a <math>A(n)=2</math> and <math>B(n)=5</math>, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of <math>\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}</math>, totaling <math>4 \cdot \frac{1}{36} = \frac{1}{9}</math>. Subtracting all these probabilities from <math>\frac{47}{288}</math> leaves <math>\frac{15}{288}=\frac{5}{96}</math> chance of getting a 1 on die <math>A</math> and a 6 on die <math>B</math> or a 6 on die <math>A</math> and a 1 on die <math>B</math>:
+
1: <math> 2ab\sqrt{6} = 104\sqrt{6} </math>  
  
−
<math>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</math>
+
2: <math> 2ac\sqrt{10} = 468\sqrt{10} </math>
  
−
Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so
+
3: <math> 2bc\sqrt{15} = 144\sqrt{15} </math>  
  
−
:<math>A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}</math>
+
4: <math> 2a^2 + 3b^2 + 5c^2 = 2006 </math>  
−
:<math>A(1)\cdot A(6)=\frac{5}{192}</math>
 
  
−
Also, we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that the total probability must be <math>1</math>, so:
+
Solving the first three equations gives:  
  
−
:<math>A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}</math>
+
<math> ab = 52 </math>  
−
:<math>A(1)+A(6)=\frac{1}{3}</math>
 
  
−
Combining the equations:
+
<math> ac = 234 </math>
  
−
:<math>A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}</math>
+
<math> bc = 72 </math>  
  
−
:<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
+
Multiplying these equations gives:  
  
−
:<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
+
<math> (abc)^2 = 52 \cdot 234 \cdot 72</math>
  
−
:<math>192 A(6)^2 - 64 A(6) + 5 = 0</math>
+
<math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
  
−
:<math>A(6)=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192}</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 yield the third variable. Doing so yields:  
  
−
:<math>A(6)=\frac{64\pm16}{384}</math>
+
<math>a=13</math>
  
−
:<math>A(6)=\frac{5}{24}, \frac{1}{8}</math>
+
<math>b=4</math>
  
−
We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>.  Therefore, it has to be <math>\frac{5}{24}</math> and the answer is <math>5+24=29</math>.
+
<math>c=18</math>
  
−
Note also that the initial assumption that face <math>F</math> was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was.
+
Which clearly fits the fourth equation:
 +
<math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math>
  
 
== See also ==
 
== See also ==

Revision as of 18:28, 25 September 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:

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

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

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

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=16994"