Difference between revisions of "2006 AIME A Problems/Problem 5"
(cut down on some of the extraneous mathwork, +box) |
(→Solution) |
||
Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
− | + | We begin by [[equate | equating]] the two expressions: | |
− | <math> | + | <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: | |
− | :<math> | + | 1: <math> 2ab\sqrt{6} = 104\sqrt{6} </math> |
− | |||
− | + | 2: <math> 2ac\sqrt{10} = 468\sqrt{10} </math> | |
− | :<math> | + | 3: <math> 2bc\sqrt{15} = 144\sqrt{15} </math> |
− | :<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 yield the third variable. Doing so yields: | ||
+ | |||
+ | <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> | ||
== See also == | == See also == |
Revision as of 14:08, 25 September 2007
Problem
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face 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
. Given that the probability of obtaining face
is
where
and
are relatively prime positive integers, find
Solution
We begin by equating the two expressions:
Squaring both sides yeilds:
Since ,
, and
are integers:
1:
2:
3:
4:
Solving the first three equations gives:
Multiplying these equations gives:
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:
Which clearly fits the fourth equation:
See also
2006 AIME II (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 |