Difference between revisions of "2006 AIME I Problems/Problem 5"
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Which clearly fits the fourth equation: | Which clearly fits the fourth equation: | ||
| − | <math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math> | + | <math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math> |
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== See also == | == See also == | ||
* [[2006 AIME I Problems]] | * [[2006 AIME I Problems]] | ||
Revision as of 23:19, 30 June 2006
Problem
The number 
can be written as 
where 
and 
are positive integers. Find 
Solution
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 yeild the third variable. Doing so yeilds:
Which clearly fits the fourth equation:
