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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| + | <math>abc=\boxed{936}</math> | ||
== See also == | == See also == | ||
Revision as of 09:01, 13 November 2007
Problem
The number 
can be written as 
where 
and 
are 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 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 | ||
