Difference between revisions of "2015 AMC 12B Problems/Problem 23"
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The surface area is <math>2(ab+bc+ca)</math>, the volumn is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>. | The surface area is <math>2(ab+bc+ca)</math>, the volumn is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>. | ||
Revision as of 23:40, 3 March 2015
Problem
A rectangular box measures , where , , and are integers and . The volume and the surface area of the box are numerically equal. How many ordered triples are possible?
Solution
The surface area is , the volumn is , so .
Divide both sides by , we get that
First consider the bound of the variable . Since we have , or .
Also note that , we have . Thus, , so .
So we have or .
Before the casework, let's consider the possible range for if .
From , we have . From , we have . Thus
When , , so . The solutions we find are , for a total of solutions.
When , , so . The solutions we find are , for a total of solutions.
When , , so . The only solution in this case is .
When , is forced to be , and thus .
Thus, there are solutions
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.