Difference between revisions of "2015 AMC 12B Problems/Problem 23"
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+ | {{duplicate|[[2015 AMC 12B Problems|2015 AMC 12B #23]] and [[2015 AMC 10B Problems|2015 AMC 10B #25]]}} | ||
==Problem== | ==Problem== | ||
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A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible? | A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible? | ||
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This is also AMC 10B Problem 25, but the pages are separate. Someone should fix this. | This is also AMC 10B Problem 25, but the pages are separate. Someone should fix this. | ||
EDIT: fixed it, but someone help with the link | EDIT: fixed it, but someone help with the link | ||
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EDIT #2: fixed all | EDIT #2: fixed all | ||
Revision as of 23:17, 17 January 2021
- The following problem is from both the 2015 AMC 12B #23 and 2015 AMC 10B #25, so both problems redirect to this page.
Contents
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
We need Since , we get . Thus . From the second equation we see that . Thus .
- If we need . We get five roots
- If we need . We get three roots .
- If we need , which is the same as . We get only one root (corresponding to ) .
- If we need . Then . We get one root .
Thus, there are solutions.
Solution 2
The surface area is , and the volume is , so equating the two yields
Divide both sides by to obtain
First consider the bound of the variable . Since we have , or .
Also note that , hence . 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 , we get , so . We find the solutions , , , , , for a total of solutions.
When , we get , so . We find the solutions , , , for a total of solutions.
When , we get , so . The only solution in this case is .
When , is forced to be , and thus .
Thus, there are solutions.
Note
This is also AMC 10B Problem 25, but the pages are separate. Someone should fix this. EDIT: fixed it, but someone help with the link
EDIT #2: fixed all
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 10 Problems and Solutions |
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.