Difference between revisions of "2015 AMC 10B Problems/Problem 25"
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When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>. | When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>. | ||
− | Thus, | + | Thus, our answer is <math>\boxed{\textbf{(B)}\;10}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|after=Last Problem|num-b=24}} | {{AMC10 box|year=2015|ab=B|after=Last Problem|num-b=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:00, 4 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, our answer is
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
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