Difference between revisions of "2015 AMC 10B Problems/Problem 25"
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When <math>a=5</math>, we get <math>\frac{1}{b}+\frac{1}{c}=\frac{3}{10}</math>, so <math>b=5, 6</math>. The only solution in this case is <math>(a, b, c)=(5, 5, 10)</math>. | When <math>a=5</math>, we get <math>\frac{1}{b}+\frac{1}{c}=\frac{3}{10}</math>, so <math>b=5, 6</math>. The only solution in this case is <math>(a, b, c)=(5, 5, 10)</math>. | ||
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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, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions. | Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions. | ||
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+ | Minor Edit by Snow52 | ||
==See Also== | ==See Also== |
Revision as of 16:41, 12 September 2020
Contents
[hide]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 1
The surface area is , the volume is , so .
Divide both sides by , we have:
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
Simplification of Solution 1
The surface area is , the volume is , so .
Divide both sides by , we have: First consider the bound of the variable . Since we have , or .
Also note that , we have . Thus, , so .
So we have or .
We can say , where .
Notice that This is our key step. Then we can say , . If we clear the fraction about b and c (do the math), our immediate result is that . Realize also that .
Now go through cases for and you end up with the same result. However, now you don't have to guess solutions. For example, when , then and .
- minor edit by Williamgolly, minor edit by Tiblis
Solution 2
We needSince , 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 3 (Basically the exact same as Solution 1)
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.
Minor Edit by Snow52
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 |
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