Difference between revisions of "2018 AMC 10B Problems/Problem 4"
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==Solution 2== | ==Solution 2== | ||
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+ | Simply use guess and check to find that the dimensions are <math>4</math> by <math>6</math> by <math>12</math>. Therefore, the answer is <math>4 + 6 + 12 = 22</math>. <math>\boxed{B}</math> | ||
==Solution 3== | ==Solution 3== | ||
− | + | If you find the GCD of <math>24</math>, <math>48</math>, and <math>72</math> you get your first number, <math>12</math>. After this, do <math>48 \div 12</math> and <math>72 \div 12</math> to get <math>4</math> and <math>6</math>, the other 2 numbers. When you add up your <math>3</math> numbers, you get <math>22</math> which is <math>\boxed{B}</math>. | |
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==See Also== | ==See Also== |
Revision as of 09:42, 3 March 2019
Problem
A three-dimensional rectangular box with dimensions , , and has faces whose surface areas are , , , , , and square units. What is + + ?
Solution 1
Let be the length of the shortest dimension and be the length of the longest dimension. Thus, , , and . Divide the first two equations to get . Then, multiply by the last equation to get , giving . Following, and .
The final answer is .
Solution 2
Simply use guess and check to find that the dimensions are by by . Therefore, the answer is .
Solution 3
If you find the GCD of , , and you get your first number, . After this, do and to get and , the other 2 numbers. When you add up your numbers, you get which is .
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.