Difference between revisions of "2018 AMC 10B Problems/Problem 4"
(→Solution 3) |
(→Solution 3) |
||
| Line 23: | Line 23: | ||
==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 | + | 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>. |
==See Also== | ==See Also== | ||
Revision as of 23:42, 13 March 2018
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. 
