Difference between revisions of "2002 AMC 12P Problems/Problem 3"
(→See also) |
(→Problem) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | + | The dimensions of a rectangular box in inches are all positive integers and the volume of the box is <math>2002</math> in<math>^3</math>. Find the minimum possible sum of the three dimensions. | |
− | <math> \ | + | <math>\text{(A) }36 \qquad \text{(B) }38 \qquad \text{(C) }42 \qquad \text{(D) }44 \qquad \text{(E) }92</math> |
== Solution == | == Solution == |
Revision as of 23:40, 29 December 2023
Problem
The dimensions of a rectangular box in inches are all positive integers and the volume of the box is in. Find the minimum possible sum of the three dimensions.
Solution
If , then . Since , must be to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of .
See also
2002 AMC 12P (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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.