Difference between revisions of "2016 AIME II Problems/Problem 4"
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Revision as of 19:16, 18 August 2020
Contents
Problem
An 
rectangular box is built from 
unit cubes. Each unit cube is colored red, green, or yellow. Each of the 
layers of size 
parallel to the 
faces of the box contains exactly 
red cubes, exactly 
green cubes, and some yellow cubes. Each of the 
layers of size 
parallel to the 
faces of the box contains exactly 
green cubes, exactly 
yellow cubes, and some red cubes. Find the smallest possible volume of the box.
Solution
By counting the number of green cubes 
different ways, we have 
, or 
. Notice that there are only 
possible colors for unit cubes, so for each of the layers, there are 
yellow cubes, and similarly there are 
red cubes in each of the 
layers. Therefore, we have 
and 
. We check a few small values of 
and solve for 
, checking 
gives 
with a volume of 
, 
gives 
with a volume of 
, and 
gives 
, with a volume of 
. Any higher 
will 
, so therefore, the minimum volume is 
.
Solution by Shaddoll
Solution 2
The total number of green cubes is given by 
.
Let 
be the number of red cubes on each one of the layers then the total number of red cubes is 
. Substitute 
gives 
.
Repeating the procedure on the number of yellow cubes 
on each of the layers gives 
.
Therefore 
and 
. Multiplying yields 
.
Since 
is fixed, 
is minimized when is maximized, which occurs when , are minimized (since each of 
, 
is fixed). Thus 
~ Nafer
See also
| 2016 AIME II (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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
