Difference between revisions of "2016 AIME II Problems/Problem 4"
(→Solution) |
m (→Solution) |
||
Line 2: | Line 2: | ||
==Solution== | ==Solution== | ||
− | By counting the number of green cubes <math>2</math> different ways, we have <math>12a=20b</math>, or <math>a=\dfrac{5}{3} b</math>. Notice that there are only <math>3</math> possible colors for unit cubes, so for each of the <math>1 \times b \times c</math> layers, there are <math>bc-21</math> yellow cubes, and similarly there are <math>ac-45</math> red cubes in each of the <math>1 \times a \times | + | By counting the number of green cubes <math>2</math> different ways, we have <math>12a=20b</math>, or <math>a=\dfrac{5}{3} b</math>. Notice that there are only <math>3</math> possible colors for unit cubes, so for each of the <math>1 \times b \times c</math> layers, there are <math>bc-21</math> yellow cubes, and similarly there are <math>ac-45</math> red cubes in each of the <math>1 \times a \times b</math> layers. Therefore, we have <math>a(bc-21)=25b</math> and <math>b(ac-45)=9a</math>. We check a few small values of <math>a,b</math> and solve for <math>c</math>, checking <math>(a,b)=(5,3)</math> gives <math>c=12</math> with a volume of <math>180</math>, <math>(a,b)=(10,6)</math> gives <math>c=6</math> with a volume of <math>360</math>, and <math>(a,b)=(15,9)</math> gives <math>c=4</math>, with a volume of <math>540</math>. Any higher <math>(a,b)</math> will <math>ab>180</math>, so therefore, the minimum volume is <math>\boxed{180}</math>. |
Solution by Shaddoll | Solution by Shaddoll |
Revision as of 13:06, 15 January 2017
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
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 |