Difference between revisions of "2016 AIME II Problems/Problem 4"
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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 c</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>. | 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 c</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 | + | == Solution 2 == |
+ | |||
+ | The total number of green cubes is given by <math>12a=20b\Longrightarrow a=\frac{5}{3}b</math>. | ||
+ | |||
+ | Let <math>r</math> be the number of red cubes on each one of the <math>b</math> layers then the total number of red cubes is <math>9a=br</math>. Substitute <math>a=\frac{5}{3}b</math> gives <math>r=15</math>. | ||
− | == | + | Repeating the procedure on the number of yellow cubes <math>y</math> on each of the <math>a</math> layers gives <math>y=15</math>. |
+ | |||
+ | Therefore <math>bc=9+12+15=36</math> and <math>ac=15+20+25=60</math>. Multiplying yields <math>abc^2=2160</math>. | ||
+ | |||
+ | Since <math>abc^2</math> is fixed, <math>abc</math> is minimized when <math>c</math> is maximized, which occurs when <math>a</math>, <math>b</math> are minimized (since each of <math>ac</math>, <math>bc</math> is fixed). Thus <math>(a,b,c)=(3,5,12)\Longrightarrow abc=\boxed{180}</math> | ||
+ | |||
+ | ~ Nafer | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=3|num-a=5}} | {{AIME box|year=2016|n=II|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:55, 18 October 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 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.