Difference between revisions of "2018 AMC 8 Problems/Problem 19"
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− | ==Problem | + | ==Problem== |
In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid? | In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid? | ||
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<math>\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16</math> | <math>\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns: | ||
+ | |||
+ | +−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is <math>\boxed{\textbf{(C) } 8}</math>. | ||
+ | |||
+ | -NinjaBoi2000 | ||
+ | |||
+ | ==Solution 2== | ||
+ | The top box is fixed by the problem. | ||
+ | |||
+ | Choose the left 3 bottom-row boxes freely. There are <math>2^3=8</math> ways. | ||
+ | |||
+ | Then the left 2 boxes on the row above are determined. | ||
+ | |||
+ | Then the left 1 box on the row above that is determined | ||
+ | |||
+ | Then the right 1 box on that row is determined. | ||
+ | |||
+ | Then the right 1 box on the row below is determined. | ||
+ | |||
+ | Then the right 1 box on the bottom row is determined, completing the diagram. | ||
+ | |||
+ | So the answer is <math>\boxed{\textbf{(C) } 8}</math>. | ||
+ | |||
+ | |||
+ | ~BraveCobra22aops | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let the plus sign represent 1 and the negative sign represent -1. | ||
+ | |||
+ | The four numbers on the bottom are <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, which are either 1 or -1. | ||
+ | |||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; | ||
+ | draw(box); label("$a$",(0,0)); | ||
+ | draw(shift(1,0)*box); label("$b$",(1,0)); | ||
+ | draw(shift(2,0)*box); label("$c$",(2,0)); | ||
+ | draw(shift(3,0)*box); label("$d$",(3,0)); | ||
+ | draw(shift(0.5,0.4)*box); label("$ab$",(0.5,0.4)); | ||
+ | draw(shift(1.5,0.4)*box); label("$bc$",(1.5,0.4)); | ||
+ | draw(shift(2.5,0.4)*box); label("$cd$",(2.5,0.4)); | ||
+ | draw(shift(1,0.8)*box); label("$ab^2c$",(1,0.8)); | ||
+ | draw(shift(2,0.8)*box); label("$bc^2d$",(2,0.8)); | ||
+ | draw(shift(1.5,1.2)*box); label("$ab^3c^3d$",(1.5,1.2)); | ||
+ | </asy> | ||
+ | |||
+ | Which means <math>ab^3c^3d</math> = 1. Since <math>b</math> and <math>c</math> are either 1 or -1, <math>b^3 = b</math> and <math>c^3 = c</math>. This shows that <math>abcd</math> = 1. | ||
+ | |||
+ | Therefore either <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are all positive or negative, or 2 are positive and 2 are negative. | ||
+ | |||
+ | There are 2 ways where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are 1 (1, 1, 1, 1) and (-1, -1, -1, -1) | ||
+ | |||
+ | There are 6 ways where 2 variables are positive and 2 are negative: (1, 1, -1, -1), (1, -1, 1, -1), (-1, 1, 1, -1), (-1, -1, 1, 1), (-1, 1, -1, 1), and (-1, -1, 1, 1). | ||
+ | |||
+ | So the answer is <math>\boxed{\textbf{(C) } 8}</math>. | ||
+ | |||
+ | ~atharvd | ||
+ | |||
+ | ==Video Solution (CREATIVE ANALYSIS!!!)== | ||
+ | https://youtu.be/29RtYSU89vA | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/j8wm3gfOYvU | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC8 box|year=2018|num-b=18|num-a=20}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 14:16, 30 March 2023
Contents
Problem
In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
Solution 1
You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:
+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is .
-NinjaBoi2000
Solution 2
The top box is fixed by the problem.
Choose the left 3 bottom-row boxes freely. There are ways.
Then the left 2 boxes on the row above are determined.
Then the left 1 box on the row above that is determined
Then the right 1 box on that row is determined.
Then the right 1 box on the row below is determined.
Then the right 1 box on the bottom row is determined, completing the diagram.
So the answer is .
~BraveCobra22aops
Solution 3
Let the plus sign represent 1 and the negative sign represent -1.
The four numbers on the bottom are , , , and , which are either 1 or -1.
Which means = 1. Since and are either 1 or -1, and . This shows that = 1.
Therefore either , , , and are all positive or negative, or 2 are positive and 2 are negative.
There are 2 ways where , , , and are 1 (1, 1, 1, 1) and (-1, -1, -1, -1)
There are 6 ways where 2 variables are positive and 2 are negative: (1, 1, -1, -1), (1, -1, 1, -1), (-1, 1, 1, -1), (-1, -1, 1, 1), (-1, 1, -1, 1), and (-1, -1, 1, 1).
So the answer is .
~atharvd
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.