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Difference between revisions of "2018 AMC 8 Problems/Problem 19"

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You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:  
 
You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:  
  
−
<math>+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++</math>. There are 8 patterns and so the answer is <math>\boxed{\textbf{(C) } 8}</math>.
+
<math>+<cmath>−</cmath>−</math><math>+</math>, <math>−+</math><math>+−</math>, <math>−<cmath>−</cmath>−</math><math>−</math>, <math>+<cmath>+</cmath>+</math><math>+</math>, <math>−<cmath>+</cmath>−</math><math>+</math>, <math>+<cmath>−</cmath>+</math><math>−</math>, <math>+<cmath>+</cmath>−</math><math>−</math>, <math>−<cmath>−</cmath>+</math><math>+</math>. There are 8 patterns and so the answer is <math>\boxed{\textbf{(C) } 8}</math>.
  
 
-NinjaBoi2000
 
-NinjaBoi2000

Revision as of 21:54, 26 November 2023

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?

[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("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]

$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

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:

$+<cmath>−</cmath>−$ (Error compiling LaTeX. Unknown error_msg)$+$, $−+$ (Error compiling LaTeX. Unknown error_msg)$+−$ (Error compiling LaTeX. Unknown error_msg), $−<cmath>−</cmath>−$ (Error compiling LaTeX. Unknown error_msg)$−$ (Error compiling LaTeX. Unknown error_msg), $+<cmath>+</cmath>+$$+$, $−<cmath>+</cmath>−$ (Error compiling LaTeX. Unknown error_msg)$+$, $+<cmath>−</cmath>+$ (Error compiling LaTeX. Unknown error_msg)$−$ (Error compiling LaTeX. Unknown error_msg), $+<cmath>+</cmath>−$ (Error compiling LaTeX. Unknown error_msg)$−$ (Error compiling LaTeX. Unknown error_msg), $−<cmath>−</cmath>+$ (Error compiling LaTeX. Unknown error_msg)$+$. There are 8 patterns and so the answer is $\boxed{\textbf{(C) } 8}$.

-NinjaBoi2000

Solution 2

The top box is fixed by the problem.

Choose the left 3 bottom-row boxes freely. There are $2^3=8$ 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 $\boxed{\textbf{(C) } 8}$.


~BraveCobra22aops

Solution 3

Let the plus sign represent 1 and the negative sign represent -1.

The four numbers on the bottom are $a$, $b$, $c$, and $d$, 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 $ab^3c^3d$ = 1. Since $b$ and $c$ are either 1 or -1, $b^3 = b$ and $c^3 = c$. This shows that $abcd$ = 1.

Therefore either $a$, $b$, $c$, and $d$ are all positive or negative, or 2 are positive and 2 are negative.

There are 2 ways where $a$, $b$, $c$, and $d$ 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 $\boxed{\textbf{(C) } 8}$.

~atharvd

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/29RtYSU89vA

~Education, the Study of Everything

Video Solution

https://youtu.be/j8wm3gfOYvU

~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

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