2018 AMC 8 Problems/Problem 19

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:

+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. 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 $\boxed{\textbf{(C) } 8}$ .

Note: This result can also be achieved by realizing that there are $4! / 2! 2! = 6$ ways to arrange 2 negatives and positives and 1 way each to arrange four of one sign.

~atharvd ~cxsmi (Note)

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/29RtYSU89vA

~Education, the Study of Everything

Video Solution

https://youtu.be/j8wm3gfOYvU

~savannahsolver

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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