# 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 $\boxed{\textbf{(C) } 8}$.

~atharvd

## Video Solution (CREATIVE ANALYSIS!!!)

~Education, the Study of Everything

## Video Solution

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 