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

(Problem 19)
(Problem 19)
Line 18: Line 18:
  
 
<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==
 +
 +
There are a total of <math>2^4=16</math> total arrangements for the bottom row, half of which result in a <math>+</math> for the top row, so the answer is <math>\frac{16}{2}=\boxed{\textbf{(C) } 8}</math>
 +
 +
Solution by mathleticguyyy
  
 
{{AMC8 box|year=2018|num-b=18|num-a=20}}
 
{{AMC8 box|year=2018|num-b=18|num-a=20}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:00, 21 November 2018

Problem 19

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

There are a total of $2^4=16$ total arrangements for the bottom row, half of which result in a $+$ for the top row, so the answer is $\frac{16}{2}=\boxed{\textbf{(C) } 8}$

Solution by mathleticguyyy

2018 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png