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

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=Problem 19=
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==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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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:  
  
+--+, -++-, ----, ++++, -+-+, +-+-, ++--, --++. There are 8 patterns and so the answer is <math>\boxed{\textbf{(C) } 8}</math>
+
+−−+, ++, −−−−, ++++, ++, ++, ++−−, −−++. There are 8 patterns and so the answer is <math>\boxed{\textbf{(C) } 8}</math>
  
 
-NinjaBoi2000
 
-NinjaBoi2000
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==Solution 3==  
 
==Solution 3==  
There is also a pretty simple approach to this problem. Since in the bottom row you can either have 4 of the same signs, 3 of the same signs and one of another, and 2 of the same signs and one of the other, this can be thought of as the 4th Row of the Pascal’s Triangle, which is <math>1 4 6 4 1</math>. Since 3 of one sign and 1 of the other doesn’t work, all you need to add is <math>1 + 6 + 1 = 8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math>
+
There is also a pretty simple approach to this problem. Since in the bottom row you can either have 4 of the same signs, 3 of the same signs and one of another, and 2 of the same signs and two of the other, this can be thought of as the 4th Row of the Pascal’s Triangle, which is <math>1 4 6 4 1</math>. Since 3 of one sign and 1 of the other doesn’t work, all you need to add is <math>1 + 6 + 1 = 8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 13:56, 18 January 2021

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 sign of the next row on the pyramid depends on previous row. There are two options for the previous row, - or +. There are three rows to the pyramid that depend on what the top row is. Therefore, the ways you can make a + on the top is $2^3=8$, so the answer is $\boxed{\textbf{(C) } 8}$

Solution 3

There is also a pretty simple approach to this problem. Since in the bottom row you can either have 4 of the same signs, 3 of the same signs and one of another, and 2 of the same signs and two of the other, this can be thought of as the 4th Row of the Pascal’s Triangle, which is $1 4 6 4 1$. Since 3 of one sign and 1 of the other doesn’t work, all you need to add is $1 + 6 + 1 = 8$, so the answer is $\boxed{\textbf{(C) } 8}$.

See Also

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

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