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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==Solution 1==
 
==Solution 1==
Instead of + and -, let us use 1 and 0, respectively. If we let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be the values of the four cells on the bottom row, then the three cells on the next row are equal to <math>a+b</math>, <math>b+c</math>, and <math>c+d</math> taken modulo 2 (this is exactly the same as finding <math>a \text{ XOR } b</math>, and so on). The two cells on the next row are <math>a+2b+c</math> and <math>b+2c+d</math> taken modulo 2, and lastly, the cell on the top row gets <math>a+3b+3c+d \pmod{2}</math>.
+
You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:
  
Thus, we are looking for the number of assignments of 0's and 1's for <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> such that <math>a+3b+3c+d \equiv 1 \pmod{2}</math>, or in other words, is odd. As <math>3 \equiv 1 \pmod{2}</math>, this is the same as finding the number of assignments such that <math>a+b+c+d \equiv 1 \pmod{2}</math>. Notice that, no matter what <math>a</math>, <math>b</math>, and <math>c</math> are, this uniquely determines <math>d</math>. There are <math>2^3 = 8</math> ways to assign 0's and 1's arbitrarily to <math>a</math>, <math>b</math>, and <math>c</math>, 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
  
 
==Solution 2==
 
==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 <math>2^3=8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math>
  
Each row is fully determined by its leftmost cell (the other cells in the row are restricted from the cells above). Since we are given the first row already, we still need to decide the other <math>3</math> rows. The first cell in each row has only <math>2</math> possibilities (+ and -), so we have <math>2^3=\boxed{\textbf{(C) }8}</math> ways.
+
==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 <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==

Latest 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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