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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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<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==
+
==Solution 1==
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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:
  
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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>.
  
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==Easy Solution==
+
-NinjaBoi2000
  
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Each row is decided by the first cells (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 2==
 +
The top box is fixed by the problem.
 +
 
 +
Choose the left 3 bottom-row boxes freely. There are <math>2^3=8</math> 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 <math>\boxed{\textbf{(C) } 8}</math>.
 +
 
 +
 
 +
~BraveCobra22aops
 +
 
 +
==Solution 3==
 +
Let the plus sign represent 1 and the negative sign represent -1.
 +
 
 +
The four numbers on the bottom are <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, 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));
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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 <math>ab^3c^3d</math> = 1. Since <math>b</math> and <math>c</math> are either 1 or -1, <math>b^3 = b</math> and <math>c^3 = c</math>. This shows that <math>abcd</math> = 1.
 +
 
 +
Therefore either <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are all positive or negative, or 2 are positive and 2 are negative.
 +
 
 +
There are 2 ways where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> 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 <math>\boxed{\textbf{(C) } 8}</math>.
 +
 
 +
Note: This result can also be achieved by realizing that there are <math>4! / 2! 2! = 6</math> ways to arrange <math>2</math> negatives and <math>2</math> positives and <math>1</math> way each to arrange four of one sign.
 +
 
 +
~atharvd
 +
 
 +
~cxsmi (Note)
 +
 
 +
==Solution 4==
 +
The pyramid is built on the basic 3 blocks pattern: one above and two below. The basic pattern have four possible symbols and half of them have a <math>+</math> on the above, half of them have a <math>-</math> above. So, For the lowest layer with <math>4</math> blocks, there are <math>2^4=16</math> possible combination and half of them will lead a <math>+</math> (or <math>-</math>) on the top. The answer is <math>16/2=\boxed{\textbf{(C) } 8}</math>.
 +
 
 +
If you notice this rule, you can give the answer whatever how many layers you have. The answer will be <math>2^{n-1}</math> for the layer with <math>n</math> blocks.
 +
 
 +
==Solution 5==
 +
 
 +
We can use casework to solve this problem.
 +
The only way for the top cell to have a <math>+</math> in it is if the third row of the pyramid (the one with <math>2</math> cells) is either <math>--</math> or <math>++</math>. First, let's pretend that the third row of the pyramid is <math>++</math>. The only way for that to happen is if the second row (the one with <math>3</math> cells) is --- or <math>+++</math>. Now, let's pretend that the second is <math>+++</math>. That would have <math>2</math> possibilities for the first row (the one with <math>4</math> cells), <math>++++</math> and <math>----</math>. Next, let's pretend that the second row is ---. That makes two more possibilities for the first row, <math>-+-+</math> and <math>+-+-</math>. Now, let's pretend that the 3rd row is <math>--</math>, which means that the second row is either <math>-+-</math> or <math>+-+</math>. You will soon find that <math>-+-</math> find <math>2</math> possibilities for the first row, <math>-++-</math> or <math>-++-</math>, and <math>2</math> possibilities for <math>+-+</math>, <math>--++</math> and <math>++--</math>.
 +
Together, we find that the answer is <math>2+2+2+2=\boxed{\textbf{(C) } 8}</math>$
 +
 
 +
==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
  
 
==See Also==
 
==See Also==

Revision as of 13:57, 18 June 2024

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 $2$ positives and $1$ way each to arrange four of one sign.

~atharvd

~cxsmi (Note)

Solution 4

The pyramid is built on the basic 3 blocks pattern: one above and two below. The basic pattern have four possible symbols and half of them have a $+$ on the above, half of them have a $-$ above. So, For the lowest layer with $4$ blocks, there are $2^4=16$ possible combination and half of them will lead a $+$ (or $-$) on the top. The answer is $16/2=\boxed{\textbf{(C) } 8}$.

If you notice this rule, you can give the answer whatever how many layers you have. The answer will be $2^{n-1}$ for the layer with $n$ blocks.

Solution 5

We can use casework to solve this problem. The only way for the top cell to have a $+$ in it is if the third row of the pyramid (the one with $2$ cells) is either $--$ or $++$. First, let's pretend that the third row of the pyramid is $++$. The only way for that to happen is if the second row (the one with $3$ cells) is --- or $+++$. Now, let's pretend that the second is $+++$. That would have $2$ possibilities for the first row (the one with $4$ cells), $++++$ and $----$. Next, let's pretend that the second row is ---. That makes two more possibilities for the first row, $-+-+$ and $+-+-$. Now, let's pretend that the 3rd row is $--$, which means that the second row is either $-+-$ or $+-+$. You will soon find that $-+-$ find $2$ possibilities for the first row, $-++-$ or $-++-$, and $2$ possibilities for $+-+$, $--++$ and $++--$. Together, we find that the answer is $2+2+2+2=\boxed{\textbf{(C) } 8}$$

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

See Also

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

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