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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?
  
Line 20: Line 20:
  
 
==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 (mod) 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 (mod) 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>, which is C.
+
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));
 +
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==

Latest 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 18 		Followed by
Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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