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Difference between revisions of "2018 AMC 8 Problems/Problem 19"
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==Solution 6 (Non-rigorous)==
 
==Solution 6 (Non-rigorous)==
 
There are two choices for the uppermost: <math>+</math> or <math>-</math>. There are 16 cases in total (<math>2^4</math>). Half of 16 is 8. The answer is <math>\boxed{\textbf{(C) } 8}</math>.
 
There are two choices for the uppermost: <math>+</math> or <math>-</math>. There are 16 cases in total (<math>2^4</math>). Half of 16 is 8. The answer is <math>\boxed{\textbf{(C) } 8}</math>.
 +
 
(Note: this is not recommended for solving, just an idea after solving.)
 
(Note: this is not recommended for solving, just an idea after solving.)
 +
 +
~nolypoly11
 +
 +
==Solution 7 (Using probability)==
 +
If you pick the signs for two boxes at random, there is equal probability that the one above them will be either a <math>+</math> or a <math>-</math>. There are 16(<math>2^4</math>) cases possible for the bottom four, so half of that, <math>\boxed{\textbf{(C) } 8}</math>, is the answer.
  
 
==Video Solution==
 
==Video Solution==
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[[Category:Introductory Combinatorics Problems]]

Latest revision as of 18:11, 6 June 2025

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 $2+2+2+2=\boxed{\textbf{(C) } 8}$.

Solution 6 (Non-rigorous)

There are two choices for the uppermost: $+$ or $-$. There are 16 cases in total ($2^4$). Half of 16 is 8. The answer is $\boxed{\textbf{(C) } 8}$.

(Note: this is not recommended for solving, just an idea after solving.)

~nolypoly11

Solution 7 (Using probability)

If you pick the signs for two boxes at random, there is equal probability that the one above them will be either a $+$ or a $-$. There are 16($2^4$) cases possible for the bottom four, so half of that, $\boxed{\textbf{(C) } 8}$, is the answer.

Video Solution

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)
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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