Difference between revisions of "2018 AMC 8 Problems/Problem 19"
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Together, we find that the answer is <math>2+2+2+2=\boxed{\textbf{(C) } 8}</math>. | Together, we find that the answer is <math>2+2+2+2=\boxed{\textbf{(C) } 8}</math>. | ||
| − | ==Video Solution | + | ==Video Solution== |
https://youtu.be/29RtYSU89vA | https://youtu.be/29RtYSU89vA | ||
Latest revision as of 20:10, 4 December 2024
Contents
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]](http://latex.artofproblemsolving.com/f/5/9/f59b1af815a0ba23baf250a48b1029ee9d1bd5e4.png)
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 
.
-NinjaBoi2000
Solution 2
The top box is fixed by the problem.
Choose the left 3 bottom-row boxes freely. There are 
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 .
~BraveCobra22aops
Solution 3
Let the plus sign represent 1 and the negative sign represent -1.
The four numbers on the bottom are 
, 
, 
, and 
, 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]](http://latex.artofproblemsolving.com/a/9/a/a9abfa418becfdb1bad047a9bceadea7cf56328d.png)
Which means 
= 1. Since and are either 1 or -1, 
and 
. This shows that 
= 1.
Therefore either , , , and are all positive or negative, or 2 are positive and 2 are negative.
There are 2 ways where , , , and 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 .
Note: This result can also be achieved by realizing that there are 
ways to arrange 
negatives and positives and 
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 
blocks, there are 
possible combination and half of them will lead a (or ) on the top. The answer is 
.
If you notice this rule, you can give the answer whatever how many layers you have. The answer will be 
for the layer with 
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 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 
cells) is --- or 
. Now, let's pretend that the second is . That would have possibilities for the first row (the one with 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 possibilities for the first row, 
or , and possibilities for , 
and 
.
Together, we find that the answer is 
.
Video Solution
~Education, the Study of Everything
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
