Difference between revisions of "2018 AMC 8 Problems/Problem 19"
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==Solution 3== | ==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 3 3 1</math>. Since putting 3 of one sign and 1 of the other in the 4th row doesn’t work, all you need to add is <math>1 + 3 + 3 + 1 = 8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math>. | 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 3 3 1</math>. Since putting 3 of one sign and 1 of the other in the 4th row doesn’t work, all you need to add is <math>1 + 3 + 3 + 1 = 8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math>. | ||
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| + | ==Solution 4== | ||
| + | Note that there are <math>2^4 = 16</math> possible arrangements of the bottom row, and by symmetry, half of them result in a <math>+</math> sign at the top and half will result in a <math>-</math> sign at the top. Therefore, the answer is <math>16/2 = \boxed{\textbf{(C) }8}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 14:31, 16 January 2022
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 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 
, so the answer is
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 
. Since putting 3 of one sign and 1 of the other in the 4th row doesn’t work, all you need to add is 
, so the answer is .
Solution 4
Note that there are 
possible arrangements of the bottom row, and by symmetry, half of them result in a 
sign at the top and half will result in a 
sign at the top. Therefore, the answer is 
.
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. 
