Difference between revisions of "2018 AMC 8 Problems/Problem 19"
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− | ==Problem | + | ==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== |
+ | 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 <math>\boxed{\textbf{(C) } 8}</math> | |
− | Solution | + | -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 <math>2^3=8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math> | ||
+ | |||
+ | ==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 4 6 4 1</math>. Since 3 of one sign and 1 of the other doesn’t work, all you need to add is <math>1 + 6 + 1 = 8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math>. | ||
+ | |||
+ | ==See Also== | ||
{{AMC8 box|year=2018|num-b=18|num-a=20}} | {{AMC8 box|year=2018|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:56, 18 January 2021
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?
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 3 of one sign and 1 of the other doesn’t work, all you need to add is , so 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 |
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