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Difference between revisions of "2018 AMC 8 Problems/Problem 19"

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==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>, so the answer is <math>\boxed{\textbf{(C) } 8}</math>
 
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 one 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==
 
==See Also==

Revision as of 22:10, 9 November 2019

Problem 19

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

Instead of + and -, let us use 1 and 0, respectively. If we let $a$, $b$, $c$, and $d$ be the values of the four cells on the bottom row, then the three cells on the next row are equal to $a+b$, $b+c$, and $c+d$ taken modulo (mod) 2 (this is exactly the same as finding $a \text{ XOR } b$, and so on). The two cells on the next row are $a+2b+c$ and $b+2c+d$ taken modulo (mod) 2, and lastly, the cell on the top row gets $a+3b+3c+d \pmod{2}$.

Thus, we are looking for the number of assignments of 0's and 1's for $a$, $b$, $c$, $d$ such that $a+3b+3c+d \equiv 1 \pmod{2}$, or in other words, is odd. As $3 \equiv 1 \pmod{2}$, this is the same as finding the number of assignments such that $a+b+c+d \equiv 1 \pmod{2}$. Notice that, no matter what $a$, $b$, and $c$ are, this uniquely determines $d$. There are $2^3 = 8$ ways to assign 0's and 1's arbitrarily to $a$, $b$, and $c$, so the answer is $\boxed{\textbf{(C) } 8}$

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 $2^3=8$, so the answer is $\boxed{\textbf{(C) } 8}$

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 one of the other, this can be thought of as the 4th Row of the Pascal’s Triangle, which is $1 4 6 4 1$. Since 3 of one sign and 1 of the other doesn’t work, all you need to add is $1 + 6 + 1 = 8$, so the answer is $\boxed{\textbf{(C) } 8}$

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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