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

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==Solution 2==
 
==Solution 2==
  
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Each row is decided by the first cells (the other cells in the row are restricted from the cells above). Since we are given the first row already, we still need to decide the other <math>3</math> rows. The first cell in each row has only <math>2</math> possibilities (+ and -), so we have <math>2^3=\boxed{\textbf{(C) }8}</math> ways.
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Each row is fully determined by its leftmost cell (the other cells in the row are restricted from the cells above). Since we are given the first row already, we still need to decide the other <math>3</math> rows. The first cell in each row has only <math>2</math> possibilities (+ and -), so we have <math>2^3=\boxed{\textbf{(C) }8}</math> ways.
  
 
==See Also==
 
==See Also==

Revision as of 22:16, 1 December 2018

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 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 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

Each row is fully determined by its leftmost cell (the other cells in the row are restricted from the cells above). Since we are given the first row already, we still need to decide the other $3$ rows. The first cell in each row has only $2$ possibilities (+ and -), so we have $2^3=\boxed{\textbf{(C) }8}$ ways.

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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