Difference between revisions of "2020 AMC 8 Problems/Problem 21"

(Undo revision 137613 by Franzliszt (talk) this is the exact same as solution 1.)
(Tag: Undo)
(Video Solution by WhyMath)
 
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==Problem 21==
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==Problem==
 
A game board consists of <math>64</math> squares that alternate in color between black and white. The figure below shows square <math>P</math> in the bottom row and square <math>Q</math> in the top row. A marker is placed at <math>P.</math> A step consists of moving the marker onto one of the adjoining white squares in the row above. How many <math>7</math>-step paths are there from <math>P</math> to <math>Q?</math> (The figure shows a sample path.)
 
A game board consists of <math>64</math> squares that alternate in color between black and white. The figure below shows square <math>P</math> in the bottom row and square <math>Q</math> in the top row. A marker is placed at <math>P.</math> A step consists of moving the marker onto one of the adjoining white squares in the row above. How many <math>7</math>-step paths are there from <math>P</math> to <math>Q?</math> (The figure shows a sample path.)
  
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<math>\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35</math>
 
<math>\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35</math>
  
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==Solution 1==
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Notice that, in order to step onto any particular white square, the marker must have come from one of the <math>1</math> or <math>2</math> white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from <math>P</math> to that square is the sum of the numbers of ways to move from <math>P</math> to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from <math>P</math> to that square, as already stated).
  
==Solution==
 
We count paths. Noticing that we can only go along white squares, to get to a white square we can only go from the two whites beneath it. Here is a diagram:
 
 
<asy>
 
<asy>
 
int N = 7;
 
int N = 7;
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label("$28$", (6.5, 7.5));
 
label("$28$", (6.5, 7.5));
 
</asy>
 
</asy>
So the answer is <math>\boxed{\textbf{(A)}28}</math>
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~yofro (Diagram credits to franzliszt)
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The answer is therefore <math>\boxed{\textbf{(A) }28}</math>.
  
 
==Solution 2==
 
==Solution 2==
Suppose we "extend" the chessboard indefinitely to the right:
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Suppose we "extend" the chessboard infinitely with <math>2</math> additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.
  
 
<asy>
 
<asy>
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label("$Y$", (8.5,5.5));
 
label("$Y$", (8.5,5.5));
 
</asy>
 
</asy>
The total number of paths from <math>P</math> to <math>Q</math> (including invalid paths which cross over the red line) is <math>\binom{7}{3} = 35</math>. We subtract the number of invalid paths that pass through <math>X</math> or <math>Y</math>. The number of paths that pass through <math>X</math> is <math>\binom{3}{0}\binom{4}{3} = 4</math> and the number of paths that pass through <math>Y</math> is <math>\binom{5}{1}\binom{2}{2} = 5</math>. However, we overcounted the invalid paths which pass through both <math>X</math> and <math>Y</math>, of which there are 2 paths. Hence, the number of invalid paths is <math>4+5-2=7</math> and the number of valid paths from <math>P</math> to <math>Q</math> is <math>35-7 = \boxed{\textbf{(A)} 28}</math>. -scrabbler94
 
  
==See also==  
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The total number of paths from <math>P</math> to <math>Q</math>, including invalid paths which cross over the red line, is then the number of paths which make <math>4</math> steps up-and-right and <math>3</math> steps up-and-left, which is <math>\binom{4+3}{3} = \binom{7}{3} = 35</math>. We need to subtract the number of invalid paths, i.e. the number of paths that pass through <math>X</math> or <math>Y</math>. To get to <math>X</math>, the marker has to make <math>3</math> up-and-right steps, after which it can proceed to <math>Q</math> with <math>3</math> steps up-and-left and <math>1</math> step up-and-right. Thus, the number of paths from <math>P</math> to <math>Q</math> that pass through <math>X</math> is <math>1 \cdot \binom{3+1}{3} = 4</math>. Similarly, the number of paths that pass through <math>Y</math> is <math>\binom{4+1}{1}\cdot 1 = 5</math>. However, we have now double-counted the invalid paths which pass through both <math>X</math> and <math>Y</math>; from the diagram, it is clear that there are only <math>2</math> of these (as the marker can get from <math>X</math> to <math>Y</math> by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is <math>4+5-2=7</math>, and the number of valid paths from <math>P</math> to <math>Q</math> is <math>35-7 = \boxed{\textbf{(A) }28}</math>.
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==Video Solution by North America Math Contest Go Go Go==
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https://www.youtube.com/watch?v=rWxhlItMAN0
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 +
~North America Math Contest Go Go Go
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/YR3oghbziBA
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 +
~savannahsolver
 +
 
 +
==Video Solutions==
 +
https://youtu.be/hGCxt8G9g-s
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=1247
 +
 
 +
~Interstigation
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 +
==See also==
 
{{AMC8 box|year=2020|num-b=20|num-a=22}}
 
{{AMC8 box|year=2020|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:04, 19 June 2021

Problem

A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$-step paths are there from $P$ to $Q?$ (The figure shows a sample path.)

[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]

$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$

Solution 1

Notice that, in order to step onto any particular white square, the marker must have come from one of the $1$ or $2$ white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated).

[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label("$1$", (5.5, .5)); label("$1$", (4.5, 1.5)); label("$1$", (6.5, 1.5)); label("$1$", (3.5, 2.5)); label("$1$", (7.5, 2.5)); label("$2$", (5.5, 2.5)); label("$1$", (2.5, 3.5)); label("$3$", (6.5, 3.5)); label("$3$", (4.5, 3.5)); label("$4$", (3.5, 4.5)); label("$3$", (7.5, 4.5)); label("$6$", (5.5, 4.5)); label("$10$", (4.5, 5.5)); label("$9$", (6.5, 5.5)); label("$19$", (5.5, 6.5)); label("$9$", (7.5, 6.5)); label("$28$", (6.5, 7.5)); [/asy]

The answer is therefore $\boxed{\textbf{(A) }28}$.

Solution 2

Suppose we "extend" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.

[asy] int N = 7; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } draw((8,0) -- (8,8),red); label("$P$", (5.5,.5)); label("$Q$", (6.5,7.5)); label("$X$", (8.5,3.5)); label("$Y$", (8.5,5.5)); [/asy]

The total number of paths from $P$ to $Q$, including invalid paths which cross over the red line, is then the number of paths which make $4$ steps up-and-right and $3$ steps up-and-left, which is $\binom{4+3}{3} = \binom{7}{3} = 35$. We need to subtract the number of invalid paths, i.e. the number of paths that pass through $X$ or $Y$. To get to $X$, the marker has to make $3$ up-and-right steps, after which it can proceed to $Q$ with $3$ steps up-and-left and $1$ step up-and-right. Thus, the number of paths from $P$ to $Q$ that pass through $X$ is $1 \cdot \binom{3+1}{3} = 4$. Similarly, the number of paths that pass through $Y$ is $\binom{4+1}{1}\cdot 1 = 5$. However, we have now double-counted the invalid paths which pass through both $X$ and $Y$; from the diagram, it is clear that there are only $2$ of these (as the marker can get from $X$ to $Y$ by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is $4+5-2=7$, and the number of valid paths from $P$ to $Q$ is $35-7 = \boxed{\textbf{(A) }28}$.

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=rWxhlItMAN0

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/YR3oghbziBA

~savannahsolver

Video Solutions

https://youtu.be/hGCxt8G9g-s

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=1247

~Interstigation

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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