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Difference between revisions of "2017 AMC 8 Problems/Problem 15"

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<math>\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36</math>
 
<math>\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36</math>
  
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==Solution==
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==Solution 1 (Not Very Efficient)==
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===Solution 1===
 
 
Notice that the upper-most section contains a 3 by 3 square that looks like:
 
Notice that the upper-most section contains a 3 by 3 square that looks like:
−
<asy>label("$8$", (1, 0)); label("$C$", (2, 0)); label("$8$", (3, 0)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$M$", (1, 2)); label("$A$", (2, 2)); label("$M$", (3, 2));</asy>
 
  
 +
<asy>label("$8$", (1, 2)); label("$C$", (2, 2)); label("$8$", (3, 2)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$M$", (1, 0)); label("$A$", (2, 0)); label("$M$", (3, 0));</asy>
  
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It has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply 6*4 to get <math>\boxed{24}</math> total paths.
 
  
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===Solution 2===
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It has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply <math>{6 \cdot 4}</math> to get <math>\boxed{\textbf{(D)}\ 24}</math> total paths.
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There are three different kinds of paths that are on this diagram. The first kind is when you directly count A, M, C in a straight line. The second is when you count A, turn left or right to get M, then go straight to count M and C. The third is the one where you start with A, move forward to count M, turn left or right to count C, then move straight again to get 8.
 
  
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There are 8 paths for each kind of path, making for <math>8 \cdot 3=\boxed{24}</math> paths.
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==Solution 2==
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There are three different kinds of paths that are on this diagram. The first kind is when you directly count <math>A</math>, <math>M</math>, <math>C</math> in a straight line. The second is when you count <math>A</math>, turn left or right to get <math>M</math>, then go up or down to count <math>8</math> and <math>C</math>. The third is the one where you start with <math>A</math>, move up or down to count <math>M</math>, turn left or right to count <math>C</math>, then move straight again to get <math>8</math>.
  
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===Solution 3===
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There are 8 paths for each kind of path, making for <math>8 \cdot 3=\boxed{\textbf{(D)}\ 24}</math> paths.
−
Notice that the upper-most section contains a 3 by 3 square that looks like:
+
 
−
<asy>label("$8$", (1, 0)); label("$C$", (2, 0)); label("$8$", (3, 0)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$M$", (1, 2)); label("$A$", (2, 2)); label("$M$", (3, 2));</asy>
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==Solution 3 (easy and recommended)==
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Notice that the <math>A</math> is adjacent to <math>4</math> <math>M</math>s, each <math>M</math> is adjacent to <math>3</math> <math>C</math>s, and each <math>C</math> is adjacent to <math>2</math> <math>8</math>'s. So for each <math>A</math>, there are <math>4</math> <math>M</math>s, and for each <math>M</math>, there are <math>3</math> <math>C</math>s, and for each <math>C</math>, there are <math>2</math> <math>8</math>s. Thus, the answer is <math>1\cdot 4\cdot 3\cdot 2 = \boxed{\textbf{(D)}\ 24}.</math>
 +
 
 +
==Solution 4==
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We can do this problem by computing how many ways there are to get to each letter (in order). There is <math>1</math> way to get to the <math>A</math> in the center. We can only get to each of the other <math>M</math>s by going there from the <math>A</math>, so there is <math>1</math> way to get to each of the four <math>M</math>s. For the <math>C</math>s, we notice that four <math>C</math>s are surrounded by one <math>M</math>, and four <math>C</math>s are surrounded by two <math>M</math>s. Finally, each of the <math>8</math>s is surrounded by one <math>C</math> with one way to get there and one <math>C</math> with two ways to get there. Therefore, there are three paths to any of the <math>8</math>s. Since there are eight <math>8</math>s, the answer is <math>3 \cdot 8 = \boxed{\textbf{(D)}\ 24}.</math>
 +
 
 +
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 +
https://youtu.be/oWMU6ph6LBM
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/GIo37KUPQ5o
  
 +
https://youtu.be/CTnICuhvbAk
  
−
It has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply 6*4 to get <math>\boxed{24}</math> total paths.
+
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2017|num-b=14|num-a=16}}
 
{{AMC8 box|year=2017|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:36, 24 December 2023

Problem

In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.

[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label("$8$", (1, 0)); label("$C$", (2, 0)); label("$8$", (3, 0)); label("$8$", (0, 1)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$8$", (4, 1)); label("$C$", (0, 2)); label("$M$", (1, 2)); label("$A$", (2, 2)); label("$M$", (3, 2)); label("$C$", (4, 2)); label("$8$", (0, 3)); label("$C$", (1, 3)); label("$M$", (2, 3)); label("$C$", (3, 3)); label("$8$", (4, 3)); label("$8$", (1, 4)); label("$C$", (2, 4)); label("$8$", (3, 4));[/asy]

$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$

Solution 1 (Not Very Efficient)

Notice that the upper-most section contains a 3 by 3 square that looks like:

[asy]label("$8$", (1, 2)); label("$C$", (2, 2)); label("$8$", (3, 2)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$M$", (1, 0)); label("$A$", (2, 0)); label("$M$", (3, 0));[/asy]


It has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply ${6 \cdot 4}$ to get $\boxed{\textbf{(D)}\ 24}$ total paths.

Solution 2

There are three different kinds of paths that are on this diagram. The first kind is when you directly count $A$, $M$, $C$ in a straight line. The second is when you count $A$, turn left or right to get $M$, then go up or down to count $8$ and $C$. The third is the one where you start with $A$, move up or down to count $M$, turn left or right to count $C$, then move straight again to get $8$.

There are 8 paths for each kind of path, making for $8 \cdot 3=\boxed{\textbf{(D)}\ 24}$ paths.

Solution 3 (easy and recommended)

Notice that the $A$ is adjacent to $4$ $M$s, each $M$ is adjacent to $3$ $C$s, and each $C$ is adjacent to $2$ $8$'s. So for each $A$, there are $4$ $M$s, and for each $M$, there are $3$ $C$s, and for each $C$, there are $2$ $8$s. Thus, the answer is $1\cdot 4\cdot 3\cdot 2 = \boxed{\textbf{(D)}\ 24}.$

Solution 4

We can do this problem by computing how many ways there are to get to each letter (in order). There is $1$ way to get to the $A$ in the center. We can only get to each of the other $M$s by going there from the $A$, so there is $1$ way to get to each of the four $M$s. For the $C$s, we notice that four $C$s are surrounded by one $M$, and four $C$s are surrounded by two $M$s. Finally, each of the $8$s is surrounded by one $C$ with one way to get there and one $C$ with two ways to get there. Therefore, there are three paths to any of the $8$s. Since there are eight $8$s, the answer is $3 \cdot 8 = \boxed{\textbf{(D)}\ 24}.$

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/oWMU6ph6LBM

~Education, the Study of Everything

Video Solution

https://youtu.be/GIo37KUPQ5o

https://youtu.be/CTnICuhvbAk

~savannahsolver

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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. AMC logo.png

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