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

Revision as of 18:02, 1 April 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

@@ Line 24: / Line 24: @@
 ==Solution 4==
 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==

2017 AMC 8 (Problems • Answer Key • Resources)
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