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

Revision as of 18:58, 29 March 2021

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

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*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

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. Thus, the answer is $4\cdot 3\cdot 2 = \boxed{\textbf{(D)}\ 24}.$

Video Solution

https://youtu.be/GIo37KUPQ5o

@@ Line 14: / Line 14: @@
 ==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 straight to count 8 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.
+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>.
 There are 8 paths for each kind of path, making for <math>8 \cdot 3=\boxed{\textbf{(D)}\ 24}</math> paths.

2017 AMC 8 (Problems • Answer Key • Resources)
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