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> | ||
− | + | ==Solution 1== | |
− | + | Notice that the upper-most section contains a 3 by 3 square that looks like: | |
− | Notice that the < | + | <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 <math>{6 \cdot 4}</math> to get <math>\boxed{\textbf{(D)}\ 24}</math> 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 | + | 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{24}</math> paths. | + | There are 8 paths for each kind of path, making for <math>8 \cdot 3=\boxed{\textbf{(D)}\ 24}</math> paths. |
− | + | ==Solution 3== | |
− | Notice that the | + | 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. Thus, the answer is <math>4\cdot 3\cdot 2 = \boxed{\textbf{(D)}\ 24}.</math> |
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− | + | ==Video Solution== | |
+ | https://youtu.be/GIo37KUPQ5o | ||
==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 20:01, 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.
Solution 1
Notice that the upper-most section contains a 3 by 3 square that looks like:
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 to get total paths.
Solution 2
There are three different kinds of paths that are on this diagram. The first kind is when you directly count , , in a straight line. The second is when you count , turn left or right to get , then go up or down to count and . The third is the one where you start with , move up or down to count , turn left or right to count , then move straight again to get .
There are 8 paths for each kind of path, making for paths.
Solution 3
Notice that the is adjacent to s, each is adjacent to s, and each is adjacent to 's. Thus, the answer is
Video Solution
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.