Difference between revisions of "2017 AMC 8 Problems/Problem 15"
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==Solution 2== | ==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 | + | 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 8 paths for each kind of path, making for <math>8 \cdot 3=\boxed{\textbf{(D)}\ 24}</math> paths. | There are 8 paths for each kind of path, making for <math>8 \cdot 3=\boxed{\textbf{(D)}\ 24}</math> paths. | ||
Revision as of 18:56, 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]](http://latex.artofproblemsolving.com/c/3/b/c3b81f82e730c8fce3fba2c0d1e893252c432d58.png)
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]](http://latex.artofproblemsolving.com/4/e/7/4e757a2a293bbc0f2888163537a04662facfe134.png)
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 
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 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 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. 
