Difference between revisions of "2017 AMC 8 Problems/Problem 15"
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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 | + | 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== | ==Solution 2== |
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.