Difference between revisions of "2012 AMC 10B Problems/Problem 25"
(→Solution) |
(→Solution) |
||
Line 96: | Line 96: | ||
Finally, from each of the first and second green arrows, there is respectively <math>2</math> ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are <math>3</math> ways to get to the first orange arrow. Therefore there are <math>120 \cdot (2+2+3+3) = 1200</math> ways to get to each of the orange arrows, hence <math>2400</math> ways to get to the point <math>B</math>. <math>\framebox{E}</math> | Finally, from each of the first and second green arrows, there is respectively <math>2</math> ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are <math>3</math> ways to get to the first orange arrow. Therefore there are <math>120 \cdot (2+2+3+3) = 1200</math> ways to get to each of the orange arrows, hence <math>2400</math> ways to get to the point <math>B</math>. <math>\framebox{E}</math> | ||
− | ==Solution== | + | ==Solution 2== |
Suppose the bug just went through one of the green arrows. There is only <math>1</math> path it can take that goes through the remaining white arrow, depending on whether it just took one of the top two or one of the bottom two green arrows. If the bug does not take the reverse white arrow, it has <math>4</math> possibilities. Thus, the bug has <math>5</math> possible paths in total once it has crossed a green arrow. <math>\framebox{E}</math> is the only answer divisible by 5. | Suppose the bug just went through one of the green arrows. There is only <math>1</math> path it can take that goes through the remaining white arrow, depending on whether it just took one of the top two or one of the bottom two green arrows. If the bug does not take the reverse white arrow, it has <math>4</math> possibilities. Thus, the bug has <math>5</math> possible paths in total once it has crossed a green arrow. <math>\framebox{E}</math> is the only answer divisible by 5. |
Revision as of 13:22, 4 January 2014
- The following problem is from both the 2012 AMC 12B #22 and 2012 AMC 10B #25, so both problems redirect to this page.
A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
Solution 1
There is way to get to any of the red arrows. From the first red arrow, there are
ways to get to each of the first and the second blue arrows; from the second red arrow, there are
ways to get to each of the first and the second blue arrows. So there are in total
ways to get to each of the blue arrows.
From each of the first and second blue arrows, there are respectively ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively
ways to get to each of the first and the second green arrows. Therefore there are in total
ways to get to each of the green arrows.
Finally, from each of the first and second green arrows, there is respectively ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are
ways to get to the first orange arrow. Therefore there are
ways to get to each of the orange arrows, hence
ways to get to the point
.
Solution 2
Suppose the bug just went through one of the green arrows. There is only path it can take that goes through the remaining white arrow, depending on whether it just took one of the top two or one of the bottom two green arrows. If the bug does not take the reverse white arrow, it has
possibilities. Thus, the bug has
possible paths in total once it has crossed a green arrow.
is the only answer divisible by 5.
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by $\whitesquare$ (Error compiling LaTeX. Unknown error_msg) | |
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 AMC 10 Problems and Solutions |
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.