2012 AMC 10B Problems/Problem 25
- The following problem is from both the 2012 AMC 12B #22 and 2012 AMC 10B #25, so both problems redirect to this page.
Problem 25
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 (top) red arrow, there are ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) 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 are 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 (using the answer choices)
For every blue arrow, there are ways to reach it without using the reverse arrow since the bug can choose any of red arrows to pass through and black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black arrow. Otherwise, the bug will have to travel through both red segments, which is impossible because now there is no path to take after the bug emerges from the reverse arrow. Similarly, with the blue segments, the second black arrow taken must be the one that is closest to the blue arrow that was taken. Also, it is trivial that the two black arrows taken must be different. Therefore, if the reverse arrow is taken, the blue arrow taken determines the entire path and there is path for every arrow. Since the bug cannot return once it takes a blue arrow, the answer must be divisible by . is the only answer that is.
2012 AMC 10B (Problems • Answer Key • Resources) | ||
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Preceded by Problem 21 |
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