2018 AIME I Problems/Problem 14
Contents
Problem
Let be a heptagon. A frog starts jumping at vertex . From any vertex of the heptagon except , the frog may jump to either of the two adjacent vertices. When it reaches vertex , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than jumps that end at .
Solution
This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line . We can count the number of left/right (L/R) paths of length that start at and end at either or .
We can visualize the paths using the common grid counting method by starting at the origin , so that a right (R) move corresponds to moving 1 in the positive direction, and a left (L) move corresponds to moving 1 in the positive direction. Because we don't want to move more than 2 units left or more than 3 units right, our path must not cross the lines or . Letting be the number of such paths from to under these constraints, we have the following base cases:
and recursive step for .
The filled in grid will look something like this, where the lower-left corresponds to the origin:
The bolded numbers on the top diagonal represent the number of paths from to in 2, 4, 6, 8, 10 moves, and the numbers on the bottom diagonal represent the number of paths from to in 3, 5, 7, 9, 11 moves. We don't care about the blank entries or entries above the line . The total number of ways is .
(Solution by scrabbler94)
Solution 2
Let denotes the number of sequences with length that ends at . Define similarly for the other vertices. We seek for a recursive formula for . Computing a few terms we have , , , , and .
Using the formula yields , , , , , , , and .
Finally adding yields .
~ Nafer
Video Solution
https://www.youtube.com/watch?v=uWNExJc3hok
Solution 3 (One page solution)
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Video Solution
To see a comprehensive video covering this, visit : https://www.youtube.com/watch?v=uWNExJc3hok