Difference between revisions of "2018 AIME I Problems/Problem 14"
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The number of ways to jump to the ends within 12 jumps is <math>221 + 130 = \boxed{351}</math>. | The number of ways to jump to the ends within 12 jumps is <math>221 + 130 = \boxed{351}</math>. | ||
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Revision as of 13:11, 9 March 2018
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
(incomplete, someone help format this)
Make a table showing how many ways there are to get to each vertex in a certain amount of jumps. all equal the sum of their adjacent elements in the previous jump.
equals the previous
.
equals the previous
. Each
equals the previous adjacent element in the previous jump.
Jump & E & P_3 & P_2 & P_1 & S & P_5 & P_4 & E
0 0 0 0 0 1 0 0 0 \ 1 0 0 0 1 0 1 0 0 \ 2 0 0 1 0 2 0 1 0 \ 3 0 1 0 3 0 3 0 1 \ 4 1 0 4 0 6 0 3 1 \ 5 1 4 0 10 0 9 0 4 \ 6 5 0 14 0 19 0 9 4 \ 7 5 14 0 33 0 28 0 13 \ 8 19 0 47 0 61 0 28 13 \ 9 19 47 0 108 0 89 0 41 \ 10 66 0 155 0 197 0 89 41 \ 11 66 155 0 352 0 286 0 130 \ 12 221 - - - - - - 130
The number of ways to jump to the ends within 12 jumps is .
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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