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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| + | {{AIME box|year=2018|n=I|num-b=13|num-a=15}} | ||
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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) | ||
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Followed by Problem 15 | |
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