Difference between revisions of "2018 AIME I Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | (incomplete, someone help format this) | + | (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. <math>P_2, P_1, S, P_5</math> all equal the sum of their adjacent elements in the previous jump. <math>P_3</math> equals the previous <math>P_2</math>. <math>P_4</math> equals the previous <math>P_5</math>. Each <math>E</math> equals the previous adjacent element in the previous jump. | + | Make a table showing how many ways there are to get to each vertex in a certain amount of jumps. <math>P_2, P_1, S, P_5</math> all equal the sum of their adjacent elements in the previous jump. <math>P_3</math> equals the previous <math>P_2</math>. <math>P_4</math> equals the previous <math>P_5</math>. Each <math>E</math> equals the previous adjacent element in the previous jump. \\ |
Jump & E & P_3 & P_2 & P_1 & S & P_5 & P_4 & E | Jump & E & P_3 & P_2 & P_1 & S & P_5 & P_4 & E | ||
| − | 0 0 0 0 0 1 0 0 0 | + | 0 0 0 0 0 1 0 0 0 \\ |
| − | 1 0 0 0 1 0 1 0 0 | + | 1 0 0 0 1 0 1 0 0 \\ |
| − | 2 0 0 1 0 2 0 1 0 | + | 2 0 0 1 0 2 0 1 0 \\ |
| − | 3 0 1 0 3 0 3 0 1 | + | 3 0 1 0 3 0 3 0 1 \\ |
| − | 4 1 0 4 0 6 0 3 1 | + | 4 1 0 4 0 6 0 3 1 \\ |
| − | 5 1 4 0 10 0 9 0 4 | + | 5 1 4 0 10 0 9 0 4 \\ |
| − | 6 5 0 14 0 19 0 9 4 | + | 6 5 0 14 0 19 0 9 4 \\ |
| − | 7 5 14 0 33 0 28 0 13 | + | 7 5 14 0 33 0 28 0 13 \\ |
| − | 8 19 0 47 0 61 0 28 13 | + | 8 19 0 47 0 61 0 28 13 \\ |
| − | 9 19 47 0 108 0 89 0 41 | + | 9 19 47 0 108 0 89 0 41 \\ |
| − | 10 66 0 155 0 197 0 89 41 | + | 10 66 0 155 0 197 0 89 41 \\ |
| − | 11 66 155 0 352 0 286 0 130 | + | 11 66 155 0 352 0 286 0 130 \\ |
| − | 12 221 - - - - - - 130 | + | 12 221 - - - - - - 130 \\ \\ |
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>. | ||
Revision as of 11:50, 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 
.
