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2018 AIME I Problems/Problem 14: Difference between revisions

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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 $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$. From any vertex of the heptagon except $E$, the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that end at $E$.

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. $P_2, P_1, S, P_5$ all equal the sum of their adjacent elements in the previous jump. $P_3$ equals the previous $P_2$. $P_4$ equals the previous $P_5$. Each $E$ 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 $221 + 130 = \boxed{351}$.

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