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Difference between revisions of "2007 AIME I Problems/Problem 6"

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Let us keep a careful tree of the possible number of paths around every multiple of <math>13</math>.
 
Let us keep a careful tree of the possible number of paths around every multiple of <math>13</math>.
  
−
From <math>\displaystyle 0 \Rightarrow 13</math>, we can end at either <math>12</math> (mult. of 3) or <math>13</math> (mult. of 13).  
+
From <math>0 \Rightarrow 13</math>, we can end at either <math>12</math> (mult. of 3) or <math>13</math> (mult. of 13).  
  
 
*Only <math>1</math> path leads to <math>12</math>
 
*Only <math>1</math> path leads to <math>12</math>
 
**Continuing from <math>12</math>, there is <math>1 \cdot 1 = 1</math> way to continue to <math>24</math>
 
**Continuing from <math>12</math>, there is <math>1 \cdot 1 = 1</math> way to continue to <math>24</math>
−
**There are <math>\displaystyle 1 \cdot \left(\frac{24-15}{3} + 1\right) = 4</math> ways to reach <math>26</math>.
+
**There are <math>1 \cdot \left(\frac{24-15}{3} + 1\right) = 4</math> ways to reach <math>26</math>.
 
*There are <math>\frac{12 - 0}{3} + 1 = 5</math> ways to reach <math>13</math>.  
 
*There are <math>\frac{12 - 0}{3} + 1 = 5</math> ways to reach <math>13</math>.  
 
** Continuing from <math>13</math>, there are <math>5 \cdot 1 = 5</math> ways to get to <math>24</math>
 
** Continuing from <math>13</math>, there are <math>5 \cdot 1 = 5</math> ways to get to <math>24</math>
−
**There are <math>5 \cdot \left(\frac{24-15}{3} + 1 + 1\right) = 25</math> ways (the first 1 to make it inclusive, the second to also jump from <math>\displaystyle 13 \Rightarrow 26</math>) to get to <math>26</math>.  
+
**There are <math>5 \cdot \left(\frac{24-15}{3} + 1 + 1\right) = 25</math> ways (the first 1 to make it inclusive, the second to also jump from <math>13 \Rightarrow 26</math>) to get to <math>26</math>.  
  
−
Regrouping, work from <math>\displaystyle 24 | 26\Rightarrow 39</math>
+
Regrouping, work from <math>24 | 26\Rightarrow 39</math>
 
*There are <math>1 + 5 = 6</math> ways to get to <math>24</math>  
 
*There are <math>1 + 5 = 6</math> ways to get to <math>24</math>  
 
** Continuing from <math>24</math>, there are <math>6 \cdot \left(\frac{39 - 27}{3}\right) = 24</math> ways to continue to <math>39</math>.
 
** Continuing from <math>24</math>, there are <math>6 \cdot \left(\frac{39 - 27}{3}\right) = 24</math> ways to continue to <math>39</math>.
 
*There are <math>4 + 25 = 29</math> ways to reach <math>26</math>.  
 
*There are <math>4 + 25 = 29</math> ways to reach <math>26</math>.  
−
** Continuing from <math>26</math>, there are <math>29 \cdot \left(\frac{39-27}{3} + 1\right) = 145</math> (note that the 1 is not to inclusive, but to count <math>\displaystyle 26 \Rightarrow 39</math>).  
+
** Continuing from <math>26</math>, there are <math>29 \cdot \left(\frac{39-27}{3} + 1\right) = 145</math> (note that the 1 is not to inclusive, but to count <math>26 \Rightarrow 39</math>).  
  
 
In total, we get <math>145 + 26 = 169</math>.
 
In total, we get <math>145 + 26 = 169</math>.
Line 59: Line 59:
  
 
The answer is <math>6*6*5 - 11=169</math>
 
The answer is <math>6*6*5 - 11=169</math>
 +
 +
=== Solution 3 ===
 +
 +
Another way would be to use a table representing the number of ways to reach a certain number
 +
 +
<math>\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}
 +
0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\
 +
\hline
 +
1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\
 +
\end{tabular}</math>
 +
 +
How we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for <math>26</math>, we can reach it from <math>13, 15, 18, 21, 24</math>, so we add all those values to get the value for <math>26</math>. For <math>27</math>, it is only reachable from <math>24</math> or <math>26</math>, so we have <math>29 + 6 = 35</math>.
 +
 +
The answer for <math>39</math> can be computed in a similar way to get <math>35 * 4 + 29 = \boxed{169}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 23:13, 8 March 2011

Problem

A frog is placed at the origin on the number line, and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of 3, or to the closest point with a greater integer coordinate that is a multiple of 13. A move sequence is a sequence of coordinates which correspond to valid moves, beginning with 0, and ending with 39. For example, $0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39$ is a move sequence. How many move sequences are possible for the frog?

Solution

Solution 1

Let us keep a careful tree of the possible number of paths around every multiple of $13$.

From $0 \Rightarrow 13$, we can end at either $12$ (mult. of 3) or $13$ (mult. of 13).

  • Only $1$ path leads to $12$
    • Continuing from $12$, there is $1 \cdot 1 = 1$ way to continue to $24$
    • There are $1 \cdot \left(\frac{24-15}{3} + 1\right) = 4$ ways to reach $26$.
  • There are $\frac{12 - 0}{3} + 1 = 5$ ways to reach $13$.
    • Continuing from $13$, there are $5 \cdot 1 = 5$ ways to get to $24$
    • There are $5 \cdot \left(\frac{24-15}{3} + 1 + 1\right) = 25$ ways (the first 1 to make it inclusive, the second to also jump from $13 \Rightarrow 26$) to get to $26$.

Regrouping, work from $24 | 26\Rightarrow 39$

  • There are $1 + 5 = 6$ ways to get to $24$
    • Continuing from $24$, there are $6 \cdot \left(\frac{39 - 27}{3}\right) = 24$ ways to continue to $39$.
  • There are $4 + 25 = 29$ ways to reach $26$.
    • Continuing from $26$, there are $29 \cdot \left(\frac{39-27}{3} + 1\right) = 145$ (note that the 1 is not to inclusive, but to count $26 \Rightarrow 39$).

In total, we get $145 + 26 = 169$.


In summary, we can draw the following tree, where in $(x,y)$, $x$ represents the current position on the number line, and $y$ represents the number of paths to get there:

  • $(12,1)$
    • $(24,1)$
      • $(39,4)$
    • $(26,4)$
      • $(39,20)$
  • $(13,5)$
    • $(24,5)$
      • $(39,20)$
    • $(26,25)$
      • $(39,125)$

Again, this totals $4 + 20 + 20 + 125 = 169$.

Solution 2

We divide it into 3 stages. The first occurs before the frog moves past 13. The second occurs before it moves past 26, and the last is everything else.

For the first stage the possible paths are $(0,13)$, $(0,3,13)$, $(0,3,6,13)$, $(0,3,6,9,13)$, $(0,3,6,9,12,13)$, and $(0,3,6,9,12)$. That is a total of 6.

For the second stage the possible paths are $(26)$, $(15,26)$, $(15,18,26)$, $(15,18,21,26)$, $(15,18,21,24,26)$, and $(15,18,21,24)$. That is a total of 6.

For the second stage the possible paths are $(39)$, $(27,39)$, $(27,30,39)$, $(27,30,33,39)$, and $(27,30,33,36,39)$. That is a total of 5.

However, we cannot jump from $12 \Rightarrow 26$ (this eliminates 5 paths) or $24 \Rightarrow 39$ (this eliminates 6 paths), so we must subtract $6 + 5 = 11$.

The answer is $6*6*5 - 11=169$

Solution 3

Another way would be to use a table representing the number of ways to reach a certain number

$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\ \hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\ \end{tabular}$

How we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for $26$, we can reach it from $13, 15, 18, 21, 24$, so we add all those values to get the value for $26$. For $27$, it is only reachable from $24$ or $26$, so we have $29 + 6 = 35$.

The answer for $39$ can be computed in a similar way to get $35 * 4 + 29 = \boxed{169}$.

See also

2007 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AIME Problems and Solutions
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