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

Revision as of 19:01, 15 March 2007

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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

From $\displaystyle 0 \Rightarrow 13$ , we can end at either $12$ (mult. of 3) or $13$ (mult. of 13). Only $1$ path leads to the former, but there are $\frac{12 - 0}{3} + 1 = 5$ ways to reach $13$ .

Continuing from $12$ , there is $1 \cdot 1 = 1$ way to continue to $24$ , but there are $\displaystyle 1 \cdot \left(\frac{24-15}{3} + 1\right) = 4$ ways to reach $26$ . Continuing from $13$ , there are $5 \cdot 1 = 5$ ways to get to $24$ , and $5 \cdot \left(\frac{24-15}{3} + 1 + 1\right) = 25$ ways (the first 1 to make it inclusive, the second to also jump from $\displaystyle 13 \Rightarrow 26$ ) to get to $26$ . Regrouping, there are $1 + 5 = 6$ ways to get to $24$ and $4 + 25 = 29$ ways to reach $27$ .

Continuing from $24$ , there are $6 \cdot \left(\frac{39 - 27}{3}\right) = 24$ ways to continue to $39$ . 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 $\displaystyle 26 \Rightarrow 39$ ). In total, we get $145 + 26 = 169$ .

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 $12 \Rightarrow 26$ or $24 \Rightarrow 39$ , so we must subtract $6 + 6 - 1 = 11$ .

$6*6*5 - 11=169$

@@ Line 7: / Line 7: @@
 Let us keep a careful tree of the possible number of paths around every multiple of <math>13</math>.
-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 the former, but there are <math>\frac{12 - 0}{3} + 1 = 5</math> ways to reach <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). Only <math>1</math> path leads to the former, but there are <math>\frac{12 - 0}{3} + 1 = 5</math> ways to reach <math>13</math>.
-Continuing from <math>12</math>, there is <math>1 \cdot 1 = 1</math> way to continue to <math>24</math>, but there are <math>1 \cdot \left(\frac{24-15}{3} + 1\right) = 4</math> ways to reach <math>26</math>. Continuing from <math>13</math>, there are <math>5 \cdot 1 = 5</math> ways to get to <math>24</math>, and <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, there are <math>1 + 5 = 6</math> ways to get to <math>24</math> and <math>4 + 25 = 29</math> ways to reach <math>27</math>.
+Continuing from <math>12</math>, there is <math>1 \cdot 1 = 1</math> way to continue to <math>24</math>, but there are <math>\displaystyle 1 \cdot \left(\frac{24-15}{3} + 1\right) = 4</math> ways to reach <math>26</math>. Continuing from <math>13</math>, there are <math>5 \cdot 1 = 5</math> ways to get to <math>24</math>, and <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>. Regrouping, there are <math>1 + 5 = 6</math> ways to get to <math>24</math> and <math>4 + 25 = 29</math> ways to reach <math>27</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>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>.
+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>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>). In total, we get <math>145 + 26 = 169</math>.
 === Solution 2 ===

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search