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2007 AIME I Problems/Problem 6

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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?

Contents


Solution

Solution 1

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 12
    • Continuing from 12, there is 1 \cdot 1 = 1 way to continue to 24
    • There are \displaystyle 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 \displaystyle 13 \Rightarrow 26) to get to 26.

Regrouping, work from \displaystyle 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 \displaystyle 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

See also

2007 AIME I (ProblemsResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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