Difference between revisions of "2015 AIME I Problems/Problem 9"
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Note that there is an ultraspecial(Yes, I know that's not a word) case where we generate a pair of <math>a_i</math> that have difference one. This can only happen if <math>a_3 = 1</math> and <math>a_1</math> and <math>a_2</math> have difference <math>2</math>. This contributes <math>14</math> cases(<math>2 * 8</math> and then subtract <math>2</math> because of the cases <math>3,1,1</math> and <math>4,2,1</math>). | Note that there is an ultraspecial(Yes, I know that's not a word) case where we generate a pair of <math>a_i</math> that have difference one. This can only happen if <math>a_3 = 1</math> and <math>a_1</math> and <math>a_2</math> have difference <math>2</math>. This contributes <math>14</math> cases(<math>2 * 8</math> and then subtract <math>2</math> because of the cases <math>3,1,1</math> and <math>4,2,1</math>). | ||
− | Therefore, our answer is <math>190 + 290 + 14 = \boxed{ | + | Therefore, our answer is <math>190 + 290 + 14 = \boxed{494}</math>. |
Solution by hyxue | Solution by hyxue | ||
Revision as of 21:57, 11 October 2019
Contents
Problem
Let be the set of all ordered triple of integers with . Each ordered triple in generates a sequence according to the rule for all . Find the number of such sequences for which for some .
Solution
Let . First note that if any absolute value equals 0, then . Also note that if at any position, , then . Then, if any absolute value equals 1, then . Therefore, if either or is less than or equal to 1, then that ordered triple meets the criteria. Assume that to be the only way the criteria is met. To prove, let , and . Then, , , and . However, since the minimum values of and are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be , . Again assume that any other scenario will not meet criteria. To prove, divide the other scenarios into two cases: , , and ; and , , and . For the first one, , , , and , by which point we see that this function diverges. For the second one, , , , and , by which point we see that this function diverges.
Therefore, the only scenarios where is when any of the following are met:
(280 options)
(280 options, 80 of which coincide with option 1)
, . (16 options, 2 of which coincide with either option 1 or option 2)
Adding the total number of such ordered triples yields .
- Note to author:
Because , , , and doesn't mean the function diverges. What if , , and too?
Solution 2
Note that the only way for a to be produced at is if either or . Since the first one will eventually get to the first three assuming that there is no for any , that is not possible because . Therefore, we need .
If consecutive numbers out of are equal, then those cases work( and or and [b]NOT[/b] and ). This is simply by PIE.
Now, note that if any of the first three numbers have difference of , we have another working case. First, we calculate how many there are given exactly one of or have difference . Given numbers such that the first have difference , exactly permutations work(assuming the numbers are and such that ): ; ; ; and . If the two consecutive numbers are and , then the last number has possiblities: . This is symmetric for and . If the consecutive numbers are , there are possibilities( minus the numbers themselves and the numbers directly above and below). Note that we are not counting any cases already counted in the first case. Therefore, this case gives you . Now we consider the case that there both adjacent s have increments of . Therefore this gives . However, note that we have to add the case where you have consecutive numbers in arrangement such that only consecutive numbers have difference . For example, is one such triple. There are triples of consecutive numbers and ways to arrange each one(e.x: ). This adds on 32 working cases, so this case gives .
Note that there is an ultraspecial(Yes, I know that's not a word) case where we generate a pair of that have difference one. This can only happen if and and have difference . This contributes cases( and then subtract because of the cases and ).
Therefore, our answer is . Solution by hyxue
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.