Difference between revisions of "2015 AIME I Problems/Problem 9"
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However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be <math>z=1</math>, <math>|y-x|=2</math>. Again assume that any other scenario will not meet criteria. | However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be <math>z=1</math>, <math>|y-x|=2</math>. Again assume that any other scenario will not meet criteria. | ||
To prove, divide the other scenarios into two cases: <math>z>1</math>, <math>|y-x|>1</math>, and <math>|z-y|>1</math>; and <math>z=1</math>, <math>|y-x|>2</math>, and <math>|z-y|>1</math>. | To prove, divide the other scenarios into two cases: <math>z>1</math>, <math>|y-x|>1</math>, and <math>|z-y|>1</math>; and <math>z=1</math>, <math>|y-x|>2</math>, and <math>|z-y|>1</math>. | ||
− | For the first one, <math>a_4</math> | + | For the first one, <math>a_4</math> \ge 2z, <math>a_5</math> \ge 4z, <math>a_6</math> \ge 8z, and <math>a_7</math> \ge 16z, by which point we see that this function diverges. |
For the second one, <math>a_4 \ge 3</math>, <math>a_5 \ge 6</math>, <math>a_6 \ge 18</math>, and <math>a_7 \ge 54</math>, by which point we see that this function diverges. | For the second one, <math>a_4 \ge 3</math>, <math>a_5 \ge 6</math>, <math>a_6 \ge 18</math>, and <math>a_7 \ge 54</math>, by which point we see that this function diverges. | ||
Therefore, the only scenarios where <math>a_n</math>=0 is when any of the following are met: | Therefore, the only scenarios where <math>a_n</math>=0 is when any of the following are met: |
Revision as of 18:05, 10 May 2015
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 =0. Also note that if at any position, , then . Then, if any absolute value equals 1, then =0. 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, \ge 2z, \ge 4z, \ge 8z, and \ge 16z, 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 =0 is when any of the following are met: <2 (280 options) <2 (280 options, 80 of which coincide with option 1) z=1, =2. (16 options, 2 of which coincide with either option 1 or option 2) Adding the total number of such ordered triples yields .
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.