2015 AIME I Problems/Problem 9
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 .
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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