2021 AIME II Problems/Problem 15
Problem
Let and be functions satisfying and for positive integers . Find the least positive integer such that .
Solution 1
Consider what happens when we try to calculate where n is not a square. If for (positive) integer k, recursively calculating the value of the function gives us . Note that this formula also returns the correct value when , but not when . Thus for .
If , returns the same value as . This is because the recursion once again stops at . We seek a case in which , so obviously this is not what we want. We want to have a different parity, or have the same parity. When this is the case, instead returns .
Write , which simplifies to . Notice that we want the expression to be divisible by 3; as a result, . We also want n to be strictly greater than , so . The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is , giving .
Indeed - if we check our answer, it works. Therefore, the answer is .
-Ross Gao
Solution 2 (More Variables)
We restrict in which for some positive integer or for some nonnegative integer By observations, we get
~MRENTHUSIASM
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See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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