Difference between revisions of "2021 AIME II Problems/Problem 15"
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-Ross Gao | -Ross Gao | ||
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+ | ==Video Solution== | ||
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==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2021|n=II|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:04, 25 March 2021
Contents
Problem
Let and be functions satisfying and for positive integers . Find the least positive integer such that .
Solution
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
Video Solution
https://www.youtube.com/watch?v=U6WkStqOl8M
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.