Difference between revisions of "2021 AIME II Problems/Problem 15"
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f\left((k+1)^2-2\right)&=k+3, \\ | f\left((k+1)^2-2\right)&=k+3, \\ | ||
&\cdots \\ | &\cdots \\ | ||
− | f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \\ | + | f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{15mm}(2) \\ |
+ | \end{align*}</cmath> | ||
+ | If <math>n</math> and <math>(k+1)^2</math> have the same parity, then starting with <math>g\left((k+1)^2\right)=k+1,</math> we will get <math>g(n)=f(n)</math> by a similar process. This contradicts the precondition <math>\frac{f(n)}{g(n)} = \frac{4}{7}.</math> Therefore, <math>n</math> and <math>(k+1)^2</math> have different parities, from which <math>n</math> and <math>(k+2)^2</math> have the same parity. | ||
+ | |||
+ | Along with the earlier restriction, note that <math>k^2<n<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{15mm}(3)</cmath> for some positive integer <math>q.</math> By observations, we get | ||
+ | <cmath>\begin{align*} | ||
+ | g\left((k+2)^2\right)&=k+2, \\ | ||
+ | g\left((k+2)^2-2\right)&=k+4, \\ | ||
+ | g\left((k+2)^2-4\right)&=k+6, \\ | ||
+ | &\cdots \\ | ||
+ | g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15mm}(4) \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 03:45, 12 May 2021
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 If and have the same parity, then starting with we will get by a similar process. This contradicts the precondition Therefore, and have different parities, from which and have the same parity.
Along with the earlier restriction, note that or for some positive integer By observations, we get
~MRENTHUSIASM
Video Solution
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 |
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