Difference between revisions of "2021 AIME II Problems/Problem 15"
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==Solution 2 (More Variables)== | ==Solution 2 (More Variables)== | ||
− | We restrict <math>n</math> in which <math>k^2<n\leq(k+1)^2</math> for some positive integer <math>k,</math> or <cmath>n=(k+1)^2-p\hspace{40mm}(1)</cmath> for some | + | We restrict <math>n</math> in which <math>k^2<n\leq(k+1)^2</math> for some positive integer <math>k,</math> or <cmath>n=(k+1)^2-p\hspace{40mm}(1)</cmath> for some integer <math>p</math> such that <math>0\leq p<2k+1.</math> By observations, we get |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
f\left((k+1)^2\right)&=k+1, \\ | f\left((k+1)^2\right)&=k+1, \\ | ||
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If <math>n</math> and <math>(k+1)^2</math> have the same parity, then starting from <math>g\left((k+1)^2\right)=k+1,</math> we get <math>g(n)=f(n)</math> in 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> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity. | If <math>n</math> and <math>(k+1)^2</math> have the same parity, then starting from <math>g\left((k+1)^2\right)=k+1,</math> we get <math>g(n)=f(n)</math> in 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> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity. | ||
− | Along with the earlier restriction, we obtain <math>k^2<n\leq(k+1)^2<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{38.25mm}(3)</cmath> for some | + | Along with the earlier restriction, we obtain <math>k^2<n\leq(k+1)^2<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{38.25mm}(3)</cmath> for some integer <math>q</math> such that <math>0<p<2k+2.</math> By observations, we get |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
g\left((k+2)^2\right)&=k+2, \\ | g\left((k+2)^2\right)&=k+2, \\ | ||
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11(p-1)&=10q. \hspace{29mm}(7) | 11(p-1)&=10q. \hspace{29mm}(7) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Since <math>\gcd(11,10)=1,</math> we know that <math>p-1</math> must be divisible by <math>10,</math> and <math>q</math> must be divisible by <math>11.</math | + | Since <math>\gcd(11,10)=1,</math> we know that <math>p-1</math> must be divisible by <math>10,</math> and <math>q</math> must be divisible by <math>11.</math> |
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− | + | Recall that the restrictions on <math>(1)</math> and <math>(2)</math> are <math>0\leq p<2k+1</math> and <math>0<q<2k+2,</math> respectively. Substituting <math>(6)</math> into either inequality gives <math>p+1<q.</math> Combining all these results produces <cmath>0<p+1<q<2k+2. \hspace{28mm}(8)</cmath> | |
− | To minimize <math>n</math> in either <math>(1)</math> or <math>(3),</math> we should minimize <math>k,</math> so we should minimize <math>p</math> and <math>q</math> in <math>( | + | To minimize <math>n</math> in either <math>(1)</math> or <math>(3),</math> we should minimize <math>k,</math> so we should minimize <math>p</math> and <math>q</math> in <math>(8).</math> |
From <math>(6)</math> and <math>(7),</math> we construct the following table: | From <math>(6)</math> and <math>(7),</math> we construct the following table: | ||
<cmath>\begin{array}{c|c|c|c} | <cmath>\begin{array}{c|c|c|c} | ||
& & & \\ [-2.5ex] | & & & \\ [-2.5ex] | ||
− | \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{k} & \textbf{Satisfies }\boldsymbol{( | + | \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{k} & \textbf{Satisfies }\boldsymbol{(8)?} \\ [0.5ex] |
\hline | \hline | ||
& & & \\ [-2ex] | & & & \\ [-2ex] |
Revision as of 18:58, 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 integer such that By observations, we get If and have the same parity, then starting from we get in a similar process. This contradicts the precondition Therefore, and must have different parities, from which and must have the same parity.
Along with the earlier restriction, we obtain or for some integer such that By observations, we get By and we have From and equating the expressions for gives Solving for produces We substitute into then simplify, cross-multiply, and rearrange: Since we know that must be divisible by and must be divisible by
Recall that the restrictions on and are and respectively. Substituting into either inequality gives Combining all these results produces
To minimize in either or we should minimize so we should minimize and in
From and we construct the following table: Finally, we have Substituting this result into either or generates
Remark
We can verify that
~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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