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, \\ | ||
Line 31: | Line 31: | ||
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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