Difference between revisions of "2021 AIME II Problems/Problem 15"

(Solution 2 (More Variables))
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-Ross Gao
 
-Ross Gao
  
==Solution 2 (More Variables)==
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==Solution 2 (Four 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{15mm}(1)</cmath> for some nonnegative integer <math>p.</math> By observations, we get
+
We consider <math>f(n)</math> and <math>g(n)</math> separately:
 +
<ul style="list-style-type:square;">
 +
  <li><math>\boldsymbol{f(n)}</math><p>
 +
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 recursion, 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 27: Line 30:
 
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. \hspace{15mm}(2) \\
+
f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{19mm}(2) \\
\end{align*}</cmath>
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\end{align*}</cmath></li><p>
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 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> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity.
+
  <li><math>\boldsymbol{g(n)}</math><p>
 
+
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> 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> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity. <p>
Along with the earlier restriction, note that <math>k^2<n\leq(k+1)^2<(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
+
It follows that <math>k^2<n<(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<q<2k+2.</math> By recursion, 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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g\left((k+2)^2-4\right)&=k+6, \\
 
g\left((k+2)^2-4\right)&=k+6, \\
 
&\cdots \\
 
&\cdots \\
g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15mm}(4) \\
+
g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15.5mm}(4) \\
\end{align*}</cmath>
+
\end{align*}</cmath></li><p>
Together, we have <cmath>\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{15mm}(5)</cmath>
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  <li><b>Answer</b><p>
From <math>(1)</math> and <math>(3),</math> equating the expressions for <math>n</math> gives <math>(k+1)^2-p=(k+2)^2-2q.</math> Solving for <math>k</math> produces <cmath>k=\frac{2q-p-3}{2}.</cmath>
+
By <math>(2)</math> and <math>(4),</math> we have <cmath>\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{27mm}(5)</cmath>
We substitute this into <math>(5),</math> then simplify, cross-multiply, and rearrange:
+
From <math>(1)</math> and <math>(3),</math> equating the expressions for <math>n</math> gives <math>(k+1)^2-p=(k+2)^2-2q.</math> Solving for <math>k</math> produces <cmath>k=\frac{2q-p-3}{2}. \hspace{41.25mm}(6)</cmath>
 +
We substitute <math>(6)</math> into <math>(5),</math> then simplify, cross-multiply, and rearrange:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\frac{\tfrac{2q-p-3}{2}+p+1}{\tfrac{2q-p-3}{2}+2q+2}&=\frac{4}{7} \\
 
\frac{\tfrac{2q-p-3}{2}+p+1}{\tfrac{2q-p-3}{2}+2q+2}&=\frac{4}{7} \\
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7p+14q-7&=-4p+24q+4 \\
 
7p+14q-7&=-4p+24q+4 \\
 
11p-11&=10q \\
 
11p-11&=10q \\
11(p-1)&=10q.
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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> <p> 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 minimize <math>k,</math> so we 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:
 +
<cmath>\begin{array}{c|c|c|c}
 +
& & & \\ [-2.5ex]
 +
\boldsymbol{p} & \boldsymbol{q} & \boldsymbol{k} & \textbf{Satisfies }\boldsymbol{(8)?} \\ [0.5ex]
 +
\hline
 +
& & & \\ [-2ex]
 +
11 & 11 & 4 & \\
 +
21 & 22 & 10 & \\
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31 & 33 & 16 & \checkmark \\
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\geq41 & \geq44 & \geq22 & \checkmark \\
 +
\end{array}</cmath>
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Finally, we have <math>(p,q,k)=(31,33,16).</math> Substituting this result into either <math>(1)</math> or <math>(3)</math> generates <math>n=\boxed{258}.</math></li><p>
 +
  <li><b>Remark</b><p>
 +
We can verify that <cmath>\frac{f(258)}{g(258)}=\frac{1\cdot31+f(258+1\cdot31)}{2\cdot33+g(258+2\cdot33)}=\frac{31+\overbrace{f(289)}^{17}}{66+\underbrace{g(324)}_{18}}=\frac{48}{84}=\frac47.</cmath></li><p>
 +
</ul>
 +
~MRENTHUSIASM
 +
 +
==Solution 3==
 +
Since <math>n</math> isn't a perfect square, let <math>n=m^2+k</math> with <math>0<k<2m+1</math>. If <math>m</math> is odd, then <math>f(n)=g(n)</math>. If <math>m</math> is even,
 +
<cmath>f(n)=(m+1)^2-(m^2+k)+(m+1)=3m+2-k</cmath>
 +
<cmath>g(n)=(m+2)^2-(m^2+k)+(m+2)=5m+6-k</cmath>
 +
<cmath>7(3m+2-k)=4(5m+6-k)</cmath>
 +
<cmath>m=3k+10</cmath>
 +
Since <math>m</math> is even, <math>k</math> is even. Since <math>k\neq 0</math>, the smallest <math>k</math> is <math>2</math> which produces the smallest <math>n</math>. <math>k=2 \implies m=16 \implies n=16^2+2=\boxed{258}</math>
 +
 +
~Afo
 +
 +
==Solution 4 (Quadratics With Two Variables)==
 +
To begin, note that if <math>n</math> is a perfect square, <math>f(n)=g(n)</math>, so <math>f(n)/g(n)=1</math>, so we must look at values of <math>n</math> that are not perfect squares (what a surprise). First, let the distance between <math>n</math> and the first perfect square greater than or equal to it be <math>k</math>, making the values of <math>f(n+k)</math> and <math>g(n+k)</math> integers. Using this notation, we see than <math>f(n)=k+f(n+k)</math>, giving us a formula for the numerator of our ratio. However, since the function of <math>g(n)</math> does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of <math>\sqrt{n+k}</math> in <math>g(n)</math> unless <math>k</math> is an even number. However, this is impossible, since if <math>k</math> was an even number, <math>f(n)=g(n)</math>, giving a ratio of one. Thus, <math>k</math> must be an odd number.
 +
 +
Thus, since <math>k</math> must be an odd number, regardless of whether <math>n</math> is even or odd, to get an integral value in <math>g(n)</math>, we must get to the next perfect square after <math>n+k</math>. To make matters easier, let <math>z^2=n+k</math>. Thus, in <math>g(n)</math>, we want to achieve <math>(z+1)^2</math>.
 +
 +
Expanding <math>(z+1)^2</math> and substituting in the fact that <math>z=\sqrt{n+k}</math> yields:
 +
 +
<cmath>(z+1)^2=z^2+2z+1=n+k+2\sqrt{n+k}+1</cmath>
 +
 +
Thus, we must add the quantity <math>k+2z+1</math> to <math>n</math> to achieve a integral value in the function <math>g(n)</math>. Thus.
 +
 +
<cmath>g(n)=(k+2z+1)+\sqrt{n+k+2\sqrt{n+k}+1}</cmath>
 +
 +
However, note that the quantity within the square root is just <math>(z+1)^2</math>, and so:
 +
 +
<cmath>g(n)=k+3z+2</cmath>
 +
 +
Thus,
 +
<cmath>\frac{f(n)}{g(n)}=\frac{k+z}{k+3z+2}</cmath>
 +
 +
Since we want this quantity to equal <math>\frac{4}{7}</math>, we can set the above equation equal to this number and collect all the variables to one side to achieve
 +
 +
<cmath>3k-5z=8</cmath>
 +
 +
Substituting back in that <math>z=\sqrt{n+k}</math>, and then separating variables and squaring yields that
 +
 +
<cmath>9k^2-73k+64=25n</cmath>
 +
 +
Now, if we treat <math>n</math> as a constant, we can use the quadratic formula in respect to <math>k</math> to get an equation for <math>k</math> in terms of <math>n</math> (without all the squares). Doing so yields
 +
 +
<cmath>\frac{73\pm\sqrt{3025+900n}}{18}=k</cmath>
  
<b>SOLUTION IN PROGRESS. NO EDIT PLEASE--A MILLION THANKS.</b>
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Now, since <math>n</math> and <math>k</math> are integers, we want the quantity within the square root to be a perfect square. Note that <math>55^2=3025</math>. Thus, assume that the quantity within the root is equal to the perfect square, <math>m^2</math>. Thus, after using a difference of squares, we have
 +
<cmath>(m-55)(m+55)=900n</cmath>
 +
Since we want <math>n</math> to be an integer, we know that the <math>LHS</math> should be divisible by five, so, let's assume that we should have <math>m</math> divisible by five. If so, the quantity <math>18k-73</math> must be divisible by five, meaning that <math>k</math> leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!).
  
<u><b>Remark</b></u>
+
Thus, we see that to achieve integers <math>n</math> and <math>k</math> that could potentially satisfy the problem statement, we must try the values of <math>k</math> congruent to one modulo five. However, if we recall a statement made earlier in the problem, we see that we can skip all even values of <math>k</math> produced by this modulo argument.
  
We can verify that <cmath>\frac{f(258)}{g(258)}=\frac{31+\overbrace{f(289)}^{17}}{2\cdot33+\underbrace{f(324)}_{18}}=\frac{48}{84}=\frac47.</cmath>
+
Also, note that <math>k=1,6</math> won't work, as they are too small, and will give an erroneous value for <math>n</math>. After trying <math>k=11,21,31</math>, we see that <math>k=31</math> will give a value of <math>m=485</math>, which yields <math>n=\boxed{258}</math>, which, if plugged in to for our equations of <math>f(n)</math> and <math>g(n)</math>, will yield the desired ratio, and we're done.
  
~MRENTHUSIASM
+
Side Note: If any part of this solution is not rigorous, or too vague, please label it in the margin with "needs proof". If you can prove it, please add a lemma to the solution doing so :)
 +
 
 +
-mathislife52
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 19:06, 3 June 2021

Problem

Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\]and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\]for positive integers $n$. Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$.

Solution 1

Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2<n<(k+1)^2$ for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$. Note that this formula also returns the correct value when $n=(k+1)^2$, but not when $n=k^2$. Thus $f(n)=k^2+3k+2-n$ for $k^2<n \leq (k+1)^2$.

If $2 \mid (k+1)^2-n$, $g(n)$ returns the same value as $f(n)$. This is because the recursion once again stops at $(k+1)^2$. We seek a case in which $f(n)<g(n)$, so obviously this is not what we want. We want $(k+1)^2,n$ to have a different parity, or $n, k$ have the same parity. When this is the case, $g(n)$ instead returns $(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n$.

Write $7f(n)=4g(n)$, which simplifies to $3k^2+k-10=3n$. Notice that we want the $LHS$ expression to be divisible by 3; as a result, $k \equiv 1 \pmod{3}$. We also want n to be strictly greater than $k^2$, so $k-10>0, k>10$. 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 $k=16$, giving $n=258$.

Indeed - if we check our answer, it works. Therefore, the answer is $\boxed{258}$.

-Ross Gao

Solution 2 (Four Variables)

We consider $f(n)$ and $g(n)$ separately:

  • $\boldsymbol{f(n)}$

    We restrict $n$ in which $k^2<n\leq(k+1)^2$ for some positive integer $k,$ or \[n=(k+1)^2-p\hspace{40mm}(1)\] for some integer $p$ such that $0\leq p<2k+1.$ By recursion, we get \begin{align*} f\left((k+1)^2\right)&=k+1, \\ f\left((k+1)^2-1\right)&=k+2, \\ f\left((k+1)^2-2\right)&=k+3, \\ &\cdots \\ f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{19mm}(2) \\ \end{align*}

  • $\boldsymbol{g(n)}$

    If $n$ and $(k+1)^2$ have the same parity, then starting from $g\left((k+1)^2\right)=k+1,$ we get $g(n)=f(n)$ by a similar process. This contradicts the precondition $\frac{f(n)}{g(n)} = \frac{4}{7}.$ Therefore, $n$ and $(k+1)^2$ must have different parities, from which $n$ and $(k+2)^2$ must have the same parity.

    It follows that $k^2<n<(k+2)^2,$ or \[n=(k+2)^2-2q\hspace{38.25mm}(3)\] for some integer $q$ such that $0<q<2k+2.$ By recursion, we get \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{15.5mm}(4) \\ \end{align*}

  • Answer

    By $(2)$ and $(4),$ we have \[\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{27mm}(5)\] From $(1)$ and $(3),$ equating the expressions for $n$ gives $(k+1)^2-p=(k+2)^2-2q.$ Solving for $k$ produces \[k=\frac{2q-p-3}{2}. \hspace{41.25mm}(6)\] We substitute $(6)$ into $(5),$ then simplify, cross-multiply, and rearrange: \begin{align*} \frac{\tfrac{2q-p-3}{2}+p+1}{\tfrac{2q-p-3}{2}+2q+2}&=\frac{4}{7} \\ \frac{p+2q-1}{-p+6q+1}&=\frac{4}{7} \\ 7p+14q-7&=-4p+24q+4 \\ 11p-11&=10q \\ 11(p-1)&=10q. \hspace{29mm}(7) \end{align*} Since $\gcd(11,10)=1,$ we know that $p-1$ must be divisible by $10,$ and $q$ must be divisible by $11.$

    Recall that the restrictions on $(1)$ and $(2)$ are $0\leq p<2k+1$ and $0<q<2k+2,$ respectively. Substituting $(6)$ into either inequality gives $p+1<q.$ Combining all these results produces \[0<p+1<q<2k+2. \hspace{28mm}(8)\] To minimize $n$ in either $(1)$ or $(3),$ we minimize $k,$ so we minimize $p$ and $q$ in $(8).$ From $(6)$ and $(7),$ we construct the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{k} & \textbf{Satisfies }\boldsymbol{(8)?} \\ [0.5ex] \hline & & & \\ [-2ex] 11 & 11 & 4 & \\ 21 & 22 & 10 & \\ 31 & 33 & 16 & \checkmark \\ \geq41 & \geq44 & \geq22 & \checkmark \\ \end{array}\] Finally, we have $(p,q,k)=(31,33,16).$ Substituting this result into either $(1)$ or $(3)$ generates $n=\boxed{258}.$

  • Remark

    We can verify that \[\frac{f(258)}{g(258)}=\frac{1\cdot31+f(258+1\cdot31)}{2\cdot33+g(258+2\cdot33)}=\frac{31+\overbrace{f(289)}^{17}}{66+\underbrace{g(324)}_{18}}=\frac{48}{84}=\frac47.\]

~MRENTHUSIASM

Solution 3

Since $n$ isn't a perfect square, let $n=m^2+k$ with $0<k<2m+1$. If $m$ is odd, then $f(n)=g(n)$. If $m$ is even, \[f(n)=(m+1)^2-(m^2+k)+(m+1)=3m+2-k\] \[g(n)=(m+2)^2-(m^2+k)+(m+2)=5m+6-k\] \[7(3m+2-k)=4(5m+6-k)\] \[m=3k+10\] Since $m$ is even, $k$ is even. Since $k\neq 0$, the smallest $k$ is $2$ which produces the smallest $n$. $k=2 \implies m=16 \implies n=16^2+2=\boxed{258}$

~Afo

Solution 4 (Quadratics With Two Variables)

To begin, note that if $n$ is a perfect square, $f(n)=g(n)$, so $f(n)/g(n)=1$, so we must look at values of $n$ that are not perfect squares (what a surprise). First, let the distance between $n$ and the first perfect square greater than or equal to it be $k$, making the values of $f(n+k)$ and $g(n+k)$ integers. Using this notation, we see than $f(n)=k+f(n+k)$, giving us a formula for the numerator of our ratio. However, since the function of $g(n)$ does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of $\sqrt{n+k}$ in $g(n)$ unless $k$ is an even number. However, this is impossible, since if $k$ was an even number, $f(n)=g(n)$, giving a ratio of one. Thus, $k$ must be an odd number.

Thus, since $k$ must be an odd number, regardless of whether $n$ is even or odd, to get an integral value in $g(n)$, we must get to the next perfect square after $n+k$. To make matters easier, let $z^2=n+k$. Thus, in $g(n)$, we want to achieve $(z+1)^2$.

Expanding $(z+1)^2$ and substituting in the fact that $z=\sqrt{n+k}$ yields:

\[(z+1)^2=z^2+2z+1=n+k+2\sqrt{n+k}+1\]

Thus, we must add the quantity $k+2z+1$ to $n$ to achieve a integral value in the function $g(n)$. Thus.

\[g(n)=(k+2z+1)+\sqrt{n+k+2\sqrt{n+k}+1}\]

However, note that the quantity within the square root is just $(z+1)^2$, and so:

\[g(n)=k+3z+2\]

Thus, \[\frac{f(n)}{g(n)}=\frac{k+z}{k+3z+2}\]

Since we want this quantity to equal $\frac{4}{7}$, we can set the above equation equal to this number and collect all the variables to one side to achieve

\[3k-5z=8\]

Substituting back in that $z=\sqrt{n+k}$, and then separating variables and squaring yields that

\[9k^2-73k+64=25n\]

Now, if we treat $n$ as a constant, we can use the quadratic formula in respect to $k$ to get an equation for $k$ in terms of $n$ (without all the squares). Doing so yields

\[\frac{73\pm\sqrt{3025+900n}}{18}=k\]

Now, since $n$ and $k$ are integers, we want the quantity within the square root to be a perfect square. Note that $55^2=3025$. Thus, assume that the quantity within the root is equal to the perfect square, $m^2$. Thus, after using a difference of squares, we have \[(m-55)(m+55)=900n\] Since we want $n$ to be an integer, we know that the $LHS$ should be divisible by five, so, let's assume that we should have $m$ divisible by five. If so, the quantity $18k-73$ must be divisible by five, meaning that $k$ leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!).

Thus, we see that to achieve integers $n$ and $k$ that could potentially satisfy the problem statement, we must try the values of $k$ congruent to one modulo five. However, if we recall a statement made earlier in the problem, we see that we can skip all even values of $k$ produced by this modulo argument.

Also, note that $k=1,6$ won't work, as they are too small, and will give an erroneous value for $n$. After trying $k=11,21,31$, we see that $k=31$ will give a value of $m=485$, which yields $n=\boxed{258}$, which, if plugged in to for our equations of $f(n)$ and $g(n)$, will yield the desired ratio, and we're done.

Side Note: If any part of this solution is not rigorous, or too vague, please label it in the margin with "needs proof". If you can prove it, please add a lemma to the solution doing so :)

-mathislife52

Video Solution

https://youtu.be/tRVe2bKwIY8

See also

2021 AIME II (ProblemsAnswer KeyResources)
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