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

(Solution 2 (More Variables))
(8 intermediate revisions by the same user not shown)
Line 8: Line 8:
 
\end{cases}</cmath>for positive integers <math>n</math>. Find the least positive integer <math>n</math> such that <math>\tfrac{f(n)}{g(n)} = \tfrac{4}{7}</math>.
 
\end{cases}</cmath>for positive integers <math>n</math>. Find the least positive integer <math>n</math> such that <math>\tfrac{f(n)}{g(n)} = \tfrac{4}{7}</math>.
  
==Solution==
+
==Solution 1==
  
 
Consider what happens when we try to calculate <math>f(n)</math> where n is not a square. If <math>k^2<n<(k+1)^2</math> for (positive) integer k, recursively calculating the value of the function gives us <math>f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n</math>. Note that this formula also returns the correct value when <math>n=(k+1)^2</math>, but not when <math>n=k^2</math>. Thus <math>f(n)=k^2+3k+2-n</math> for <math>k^2<n \leq (k+1)^2</math>.
 
Consider what happens when we try to calculate <math>f(n)</math> where n is not a square. If <math>k^2<n<(k+1)^2</math> for (positive) integer k, recursively calculating the value of the function gives us <math>f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n</math>. Note that this formula also returns the correct value when <math>n=(k+1)^2</math>, but not when <math>n=k^2</math>. Thus <math>f(n)=k^2+3k+2-n</math> for <math>k^2<n \leq (k+1)^2</math>.
Line 16: Line 16:
 
Write <math>7f(n)=4g(n)</math>, which simplifies to <math>3k^2+k-10=3n</math>. Notice that we want the <math>LHS</math> expression to be divisible by 3; as a result, <math>k \equiv 1 \pmod{3}</math>. We also want n to be strictly greater than <math>k^2</math>, so <math>k-10>0, k>10</math>. 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 <math>k=16</math>, giving <math>n=258</math>.
 
Write <math>7f(n)=4g(n)</math>, which simplifies to <math>3k^2+k-10=3n</math>. Notice that we want the <math>LHS</math> expression to be divisible by 3; as a result, <math>k \equiv 1 \pmod{3}</math>. We also want n to be strictly greater than <math>k^2</math>, so <math>k-10>0, k>10</math>. 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 <math>k=16</math>, giving <math>n=258</math>.
  
Indeed - if we check our answer, it works. Therefore, the answer is <math>\boxed{258}</math>
+
Indeed - if we check our answer, it works. Therefore, the answer is <math>\boxed{258}</math>.
  
 
-Ross Gao
 
-Ross Gao
 +
 +
==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{15mm}(1)</cmath> for some nonnegative integer <math>p.</math> By observations, we get
 +
<cmath>\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{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 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.
 +
 +
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
 +
<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>
 +
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{15mm}(5)</cmath>
 +
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{15mm}(6)</cmath>
 +
We substitute <math>(6)</math> into <math>(5),</math> then simplify, cross-multiply, and rearrange:
 +
<cmath>\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{15mm}(7)
 +
\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> Furthermore, rewriting the restriction <math>k^2<n</math> in terms of <math>p</math> and <math>q</math> gives
 +
<cmath>\begin{align*}
 +
k^2&<\underbrace{(k+1)^2-p}_{\text{by }(1)} \\
 +
p&<2k+1 &(8)\\
 +
p&<2\biggl(\phantom{ }\underbrace{\frac{2q-p-3}{2}}_{\text{by }(6)}\phantom{ }\biggr)+1 \\
 +
p&<(2q-p-3)+1 \\
 +
p&<q-1. &(9)
 +
\end{align*}</cmath>
 +
To minimize <math>n,</math> we should minimize <math>k.</math> By <math>(6),(8),</math> and <math>(9),</math> we should minimize <math>p</math> and <math>q.</math> From <math>(7),</math> we construct the following table:
 +
<cmath>\begin{array}{c|c|c}
 +
& & \\ [-2.5ex]
 +
\boldsymbol{p} & \boldsymbol{q} & \textbf{Satisfies }\boldsymbol{(9)?} \\ [0.5ex]
 +
\hline
 +
& & \\ [-2ex]
 +
11 & 11 & \\
 +
21 & 22 & \\
 +
31 & 33 & \checkmark \\
 +
\geq41 & \geq44 & \checkmark \\
 +
\end{array}</cmath>
 +
We have <math>(p,q)=(31,33).</math> Substituting this result into <math>(6)</math> produces <math>k=16.</math> Finally, substituting everything into either <math>(1)</math> or <math>(3)</math> generates <math>n=\boxed{258}.</math>
 +
 +
<u><b>Remark</b></u>
 +
 +
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>
 +
 +
~MRENTHUSIASM
  
 
==Video Solution==
 
==Video Solution==

Revision as of 06:27, 12 May 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 (More Variables)

We restrict $n$ in which $k^2<n\leq(k+1)^2$ for some positive integer $k,$ or \[n=(k+1)^2-p\hspace{15mm}(1)\] for some nonnegative integer $p.$ By observations, 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{15mm}(2) \\ \end{align*} If $n$ and $(k+1)^2$ have the same parity, then starting with $g\left((k+1)^2\right)=k+1,$ 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.

Along with the earlier restriction, note that $k^2<n\leq(k+1)^2<(k+2)^2,$ or \[n=(k+2)^2-2q\hspace{15mm}(3)\] for some positive integer $q.$ By observations, 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{15mm}(4) \\ \end{align*} By $(2)$ and $(4),$ we have \[\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{15mm}(5)\] From $(1)$ and $(3),$ equating the expressions for $n$ gives $(k+1)^2-p=(k+2)^2-2q.$ Solving for $k$ produces \[k=\frac{2q-p-3}{2}. \hspace{15mm}(6)\] 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{15mm}(7) \end{align*} Since $\gcd(11,10)=1,$ we know that $p-1$ must be divisible by $10,$ and $q$ must be divisible by $11.$ Furthermore, rewriting the restriction $k^2<n$ in terms of $p$ and $q$ gives \begin{align*} k^2&<\underbrace{(k+1)^2-p}_{\text{by }(1)} \\ p&<2k+1 &(8)\\ p&<2\biggl(\phantom{ }\underbrace{\frac{2q-p-3}{2}}_{\text{by }(6)}\phantom{ }\biggr)+1 \\ p&<(2q-p-3)+1 \\ p&<q-1. &(9) \end{align*} To minimize $n,$ we should minimize $k.$ By $(6),(8),$ and $(9),$ we should minimize $p$ and $q.$ From $(7),$ we construct the following table: \[\begin{array}{c|c|c} & & \\ [-2.5ex] \boldsymbol{p} & \boldsymbol{q} & \textbf{Satisfies }\boldsymbol{(9)?} \\ [0.5ex] \hline & & \\ [-2ex] 11 & 11 & \\ 21 & 22 & \\ 31 & 33 & \checkmark \\ \geq41 & \geq44 & \checkmark \\ \end{array}\] We have $(p,q)=(31,33).$ Substituting this result into $(6)$ produces $k=16.$ Finally, substituting everything into either $(1)$ or $(3)$ generates $n=\boxed{258}.$

Remark

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

~MRENTHUSIASM

Video Solution

https://youtu.be/tRVe2bKwIY8

See also

2021 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png