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

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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 we get $g(n)=f(n)$ by a similar process from $g\left((k+1)^2\right)=k+1.$ 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, then $\begin{align*} f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\ g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k, \end{align*}$ from which $\begin{align*} 7(3m+2-k)&=4(5m+6-k) \\ m&=3k+10. \end{align*}$ 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

@@ Line 33: / Line 33: @@
 \end{align*}</cmath></li><p>
    <li><math>\boldsymbol{g(n)}</math><p>
-If <math>n</math> and <math>(k+1)^2</math> have the same parity, then we get <math>g(n)=f(n)</math> from <math>g\left((k+1)^2\right)=k+1</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>
+If <math>n</math> and <math>(k+1)^2</math> have the same parity, then we get <math>g(n)=f(n)</math> by a similar process from <math>g\left((k+1)^2\right)=k+1.</math> 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>
 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*}

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