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

Revision as of 01:56, 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 $\begin{align*} f\left((k+1)^2\right)&=k+1, \\ f\left((k+1)^2-1\right)&=k+2, \\ f\left((k+1)^2-2\right)&=k+3, \\ &\cdots \\ f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \\ \end{align*}$

~MRENTHUSIASM

Video Solution

https://youtu.be/tRVe2bKwIY8

@@ 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>.
-==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>.
@@ 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>.
-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
+==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. \\
+\end{align*}</cmath>
+~MRENTHUSIASM
 ==Video Solution==

2021 AIME II (Problems • Answer Key • Resources)
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