2021 AIME II Problems/Problem 15

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

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

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

Page

Toolbox

Search

2021 AIME II Problems/Problem 15

Problem

Solution

See also