# 2020 AIME II Problems/Problem 8

## Problem

Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$. Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$.

## Solution (Official MAA)

First it will be shown by induction that the zeros of $f_n$ are the integers $a, {a+2,} {a+4,} \dots, {a + n(n-1)}$, where $a = n - \frac{n(n-1)}2.$

This is certainly true for $n=1$. Suppose that it is true for $n = m-1 \ge 1$, and note that the zeros of $f_m$ are the solutions of $|x - m| = k$, where $k$ is a nonnegative zero of $f_{m-1}$. Because the zeros of $f_{m-1}$ form an arithmetic sequence with common difference $2,$ so do the zeros of $f_m$. The greatest zero of $f_{m-1}$ is $$m-1+\frac{(m-1)(m-2)}2 =\frac{m(m-1)}2,$$so the greatest zero of $f_m$ is $m+\frac{m(m-1)}2$ and the least is $m-\frac{m(m-1)}2$.

It follows that the number of zeros of $f_n$ is $\frac{n(n-1)}2+1=\frac{n^2-n+2}2$, and their average value is $n$. The sum of the zeros of $f_n$ is $$\frac{n^3-n^2+2n}2.$$Let $S(n)=n^3-n^2+2n = n(n-2)(n+1)$, so the sum of the zeros exceeds $500{,}000$ if and only if $S(n) > 1{,}000{,}000 = 100^3\!.$ Because $S(n)$ is increasing for $n > 2$, the values $S(100) = 1{,}000{,}000 - 10{,}000 + 200 = 990{,}200$ and $S(101)=1{,}030{,}301 - 10{,}201 + 202 = 1{,}020{,}302$ show that the requested value of $n$ is $\boxed{101}$.

~IceMatrix

## See Also

 2020 AIME II (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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