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Difference between revisions of "2020 AIME II Problems/Problem 8"

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==Problem==
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Define a sequence recursively by <math>f_1(x)=|x-1|</math> and <math>f_n(x)=f_{n-1}(|x-n|)</math> for integers <math>n>1</math>. Find the least value of <math>n</math> such that the sum of the zeros of <math>f_n</math> exceeds <math>500,000</math>.
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==Solution 1 (Official MAA)==
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First it will be shown by induction that the zeros of <math>f_n</math> are the integers
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<math>a, {a+2,} {a+4,} \dots, {a + n(n-1)}</math>, where <math>a = n - \frac{n(n-1)}2.</math>
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This is certainly true for <math>n=1</math>. Suppose that it is true for <math>n = m-1 \ge 1</math>, and note that the zeros of <math>f_m</math> are the solutions of <math>|x - m| = k</math>, where <math>k</math> is a nonnegative zero of <math>f_{m-1}</math>. Because the zeros of <math>f_{m-1}</math> form an arithmetic sequence with common difference <math>2,</math> so do the zeros of <math>f_m</math>. The greatest zero of <math>f_{m-1}</math> is<cmath>m-1+\frac{(m-1)(m-2)}2 =\frac{m(m-1)}2,</cmath>so the greatest zero of <math>f_m</math> is <math>m+\frac{m(m-1)}2</math> and the least is <math>m-\frac{m(m-1)}2</math>.
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It follows that the number of zeros of <math>f_n</math> is <math>\frac{n(n-1)}2+1=\frac{n^2-n+2}2</math>, and their average value is <math>n</math>. The sum of the zeros of <math>f_n</math> is<cmath>\frac{n^3-n^2+2n}2.</cmath>Let <math>S(n)=n^3-n^2+2n</math>, so the sum of the zeros exceeds <math>500{,}000</math> if and only if <math>S(n) > 1{,}000{,}000 = 100^3\!.</math> Because <math>S(n)</math> is increasing for <math>n > 2</math>, the values <math>S(100) = 1{,}000{,}000 - 10{,}000 + 200 = 990{,}200</math> and <math>S(101)=1{,}030{,}301 - 10{,}201 + 202 = 1{,}020{,}302</math> show that the requested value of <math>n</math> is <math>\boxed{101}</math>.
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==Solution 2 (Same idea, easier to see)==
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Starting from <math>f_1(x)=|x-1|</math>, we can track the solutions, the number of solutions, and their sum.
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<cmath>\begin{array}{c|c|c|c}
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n&Solutions&number&sum\\
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1&1&1&1\\ 
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2&1,3&2&4\\ 
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3&0,2,4,6&4&12\\
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4&-2,0,2...10&7&28\\
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5&-5,-3,-1...15&11&55\\
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\end{array}</cmath>
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It is clear that the solutions form arithmetic sequences with a difference of 2, and the negative solutions cancel out all but <math>n</math> of the <math>1+\frac{n(n-1)}{2}</math> solutions. Thus, the sum of the solutions is <math>n \cdot [1+\frac{n(n-1)}{2}]</math>, which is a cubic function.
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<math>n \cdot [1+\frac{n(n-1)}{2}]>500,000</math>
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Multiplying both sides by <math>2</math>,
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<math>n \cdot [2+n(n-1)]>1,000,000</math>
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1 million is <math>10^6=100^3</math>, so the solution should be close to <math>100</math>.
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100 is slightly too small, so <math>\boxed{101}</math> works.
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~ dragnin
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==Video Solution==
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https://youtu.be/EwGydCuoHNM
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~MathProblemSolvingSkills.com
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==Video Solution==
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https://youtu.be/g13o0wgj4p0
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==See Also==
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{{AIME box|year=2020|n=II|num-b=7|num-a=9}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 10:22, 20 January 2024

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 1 (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 $S(n)=n^3-n^2+2n$, 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 $\boxed{101}$.

Solution 2 (Same idea, easier to see)

Starting from $f_1(x)=|x-1|$, we can track the solutions, the number of solutions, and their sum.

\[\begin{array}{c|c|c|c} n&Solutions&number&sum\\ 1&1&1&1\\ 2&1,3&2&4\\ 3&0,2,4,6&4&12\\ 4&-2,0,2...10&7&28\\ 5&-5,-3,-1...15&11&55\\ \end{array}\]

It is clear that the solutions form arithmetic sequences with a difference of 2, and the negative solutions cancel out all but $n$ of the $1+\frac{n(n-1)}{2}$ solutions. Thus, the sum of the solutions is $n \cdot [1+\frac{n(n-1)}{2}]$, which is a cubic function.

$n \cdot [1+\frac{n(n-1)}{2}]>500,000$

Multiplying both sides by $2$,

$n \cdot [2+n(n-1)]>1,000,000$

1 million is $10^6=100^3$, so the solution should be close to $100$.

100 is slightly too small, so $\boxed{101}$ works.

~ dragnin

Video Solution

https://youtu.be/EwGydCuoHNM

~MathProblemSolvingSkills.com

Video Solution

https://youtu.be/g13o0wgj4p0

See Also

2020 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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All AIME Problems and Solutions

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