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

Revision as of 14:41, 8 June 2020

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 $101.$

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 ==See Also==
 {{AIME box|year=2020|n=II|num-b=7|num-a=9}}
+[[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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

Revision as of 14:41, 8 June 2020

Problem

Solution (Official MAA)

See Also