Difference between revisions of "2020 AIME II Problems/Problem 8"
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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 = n(n-2)(n+1)</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>101.</math> | 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 = n(n-2)(n+1)</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>101.</math> | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/g13o0wgj4p0 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 01:37, 14 June 2020
Problem
Define a sequence recursively by and
for integers
. Find the least value of
such that the sum of the zeros of
exceeds
.
Solution (Official MAA)
First it will be shown by induction that the zeros of are the integers
, where
This is certainly true for . Suppose that it is true for
, and note that the zeros of
are the solutions of
, where
is a nonnegative zero of
. Because the zeros of
form an arithmetic sequence with common difference
so do the zeros of
. The greatest zero of
is
so the greatest zero of
is
and the least is
.
It follows that the number of zeros of is
, and their average value is
. The sum of the zeros of
is
Let
, so the sum of the zeros exceeds
if and only if
Because
is increasing for
, the values
and
show that the requested value of
is
Video Solution
~IceMatrix
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.