Difference between revisions of "2020 AIME II Problems/Problem 8"
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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. | Starting from <math>f_1(x)=|x-1|</math>, we can track the solutions, the number of solutions, and their sum. | ||
− | < | + | <cmath>\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}</cmath> | |
− | \end{ | ||
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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. | 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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==Video Solution== | ==Video Solution== | ||
https://youtu.be/g13o0wgj4p0 | https://youtu.be/g13o0wgj4p0 | ||
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==See Also== | ==See Also== |
Latest revision as of 11:22, 20 January 2024
Contents
Problem
Define a sequence recursively by and
for integers
. Find the least value of
such that the sum of the zeros of
exceeds
.
Solution 1 (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
.
Solution 2 (Same idea, easier to see)
Starting from , we can track the solutions, the number of solutions, and their sum.
It is clear that the solutions form arithmetic sequences with a difference of 2, and the negative solutions cancel out all but of the
solutions. Thus, the sum of the solutions is
, which is a cubic function.
Multiplying both sides by ,
1 million is , so the solution should be close to
.
100 is slightly too small, so works.
~ dragnin
Video Solution
~MathProblemSolvingSkills.com
Video Solution
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.