Difference between revisions of "2020 AIME II Problems/Problem 14"
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+ | ==Solution 2 (Official MAA)== | ||
+ | For any nonnegative integer <math>n</math>, the function <math>f</math> increases on the interval <math>[n,n+1)</math>, with <math>f(n)=0</math> and <math>f(x)<n+1</math> for every <math>x</math> in this interval. On this interval <math>f(x)=x(x-n)</math>, which takes on every real value in the interval <math>[0,n+1)</math> exactly once. Thus for each nonnegative real number <math>y</math>, the equation <math>f(x) = y</math> has exactly one solution <math>x \in [n, n+1)</math> for every <math>n \ge \lfloor y \rfloor</math>. | ||
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+ | For each integer <math>a\geq 17</math> there is exactly one <math>x</math> with <math>\lfloor x\rfloor=a</math> such that <math>f(x)=17</math>; likewise for each integer <math>b\geq a\geq 17</math> there is exactly one <math>x</math> with <math>\lfloor f(x)\rfloor=a</math> and <math>\lfloor x\rfloor=b</math> such that <math>f(f(x))=17</math>. Finally, for each integer <math>c \ge b \ge a \ge 17</math> there is exactly one <math>x</math> with <math>\lfloor f(f(x)) \rfloor = a</math>, <math>\lfloor f(x)\rfloor=b</math>, and <math>\lfloor x\rfloor=c</math> such that <math>f(f(f(x)))=17</math>. | ||
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+ | Thus <math>f(f(f(x)))=17</math> has exactly one solution <math>x</math> with <math>0\leq x\leq 2020</math> for each triple of integers <math>(a,b,c)</math> with <math>17\leq a\leq b\leq c<2020</math>, noting that <math>x=2020</math> is not a solution. This nondecreasing ordered triple can be identified with a multiset of three elements of the set of <math>2003</math> integers <math>\{17,18,19,\ldots,2019\}</math>, which can be selected in <math>\binom{2005}3</math> ways. Thus <cmath>N=\frac{2005\cdot 2004\cdot 2003}{6}\equiv 10\hskip -.2cm \pmod{1000}.</cmath> | ||
==Video Solution== | ==Video Solution== |
Revision as of 11:40, 18 June 2020
Problem
For real number let be the greatest integer less than or equal to , and define to be the fractional part of . For example, and . Define , and let be the number of real-valued solutions to the equation for . Find the remainder when is divided by .
Solution
To solve , we need to solve where , and to solve that we need to solve where .
It is clear to see for some integer there is exactly one value of in the interval where To understand this, imagine the graph of on the interval The graph starts at , is continuous and increasing, and approaches . So as long as , there will be a solution for in the interval.
Using this logic, we can find the number of solutions to . For every interval where there will be one solution for x in that interval. However, the question states , but because doesn't work we can change it to . Therefore, , and there are solutions to .
We can solve similarly. to satisfy the bounds of , so there are solutions to , and to satisfy the bounds of .
Going back to , there is a single solution for z in the interval , where . (We now have an upper bound for because we know .) There are solutions for , and the floors of these solutions create the sequence
Lets first look at the solution of where . Then would have solutions, and the floors of these solutions would also create the sequence .
If we used the solution of where , there would be solutions for . If we used the solution of where , there would be solutions for , and so on. So for the solution of where , there will be solutions for
If we now look at the solution of where , there would be solutions for . If we looked at the solution of where , there would be solutions for , and so on.
The total number of solutions to is . Using the hockey stick theorem, we see this equals , and when we take the remainder of that number when divided by , we get the answer,
~aragornmf
Solution 2 (Official MAA)
For any nonnegative integer , the function increases on the interval , with and for every in this interval. On this interval , which takes on every real value in the interval exactly once. Thus for each nonnegative real number , the equation has exactly one solution for every .
For each integer there is exactly one with such that ; likewise for each integer there is exactly one with and such that . Finally, for each integer there is exactly one with , , and such that .
Thus has exactly one solution with for each triple of integers with , noting that is not a solution. This nondecreasing ordered triple can be identified with a multiset of three elements of the set of integers , which can be selected in ways. Thus
Video Solution
https://youtu.be/bz5N-jI2e0U?t=515
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.