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Difference between revisions of "2020 AIME II Problems/Problem 14"

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The total number of solutions to <math>x</math> is <math>\binom{2004}{2} + \binom{2003}{2} + \binom{2002}{2} + ... + \binom{3}{2} + \binom{2}{2}</math>. Using the hockey stick theorem, we see this equals <math>\binom{2005}{3}</math>, and when we take the remainder of that number when divided by <math>1000</math>, we get the answer, <math>\boxed{10}</math>
 
The total number of solutions to <math>x</math> is <math>\binom{2004}{2} + \binom{2003}{2} + \binom{2002}{2} + ... + \binom{3}{2} + \binom{2}{2}</math>. Using the hockey stick theorem, we see this equals <math>\binom{2005}{3}</math>, and when we take the remainder of that number when divided by <math>1000</math>, we get the answer, <math>\boxed{10}</math>
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~aragornmf
  
 
==Video Solution==
 
==Video Solution==

Revision as of 19:29, 12 June 2020

Problem

For real number $x$ let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$, and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x$. For example, $\{3\} = 0$ and $\{4.56\} = 0.56$. Define $f(x)=x\{x\}$, and let $N$ be the number of real-valued solutions to the equation $f(f(f(x)))=17$ for $0\leq x\leq 2020$. Find the remainder when $N$ is divided by $1000$.

Solution

To solve $f(f(f(x)))=17$, we need to solve $f(x) = y$ where $f(f(y))=17$, and to solve that we need to solve $f(y) = z$ where $f(z) = 17$.

It is clear to see for some integer $a \geq 17$ there is exactly one value of $z$ in the interval $[a, a+1)$ where $f(z) = 17$ To understand this, imagine the graph of $f(z)$ on the interval $[a, a+1)$ The graph starts at $0$, is continuous and increasing, and approaches $a+1$. So as long as $a+1 > 17$, there will be a solution for $z$ in the interval.

Using this logic, we can find the number of solutions to $f(x) = y$. For every interval $[a, a+1)$ where $a \geq \left \lfloor{y}\right \rfloor$ there will be one solution for x in that interval. However, the question states $0 \leq x \leq 2020$, but because $x=2020$ doesn't work we can change it to $0 \leq x < 2020$. Therefore, $\left \lfloor{y}\right \rfloor \leq a \leq 2019$, and there are $2020 - \left \lfloor{y}\right \rfloor$ solutions to $f(x) = y$.

We can solve $f(y) = z$ similarly. $0 \leq y < 2020$ to satisfy the bounds of $x$, so there are $2020 - \left \lfloor{z}\right \rfloor$ solutions to $f(y) = z$, and $0 \leq z < 2020$ to satisfy the bounds of $y$.

Going back to $f(z) = 17$, there is a single solution for z in the interval $[a, a+1)$, where $17 \leq a \leq 2019$. (We now have an upper bound for $a$ because we know $z < 2020$.) There are $2003$ solutions for $z$, and the floors of these solutions create the sequence $17, 18, 19, ..., 2018, 2019$

Lets first look at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 17$. Then $f(y) = z$ would have $2003$ solutions, and the floors of these solutions would also create the sequence $17, 18, 19, ..., 2018, 2019$.

If we used the solution of $y$ where $\left \lfloor{y}\right \rfloor = 17$, there would be $2003$ solutions for $f(x) = y$. If we used the solution of $y$ where $\left \lfloor{y}\right \rfloor = 18$, there would be $2002$ solutions for $x$, and so on. So for the solution of $z$ where $\left \lfloor{z}\right \rfloor = 17$, there will be $2003 + 2002 + 2001 + ... + 2 + 1 = \binom{2004}{2}$ solutions for $x$

If we now look at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 18$, there would be $\binom{2003}{2}$ solutions for $x$. If we looked at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 19$, there would be $\binom{2002}{2}$ solutions for $x$, and so on.

The total number of solutions to $x$ is $\binom{2004}{2} + \binom{2003}{2} + \binom{2002}{2} + ... + \binom{3}{2} + \binom{2}{2}$. Using the hockey stick theorem, we see this equals $\binom{2005}{3}$, and when we take the remainder of that number when divided by $1000$, we get the answer, $\boxed{10}$

~aragornmf

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

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