Difference between revisions of "2016 AMC 10B Problems/Problem 25"

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<math>\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}</math>
 
<math>\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}</math>
  
 
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==Solution 1==
==Solution==
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Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have  
 
Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have  
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<cmath>f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath>
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The function can then be simplified into
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<cmath>f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath>
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 +
which becomes
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<cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor</cmath>
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We can see that for each value of <math>k</math>, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from <math>0</math> to <math>k-1</math>.
  
<cmath>f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +\rfloor k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath>
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Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when <math>\{ x \}</math> is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>.
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So we want to count how many distinct fractions less than <math>1</math> have the form <math>\frac{m}{n}</math> where <math>n \le 10</math>. '''Explanation for this is provided below.''' We can find this easily by computing
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<cmath>\sum_{k=2}^{10} \phi(k)</cmath>
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where <math>\phi(k)</math> is the [[Euler Totient Function]]. Basically <math>\phi(k)</math> counts the number of fractions with <math>k</math> as its denominator (after simplification). This comes out to be <math>31</math>.
  
The function can then be simplified into
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Because the value of <math>f(x)</math> is at least <math>0</math> and can increase <math>31</math> times, there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>.
  
<cmath>f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath>
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===Explanation:===
  
which becomes
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Arrange all such fractions in increasing order and take a current <math>\frac{m}{n}</math> to study. Let <math>p</math> denote the previous fraction in the list and <math>x_\text{old}</math> (<math>0 \le x_\text{old} < k</math> for each <math>k</math>) be the largest so that <math>\frac{x_\text{old}}{k} \le p</math>. Since  <math>\text{ }\text{ }\frac{m}{n} > p</math>, we clearly have all <math>x_\text{new} \ge x_\text{old}</math>. Therefore, the change must be nonnegative.
  
<cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor</cmath>
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But among all numerators coprime to <math>n</math> so far, <math>m</math> is the largest. Therefore, choosing <math>\frac{m}{n}</math> as <math>{x}</math> increases the value <math>\lfloor n \{ x \} \rfloor</math>. Since the overall change in <math>f(x)</math> is positive as fractions <math>m/n</math> increase, we deduce that all such fractions correspond to different values of the function.
  
We can see that for each value of k, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from 0 to k-1.  
+
Minor Latex Edits made by MATHWIZARD2010.
  
Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when x is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>.
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==Solution 2==
  
So we want to count how many distinct fractions have the form <math>\frac{m}{n}</math> where <math>n \le 10</math>. We can find this easily by computing
+
<math>x = \lfloor x \rfloor + \{ x \}</math> so we have <cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor.</cmath> Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when <math>x</math> is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>. To get all the fractions,graphing this function gives us <math>46</math> different fractions. But on average, <math>3</math> in each of the <math>5</math> intervals don’t work. This means there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>.
<cmath>\sum_{k=2}^{10} \phi(k)</cmath>
 
where <math>\phi(k)</math> is the Euler Totient Function. Basically <math>\phi(k)</math> counts the number of fractions with <math>k</math> as its denominator (after simplification). This comes out to be <math>31</math>.
 
  
Because the value of <math>f(x)</math> is at least 0 and can increase 31 times, there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=zXJrdDtZNbw
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:07, 22 January 2021

Problem

Let $f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)$, where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to $r$. How many distinct values does $f(x)$ assume for $x \ge 0$?

$\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}$

Solution 1

Since $x = \lfloor x \rfloor + \{ x \}$, we have

\[f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +k \{ x \} \rfloor - k \lfloor x \rfloor)\]

The function can then be simplified into

\[f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)\]

which becomes

\[f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor\]

We can see that for each value of $k$, $\lfloor k \{ x \} \rfloor$ can equal integers from $0$ to $k-1$.

Clearly, the value of $\lfloor k \{ x \} \rfloor$ changes only when $\{ x \}$ is equal to any of the fractions $\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}$.

So we want to count how many distinct fractions less than $1$ have the form $\frac{m}{n}$ where $n \le 10$. Explanation for this is provided below. We can find this easily by computing

\[\sum_{k=2}^{10} \phi(k)\]

where $\phi(k)$ is the Euler Totient Function. Basically $\phi(k)$ counts the number of fractions with $k$ as its denominator (after simplification). This comes out to be $31$.

Because the value of $f(x)$ is at least $0$ and can increase $31$ times, there are a total of $\fbox{\textbf{(A)}\ 32}$ different possible values of $f(x)$.

Explanation:

Arrange all such fractions in increasing order and take a current $\frac{m}{n}$ to study. Let $p$ denote the previous fraction in the list and $x_\text{old}$ ($0 \le x_\text{old} < k$ for each $k$) be the largest so that $\frac{x_\text{old}}{k} \le p$. Since $\text{ }\text{ }\frac{m}{n} > p$, we clearly have all $x_\text{new} \ge x_\text{old}$. Therefore, the change must be nonnegative.

But among all numerators coprime to $n$ so far, $m$ is the largest. Therefore, choosing $\frac{m}{n}$ as ${x}$ increases the value $\lfloor n \{ x \} \rfloor$. Since the overall change in $f(x)$ is positive as fractions $m/n$ increase, we deduce that all such fractions correspond to different values of the function.

Minor Latex Edits made by MATHWIZARD2010.

Solution 2

$x = \lfloor x \rfloor + \{ x \}$ so we have \[f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor.\] Clearly, the value of $\lfloor k \{ x \} \rfloor$ changes only when $x$ is equal to any of the fractions $\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}$. To get all the fractions,graphing this function gives us $46$ different fractions. But on average, $3$ in each of the $5$ intervals don’t work. This means there are a total of $\fbox{\textbf{(A)}\ 32}$ different possible values of $f(x)$.

Video Solution

https://www.youtube.com/watch?v=zXJrdDtZNbw

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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All AMC 10 Problems and Solutions

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