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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>
  
==Solution==
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==Solution 1==
 
   
 
   
 
Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have  
 
Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have  
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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>.  
 
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>.  
  
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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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>.
 
 
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>. We can find this easily by computing
+
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
 
 
 
<cmath>\sum_{k=2}^{10} \phi(k)</cmath>
 
<cmath>\sum_{k=2}^{10} \phi(k)</cmath>
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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>.
 
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>.
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===Explanation:===
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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.
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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.
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 +
Minor Latex Edits made by MATHWIZARD2010.
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===Supplement===
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 +
Here are all the distinct <math>\frac{m}{n}</math> and <math>\phi(k):</math>
 +
 +
When <math>n=2</math> , <math>\frac{m}{n}=\frac{1}{2}</math> . <math>\phi(2)=1</math>
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When <math>n=3</math> , <math>\frac{m}{n}=\frac{1}{3}</math> , <math>\frac{2}{3}</math> . <math>\phi(3)=2</math>
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When <math>n=4</math> , <math>\frac{m}{n}=\frac{1}{4}</math> , <math>\frac{3}{4}</math> . <math>\phi(4)=2</math>
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When <math>n=5</math> , <math>\frac{m}{n}=\frac{1}{5}</math> , <math>\frac{2}{5}</math> , <math>\frac{3}{5}</math> , <math>\frac{4}{5}</math> . <math>\phi(5)=4</math>
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When <math>n=6</math> , <math>\frac{m}{n}=\frac{1}{6}</math> , <math>\frac{5}{6}</math> . <math>\phi(6)=2</math>
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When <math>n=7</math> , <math>\frac{m}{n}=\frac{1}{7}</math> , <math>\frac{2}{7}</math> , <math>\frac{3}{7}</math> , <math>\frac{4}{7}</math> , <math>\frac{5}{7}</math> , <math>\frac{6}{7}</math> . <math>\phi(7)=6</math>
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When <math>n=8</math> , <math>\frac{m}{n}=\frac{1}{8}</math> , <math>\frac{3}{8}</math> , <math>\frac{5}{8}</math> , <math>\frac{7}{8}</math> . <math>\phi(8)=4</math>
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When <math>n=9</math> , <math>\frac{m}{n}=\frac{1}{9}</math> , <math>\frac{2}{9}</math> , <math>\frac{4}{9}</math> , <math>\frac{5}{9}</math> , <math>\frac{7}{9}</math> , <math>\frac{8}{9}</math> . <math>\phi(9)=6</math>
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When <math>n=10</math> , <math>\frac{m}{n}=\frac{1}{10}</math> , <math>\frac{3}{10}</math> , <math>\frac{7}{10}</math> , <math>\frac{9}{10}</math> . <math>\phi(10)=4</math>
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<math>\sum_{k=2}^{10} \phi(k)=31</math>
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<math>31+1=\fbox{\textbf{(A)}\ 32}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Solution 2==
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<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>.
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==Solution 3 (Casework)==
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Solution <math>1</math> is abstract. In this solution I will give a concrete explanation.
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WLOG, for example, when <math>x</math> increases from <math>\frac{2}{3}-\epsilon</math> to <math>\frac{2}{3}</math>, <math>\lfloor 3 \{ x \} \rfloor</math> will increase from <math>1</math> to <math>2</math>, <math>\lfloor 6 \{ x \} \rfloor</math> will increase from <math>3</math> to <math>4</math>, <math>\lfloor 9 \{ x \} \rfloor</math> will increase from <math>5</math> to <math>6</math>. In total, <math>f(x)</math> will increase by <math>3</math>. Because <math>\frac{1}{3}=\frac{2}{6}=\frac{3}{9}</math>, these <math>3</math> numbers are actually <math>1</math> distinct number to cause <math>f(x)</math> to change. In general, when <math>x</math> increases from <math>\frac{m}{n}-\epsilon</math> to <math>\frac{m}{n}</math>, <math>\lfloor k \{ x \} \rfloor</math> will increse from <math>k \cdot \frac{m}{n} -1</math> to <math>k \cdot \frac{m}{n} </math> if <math>k \cdot \frac{m}{n} </math> is an integer, and the value of <math>f(x)</math> will change. So the total number of distinct values <math>f(x)</math> could take is equal to the number of distinct values of <math>\frac{m}{n}</math>, where <math>0 < \frac{m}{n}<1</math> and <math>2 \le n \le 10</math>.
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Solution <math>1</math> uses Euler Totient Function to count the distinct number of <math>\frac{m}{n}</math>, I am going to use casework to count the distinct values of <math>\frac{m}{n}</math> by not counting the duplicate ones.
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When <math>n=10</math> , <math>\frac{m}{n}=\frac{1}{10}</math> , <math>\frac{2}{10}</math> , <math>...</math> , <math>\frac{9}{10}</math> <math>\Longrightarrow 9</math>
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When <math>n=9</math> , <math>\frac{m}{n}=\frac{1}{9}</math> , <math>\frac{2}{9}</math> , <math>...</math> , <math>\frac{8}{9}</math> <math>\Longrightarrow 8</math>
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When <math>n=8</math> , <math>\frac{m}{n}=\frac{1}{8}</math> , <math>\frac{2}{8}</math> , <math>...</math> , <math>\frac{7}{8}</math>  <math>\Longrightarrow 6</math> (  <math>\frac{4}{8}</math> is duplicate)
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When <math>n=7</math> , <math>\frac{m}{n}=\frac{1}{7}</math> , <math>\frac{2}{7}</math> , <math>...</math> , <math>\frac{6}{7}</math> <math>\Longrightarrow 6</math>
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When <math>n=6</math> , <math>\frac{m}{n}=\frac{1}{6}</math> , <math>\frac{5}{6}</math>  <math>\Longrightarrow 2</math> (  <math>\frac{2}{6}</math> , <math>\frac{3}{6}</math> , and <math>\frac{4}{6}</math> is duplicate)
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When <math>n=5</math>, <math>4</math>, <math>3</math>, <math>2</math>, all the <math>\frac{m}{n}</math> is duplicate.
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<math>9+8+6+6+2=31</math>, <math>31+1=\fbox{\textbf{(A)}\ 32}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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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}}
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 +
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 10:56, 7 January 2022

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.

Supplement

Here are all the distinct $\frac{m}{n}$ and $\phi(k):$

When $n=2$ , $\frac{m}{n}=\frac{1}{2}$ . $\phi(2)=1$

When $n=3$ , $\frac{m}{n}=\frac{1}{3}$ , $\frac{2}{3}$ . $\phi(3)=2$

When $n=4$ , $\frac{m}{n}=\frac{1}{4}$ , $\frac{3}{4}$ . $\phi(4)=2$

When $n=5$ , $\frac{m}{n}=\frac{1}{5}$ , $\frac{2}{5}$ , $\frac{3}{5}$ , $\frac{4}{5}$ . $\phi(5)=4$

When $n=6$ , $\frac{m}{n}=\frac{1}{6}$ , $\frac{5}{6}$ . $\phi(6)=2$

When $n=7$ , $\frac{m}{n}=\frac{1}{7}$ , $\frac{2}{7}$ , $\frac{3}{7}$ , $\frac{4}{7}$ , $\frac{5}{7}$ , $\frac{6}{7}$ . $\phi(7)=6$

When $n=8$ , $\frac{m}{n}=\frac{1}{8}$ , $\frac{3}{8}$ , $\frac{5}{8}$ , $\frac{7}{8}$ . $\phi(8)=4$

When $n=9$ , $\frac{m}{n}=\frac{1}{9}$ , $\frac{2}{9}$ , $\frac{4}{9}$ , $\frac{5}{9}$ , $\frac{7}{9}$ , $\frac{8}{9}$ . $\phi(9)=6$

When $n=10$ , $\frac{m}{n}=\frac{1}{10}$ , $\frac{3}{10}$ , $\frac{7}{10}$ , $\frac{9}{10}$ . $\phi(10)=4$

$\sum_{k=2}^{10} \phi(k)=31$

$31+1=\fbox{\textbf{(A)}\ 32}$

~isabelchen

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 $f(x)$.

Solution 3 (Casework)

Solution $1$ is abstract. In this solution I will give a concrete explanation.

WLOG, for example, when $x$ increases from $\frac{2}{3}-\epsilon$ to $\frac{2}{3}$, $\lfloor 3 \{ x \} \rfloor$ will increase from $1$ to $2$, $\lfloor 6 \{ x \} \rfloor$ will increase from $3$ to $4$, $\lfloor 9 \{ x \} \rfloor$ will increase from $5$ to $6$. In total, $f(x)$ will increase by $3$. Because $\frac{1}{3}=\frac{2}{6}=\frac{3}{9}$, these $3$ numbers are actually $1$ distinct number to cause $f(x)$ to change. In general, when $x$ increases from $\frac{m}{n}-\epsilon$ to $\frac{m}{n}$, $\lfloor k \{ x \} \rfloor$ will increse from $k \cdot \frac{m}{n} -1$ to $k \cdot \frac{m}{n}$ if $k \cdot \frac{m}{n}$ is an integer, and the value of $f(x)$ will change. So the total number of distinct values $f(x)$ could take is equal to the number of distinct values of $\frac{m}{n}$, where $0 < \frac{m}{n}<1$ and $2 \le n \le 10$.

Solution $1$ uses Euler Totient Function to count the distinct number of $\frac{m}{n}$, I am going to use casework to count the distinct values of $\frac{m}{n}$ by not counting the duplicate ones.

When $n=10$ , $\frac{m}{n}=\frac{1}{10}$ , $\frac{2}{10}$ , $...$ , $\frac{9}{10}$ $\Longrightarrow 9$

When $n=9$ , $\frac{m}{n}=\frac{1}{9}$ , $\frac{2}{9}$ , $...$ , $\frac{8}{9}$ $\Longrightarrow 8$

When $n=8$ , $\frac{m}{n}=\frac{1}{8}$ , $\frac{2}{8}$ , $...$ , $\frac{7}{8}$ $\Longrightarrow 6$ ( $\frac{4}{8}$ is duplicate)

When $n=7$ , $\frac{m}{n}=\frac{1}{7}$ , $\frac{2}{7}$ , $...$ , $\frac{6}{7}$ $\Longrightarrow 6$

When $n=6$ , $\frac{m}{n}=\frac{1}{6}$ , $\frac{5}{6}$ $\Longrightarrow 2$ ( $\frac{2}{6}$ , $\frac{3}{6}$ , and $\frac{4}{6}$ is duplicate)

When $n=5$, $4$, $3$, $2$, all the $\frac{m}{n}$ is duplicate.

$9+8+6+6+2=31$, $31+1=\fbox{\textbf{(A)}\ 32}$

~isabelchen

Video Solution

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

See Also

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

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