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

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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>.
 
<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==
+
==Solution 3 (Casework)==
  
 
Solution <math>1</math> is abstract. In this solution I will give a concrete explanation.
 
Solution <math>1</math> is abstract. In this solution I will give a concrete explanation.
  
−
WLOG, example: <math>x</math> increase from <math>\frac{2}{3}-\epsilon</math> to <math>\frac{2}{3}</math>.
+
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>. The total number of values <math>f(x)</math> could take is equal to the number of distinct values of <math>\frac{m}{n}</math>, where <math>0 \le \frac{m}{n}<1</math> and <math>2 \le n \le 10</math>. So we need to count the distinct values of <math>\frac{m}{n}</math>.
  
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<math>[3{x}]</math> will increase from <math>1 \Longrightarrow 2</math>, <math>[6{x}]</math> will increase from <math>3 \Longrightarrow 4</math>, <math>[9{x}]</math> will increase from <math>5 \Longrightarrow 6</math>. In total <math>f(x)</math> is increased by <math>3</math>.
+
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.
 +
 
 +
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>
 +
 
 +
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>
 +
 
 +
When <math>n=8</math> , <math>\frac{m}{n}=\frac{1}{8}</math> , <math>\frac{2}{8}</math> , <math>...</math> , <math>\frac{8}{8}</math> <math>\Longrightarrow 6</math> (  <math>\frac{4}{8}</math> is duplicate)
 +
 
 +
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>
 +
 
 +
When <math>n=6</math> , <math>\frac{m}{n}=\frac{1}{6}</math> , <math>\frac{2}{6}</math> , <math>...</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)
 +
 
 +
When <math>n=5</math>, <math>4</math>, <math>3</math>, <math>2</math>, all the <math>\frac{m}{n}</math> is duplicate.
 +
 
 +
<math>9+8+6+6+2=31</math>, <math>31+1=\fbox{\textbf{(A)}\ 32}</math>
 +
 
 +
~isabelchen
  
 
==Video Solution==
 
==Video Solution==

Revision as of 11:31, 13 October 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.

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 $3$. The total number of values $f(x)$ could take is equal to the number of distinct values of $\frac{m}{n}$, where $0 \le \frac{m}{n}<1$ and $2 \le n \le 10$. So we need to count the distinct values of $\frac{m}{n}$.

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

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

When $n=6$ , $\frac{m}{n}=\frac{1}{6}$ , $\frac{2}{6}$ , $...$ , $\frac{5}{6}$ $\Longrightarrow 2$ ( $\frac{2}{6}$ , $\frac{3}{6}$ , 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 (Problems • Answer Key • Resources)
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