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Difference between revisions of "2016 AMC 10B Problems/Problem 25"
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===Supplement===
 
===Supplement===
  
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Here are all the distinct <math>\frac{m}{n}</math> and <math>\phi(k)</math>.
+
Here are all the distinct <math>\frac{m}{n}</math> and <math>\phi(k):</math>
  
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When <math>n=2</math>, <math>\frac{m}{n}=\frac{1}{2}</math>. <math>\phi(2)=1</math>
+
When <math>n=2</math> , <math>\frac{m}{n}=\frac{1}{2}</math> . <math>\phi(2)=1</math>
  
−
When <math>n=3</math>, <math>\frac{m}{n}=\frac{1}{3}</math>, <math>\frac{2}{3}</math>. <math>\phi(3)=2</math>
+
When <math>n=3</math> , <math>\frac{m}{n}=\frac{1}{3}</math> , <math>\frac{2}{3}</math> . <math>\phi(3)=2</math>
  
−
When <math>n=4</math>, <math>\frac{m}{n}=\frac{1}{4}</math>, <math>\frac{3}{4}</math>. <math>\phi(4)=2</math>
+
When <math>n=4</math> , <math>\frac{m}{n}=\frac{1}{4}</math> , <math>\frac{3}{4}</math> . <math>\phi(4)=2</math>
  
−
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>
+
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>
  
−
When <math>n=6</math>, <math>\frac{m}{n}=\frac{1}{6}</math>, <math>\frac{5}{6}</math>. <math>\phi(6)=2</math>
+
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>
+
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>
  
−
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>
+
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>
  
−
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>
+
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>
  
−
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>
+
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>
  
−
<cmath>\sum_{k=2}^{10} \phi(k)</cmath>
+
<cmath>\sum_{k=2}^{10} \phi(k)=31</cmath>
−
<cmath>=31</cmath>
 
 
<cmath>31+1=\fbox{\textbf{(A)}\ 32}</cmath>
 
<cmath>31+1=\fbox{\textbf{(A)}\ 32}</cmath>
  

Revision as of 10:11, 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 $f(x)$.

Video Solution

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

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
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Problem 24 		Followed by
Last Problem
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All AMC 10 Problems and Solutions

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