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

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

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

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

## 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.

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

## Solution 4

\begin{align*} & \lfloor kx \rfloor = \lfloor x \rfloor + \lfloor x + \frac1k \rfloor + \lfloor x + \frac2k \rfloor + \dots + \lfloor x + \frac{k-1}{k} \rfloor\\ & \lfloor kx \rfloor -k \lfloor x \rfloor = \lfloor x + \frac1k \rfloor + \lfloor x + \frac2k \rfloor + \dots + \lfloor x + \frac{k-1}{k} \rfloor - (k-1) \lfloor x \rfloor \end{align*}

Therefore, \begin{align*} \sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor) & \\ &= \sum_{k=2}^{10}(\lfloor x + \frac1k \rfloor + \lfloor x + \frac2k \rfloor + \dots + \lfloor x + \frac{k-1}{k} \rfloor - (k-1) \lfloor x \rfloor)\\ &= \sum_{k=2}^{10}\sum_{i=1}^{k-1}( \lfloor x + \frac{i}{k} \rfloor - \lfloor x \rfloor)\\ &= \sum_{k=2}^{10}\sum_{i=1}^{k-1}( \lfloor \{ x \} + \frac{i}{k} \rfloor) \end{align*}

$0 \le \{ x \} < 1$, $0 < \frac{j}{k}<1$ $\Longrightarrow$ $0 < \lfloor \{ x \} + \frac{i}{k} \rfloor < 2$ $\Longrightarrow$ $\lfloor \{ x \} + \frac{i}{k} \rfloor = 0 \text{ or } 1$

$\{ x \} + \frac{i}{k} \ge 1$ $\Longrightarrow$ $\{ x \} \ge 1 - \frac{j}{k}$

Arrange $1 - \frac{i_j}{k_j}$ from small to large, $\{ x \}$ must fall in one interval. WLOG, suppose $1 - \frac{i_n}{k_n} \le \{ x \} < 1- \frac{i_{n+1}}{k_{n+1}}$.

if $j \le n$, $$\lfloor \{ x \} + \frac{i_j}{k_j} \rfloor = 1$$

if $j > n$, $$\lfloor \{ x \} + \frac{i_j}{k_j} \rfloor = 0$$

Therefore, every distinct value of $\sum_{k=2}^{10}\sum_{i=1}^{k-1}( \lfloor \{ x \} + \frac{i}{k} \rfloor)$ has one to one correspondence with a distinct value of $\frac{i}{k}$, $\frac{i}{k}$ is not reducible, $(i, k) = 1$.

Using the Euler Totient Function as in Solution 1's Supplement, the answer is $\fbox{\textbf{(A)}\ 32}$

## Remark

This problem is similar to 1985 AIME Problem 10. Both problems use the Euler Totient Function to find the number of distinct values of $\lfloor k \{ x \} \rfloor$.