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

Revision as of 20:04, 28 December 2020

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 $\frac{m}{n} > p$ , 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.

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 an 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 $f(x)$ .

@@ Line 31: / Line 31: @@
 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>.
-'''Explanation:'''
+===Explanation:===
 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>\frac{m}{n} > p</math>, we clearly have all <math>x_\text{new} \ge x_\text{old}</math>. Therefore, the change must be nonnegative.

2016 AMC 10B (Problems • Answer Key • Resources)
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