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

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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>.
 
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
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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>. [b]Explanation for this is provided below.[/b] 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> be the largest so <math>\frac{x_\text{old}}{k} < p</math>. Since <math>\frac{m}{n} > p</math>, we clearly have that all <math>x_\text{new} >= x_\text{old}</math>. Therefore, the change must be nonnegative.
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But among all numerators rel prime to <math>n</math>, <math>m</math> is the largest. Therefore, choosing <math>\frac{m}{n}</math> as <math>{x}</math> creates a positive change in the term <math>\lfloor n \{ x \} \rfloor</math>. Since the overall change in <math>f(x)</math> increases as the fractions <math>m/n</math> increase, we deduce that all such fractions correspond to different values of the function.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 23:47, 27 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 $n \le 10$. [b]Explanation for this is provided below.[/b] 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 $x_\text{old}$ ($0 \le x_\text{old} < k$ be the largest so $\frac{x_\text{old}}{k} < p$. Since $\frac{m}{n} > p$, we clearly have that all $x_\text{new} >= x_\text{old}$. Therefore, the change must be nonnegative.

But among all numerators rel prime to $n$, $m$ is the largest. Therefore, choosing $\frac{m}{n}$ as ${x}$ creates a positive change in the term $\lfloor n \{ x \} \rfloor$. Since the overall change in $f(x)$ increases as the 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 $f(x)$.

See Also

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