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

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We can see that for each value of <math>k</math>, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from <math>0</math> to <math>k-1</math>.  
 
We can see that for each value of <math>k</math>, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from <math>0</math> to <math>k-1</math>.  
  
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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>.
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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>.
 
 
 
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
 
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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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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==Solution 2==
 
==Solution 2==

Revision as of 00:54, 9 August 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 $n \le 10$. 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 $f(x)$.

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)
Preceded by
Problem 24 		Followed by
Last Problem
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All AMC 10 Problems and Solutions

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