2016 AMC 10B Problems/Problem 25

Revision as of 22:23, 27 March 2017 by Drunkenninja (talk | contribs) (Solution)

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

Since $x = \lfloor x \rfloor + \{ x \}$, we have

\[f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +\lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)\]

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 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 $f(x)$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
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