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

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 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 32 different possible values of $f(x)$ .

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2016 AMC 10B Problems/Problem 25

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