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

Revision as of 12:49, 16 March 2019

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 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 $f(x)$ .

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 <cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor</cmath>
-We can see that for each value of k, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from 0 to k-1.
+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>.
 Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when x is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>.

Page

Toolbox

Search

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

Revision as of 12:49, 16 March 2019

Problem

Solution

See Also