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

Revision as of 19:45, 26 September 2016

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 $f(x)$ .

2016 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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

@@ Line 5: / Line 5: @@
 <math>\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}</math>
+==Solution==
+Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have
+<cmath>f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor + k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath>
+The function can then be simplified into
+<cmath>f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath>
+which becomes
+<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.
+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>.
+So we want to count how many distinct fractions have the form <math>\frac{m}{n}</math> where <math>n \le 10</math>. We can find this easily by computing
+<cmath>\sum_{k=2}^{10} \phi(k)</cmath>
+where <math>\phi(k)</math> is the Euler Totient Function. Basically <math>\phi(k)</math> counts the number of fractions with <math>k</math> as its denominator (after simplification). This comes out to be <math>31</math>.
+Because the value of <math>f(x)</math> is at least 0 and can increase 31 times, there are a total of 32 different possible values of <math>f(x)</math>.
 ==See Also==
 {{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}
 {{MAA Notice}}

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

Revision as of 19:45, 26 September 2016

Problem

Solution

See Also