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

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==Solution==
 
==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>.
 +
 +
By listing out all the fractions that are changing points for the value of one or more <math>\lfloor k \{ x \} \rfloor</math>, we find that there are 31 unique fractions.
 +
 +
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>.
 +
 +
(Solution probably needs editing for clarity.)
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:13, 21 February 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 $\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}$.

By listing out all the fractions that are changing points for the value of one or more $\lfloor k \{ x \} \rfloor$, we find that there are 31 unique fractions.

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 $f(x)$.

(Solution probably needs editing for clarity.)

See Also

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

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