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How many different values can \(\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 3x \rfloor +\lfloor 4x \rfloor\)take for \(0 \leq x \leq 1\)?

 

EDIT: I go the answer, it's 7.

 Jun 28, 2018
edited by DanielCai  Jun 29, 2018
edited by DanielCai  Jun 29, 2018
 #1
avatar+21869 
0

How many different values can 

\(\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 3x \rfloor +\lfloor 4x \rfloor \) 
take for \(0 \leq x \leq 1 \)

 

lowest common multiple \(\mathbf{ \text{lcm}(1,2,3,4) = 12 }\)

 

\(\begin{array}{|c|c|c|c|} \hline x & \lfloor 1\cdot x \rfloor & \lfloor 2\cdot x \rfloor & \lfloor 3\cdot x \rfloor & \lfloor 4\cdot x \rfloor & \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 3x \rfloor +\lfloor 4x \rfloor \\ \hline \frac{0}{12} & 0 & 0 & 0 & 0 & 0 \\ \hline \frac{1}{12} & 0 & 0 & 0 & 0 & 0 \\ \hline \frac{2}{12} & 0 & 0 & 0 & 0 & 0 \\ \hline \frac{3}{12} & 0 & 0 & 0 & 1 & 1 \\ \hline \frac{4}{12} & 0 & 0 & 1 & 1 & 2 \\ \hline \frac{5}{12} & 0 & 0 & 1 & 1 & 2 \\ \hline \frac{6}{12} & 0 & 1 & 1 & 2 & 4 \\ \hline \frac{7}{12} & 0 & 1 & 1 & 2 & 4 \\ \hline \frac{8}{12} & 0 & 1 & 2 & 2 & 5 \\ \hline \frac{9}{12} & 0 & 1 & 2 & 3 & 6 \\ \hline \frac{10}{12} & 0 & 1 & 2 & 3 & 6 \\ \hline \frac{11}{12} & 0 & 1 & 2 & 3 & 6 \\ \hline \frac{12}{12} & 1 & 2 & 3 & 4 & 10 \\ \hline \end{array}\)

 

There are 7 different values: \(\mathbf{0, 1, 2, 4, 5, 6, 10}\)

 

laugh

 Jun 29, 2018
edited by heureka  Jun 29, 2018

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