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# Help

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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

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How many different values can

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

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: $$\mathbf{0, 1, 2, 4, 5, 6, 10}$$

Jun 29, 2018
edited by heureka  Jun 29, 2018