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

 Jul 14, 2023
 #1
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The function floor(x) + floor(2x) + floor(3x) + floor(4x) takes on different values for 0 <= x <= 1 if and only if the fractional part {x} of x satisfies 0 <= {x} < 1/4. The different values that {x} can take on in this range are 0, 1/4, 2/4, 3/4, and 1. Therefore, the function floor(x) + floor(2x) + floor(3x) + floor(4x) takes on 1 + 4 = 5 different values in the range 0 <= x <= 1.

 Jul 14, 2023
 #2
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This should help.

 

 Jul 15, 2023

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