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The problem

Determine the set of values of the natural number $n\in N^*$, for which the equation $\{x\}+\{2x\}+\{3x\}=\lfloor nx\rfloor$ admits nonzero real solutions., where $\{a\}$ is the fractional part of number $a$.

my idea

It's clearly that $0\leqslant\lfloor nx\rfloor<3$ so $\lfloor nx\rfloor$ can take the values $0,1,2$ because it must be an integer.

I don't know if talking it each case might help... I also tried using the identity of Hermite but got to nothing useful. Hope one of you can help! Thank you!

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Here we must break down the question into a few cases. Your previous analysis of the question is correct, and so [nx] must be either 0, 1 or 2. The case of 0 occurs when x is an integer(due to the LHS), so for [nx] to equal to zero either x is any integer and n=0 but n must be a natural number. When the LHS is equal to 1, the sum of the fractional parts must be 1, {x}+{2x}+{3x}=1 which this could only occur when

{x}=$$\frac{1}{6},\frac{1}{2},\frac{1}{3},\frac{2}{3}.$$ This means [nx] to equal to 1, when x=m.1666 or m.5 or m.333. (here m is any general positive integer); here there is only solutions when x=0.166 or when x=1.166 (no soltuion for negative values of x as n is a natural number). For x=0.166 n can have any value from [6,12) and when x=1.166 n can only be equal to 1. For x=0.5 n can take a value of 2 and 3, and for x=1.5 n can take the vakue of 1. For x=0.33, n=3,4,5. And for x=1.33, n=1. For x=0.66, n=2,3,4, x=1.66, n=1. Next case is when LHS is equal to 2, when

{x}=$$\frac{5}{6}$$ This means x=m.833 (not considering negative values of x as n is a natural number). Now we have soltuion when x=0.833,x=1.833 or x=2.833. When x=0.833 n can be equal to 3, when x=1.833 n is equal to 2, when x=2.833 n=1. There the set of values of n is $$n={1,2,3,4,5,6,7,8,9,10,11}$$ Hope this helps and if I'm wrong anywhere please do let me know. Also sorry I wasn't able to figure out how to write fractional part of x and the fraction together.

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  • $\begingroup$ Thank you for your answer! But why does ${x}+{2x}+{3x}=1$ only when x=1:6? Do you have any mathematical proof? $\endgroup$ Commented May 17 at 13:16
  • $\begingroup$ @IONELABUCIU thanks for remininding, just remembered this also true for 3 more values of {x} $\endgroup$ Commented May 17 at 14:19

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