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The number of solutions of $$x + [x] = 2013$$, where $$[x]$$ is greatest integer less than or equal to $$x$$, is
Write $$x = n + f$$ with $$n = [x]$$ and $$0 \le f < 1$$. Then $$2n + f = 2013$$, so $$f = 2013 - 2n$$ is an integer, which forces $$f = 0$$. But then $$2n = 2013$$, which is impossible since 2013 is odd. So there are no solutions.
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