Prove that $\lfloor x \rfloor +\lfloor 2x \rfloor + \dots +\lfloor 32x \rfloor =12345$ has no solution

Question

Given that $\lfloor x \rfloor +\lfloor 2x \rfloor + \dots +\lfloor 32x \rfloor =12345$ where $\lfloor x \rfloor $ denotes the floor function, prove that there is no $x$ that satisfies the equation.

Answer 1

It is not true. Write $y=\lfloor x \rfloor$ and let $z=x-y$ be the fractional part of $x$. Your equation becomes $$528 y +\sum_{i=1}^{32} \lfloor iz \rfloor =12345$$ As $z \lt 1$ the sum is less than $528$ so $y=23$ and the equation becomes $$\sum_{i=1}^{32} \lfloor iz \rfloor =201$$ and $z=0.41, x=23.41$ works perfectly. I just added it up in a spreadsheet and did a hand binary search to find which terms increase to bring the sum up to $12345$ and $12346$. Any $x \in [\frac {515}{22},\frac {398}{17})\approx [23.40909,23.41176)$ works.

Answer 2

Hint

$$\lfloor x \rfloor +\lfloor 2x \rfloor + \lfloor 4x \rfloor +\lfloor 8x \rfloor +\lfloor 16x \rfloor+\lfloor 32x \rfloor \leq 63x < \lfloor x \rfloor +\lfloor 2x \rfloor + \lfloor 4x \rfloor +\lfloor 8x \rfloor +\lfloor 16x \rfloor+\lfloor 32x \rfloor+6\\ 12345 \leq 63 x < 12351 \\ 196-\frac{1}{21} \leq x < 196+\frac{1}{21}\\ 196-\frac{2}{32} < x < 196+\frac{2}{32}\\$$

Now, split the problem in the following cases:

• $x \in (196-\frac{2}{32}, 196-\frac{1}{32})$
• $x \in [196-\frac{1}{32}, 196)$
• $x \in [196, 196+\frac{1}{32})$
• $x \in [196+\frac{1}{32}, 196+\frac{2}{32})$