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Prove that the equation

$\lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 8x\rfloor+\lfloor 16x\rfloor+\lfloor 32x\rfloor = 12345$

does not have any real solution.

($\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$).

My Attempt

Let $x=I+f$,where $I=\lfloor x\rfloor$ and $f$ denotes the fractional part of $x$. Therefore, the equation reduces to

$63I+\lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 8f\rfloor+\lfloor 16f\rfloor+\lfloor 32f\rfloor = 12345 $

$I=195+\frac{20}{21}-\frac{\lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 8f\rfloor+\lfloor 16f\rfloor+\lfloor 32f\rfloor}{63}$

Not able to proceed from here onwards

msinghal
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Maverick
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    Hint: you have $\lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 8f\rfloor+\lfloor 16f\rfloor+\lfloor 32f\rfloor = 60$ – msinghal Jul 22 '15 at 16:24
  • @MihirSinghal: I don't understand how that's true. – hexaflexagonal Jul 22 '15 at 16:25
  • Note that $12345\mod{63} = 60$. – vadim123 Jul 22 '15 at 16:26
  • You had very much the right approach. But as is often the case in number-theoretic problems, introducing fractions complicates things. It is better to write your equation as $63I=(195)(63)+60-(\lfloor 2f\rfloor+\cdots+\lfloor 32f\rfloor)$. – André Nicolas Jul 22 '15 at 16:37
  • @Mihir Singhal.I follow your hint .Now,$\lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 8f\rfloor+\lfloor 16f\rfloor<30$.Thus $\lfloor 32f\rfloor \geq 31$.Therefore,$\frac{31}{32}\leq f<1$.Finally,for all $f\in\left[\frac{31}{32},1\right)$,we have$ \lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 8f\rfloor+\lfloor 16f\rfloor+\lfloor 32f\rfloor=57<60$. Hence, no solution. – Maverick Jul 23 '15 at 01:12

2 Answers2

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Hint: The function is increasing and increases at all multiples of $\frac{1}{32}$ and remains constant at other places. But we have $f(195+\frac{31}{32})=12342$ and $f(196)=12348$.

wythagoras
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Since $\lfloor z\rfloor\in(z-1,z]$ we must have $12345\leq 63x$ and $12351>63 x$, hence: $$ x\in\left(196-\frac{1}{21},196+\frac{1}{21}\right]$$ but since $\frac{1}{21}<\frac{1}{16}$, over such interval the given function is greater than $196\cdot 63-3=12345.$

Jack D'Aurizio
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