Question

$\left[x\right]+\left[2x\right]+\left[4x\right]+\left[8x\right]+\left[16x\right]+\left[32x\right]=12345$
How many integer solution does this equation has

Answer 1

Since $\left[x\right]$ stands for greatest integer function, let us assume, $\left[x\right]=x$, that is x is an integer.
Then
$\left[x\right]+\left[2x\right]+\left[4x\right]+\left[8x\right]+\left[16x\right]+\left[32x\right]=12345$
$x+2x+4x+8x+16x+32x=12345$
$x\left(\frac{2^6-1}{2-1}\right)=12345$
$63x=12345$
$21x=4115$
$x=\frac{4115}{21}$ is not an integer.
This contradicts our initial assumption.
Hence not solution is possible.