Question

Prove that, $\left[x\right]+\left[5x\right]+\left[10x\right]+\left[20x\right]=36k+35.k\epsilon I$ does not have any real solution. Here [.] denotes greatest integer function.

Answer 1

We can see that the greatest integer function follows the following property:

$[n+r]=n+\left[r\right]$ for all $nϵI$ and $rϵR$

Now, every real number $x$ can be written as a sum of an integer and another real number between $0$ and $1$.

i.e., $x=k+r$ where $kϵI$ and $rϵ(0,1)$

Therefore, $\left[x\right]+\left[5x\right]+\left[10x\right]+\left[20x\right]$

$=[k+r]+[5k+5r]+[10k+10r]+[20k+20r]$

$=k+\left[r\right]+5k+\left[5r\right]+10k+\left[10r\right]+20k+\left[20r\right]$

$=36k+\left[r\right]+\left[5r\right]+\left[10r\right]+\left[20r\right]$

Now, we need to show that the above sum is never equal to $36k+35$.

This means that $\left[r\right]+\left[5r\right]+\left[10r\right]+\left[20r\right]$ is never equal to 35.

Here, $rϵ(0,1)$ $\Rightarrow nrϵ(0,n)$ $\Rightarrow 0\le \left[nr\right]\le n-1$ where $nϵN$.

Therefore, $0\le \left[r\right]\le 0$

$0\le \left[5r\right]\le 4$

$0\le \left[10r\right]\le 9$

$0\le \left[20r\right]\le 19$

Adding above inequalities, we get:

$\Rightarrow 0\le \left[r\right]+\left[5r\right]+\left[10r\right]+\left[20r\right]\le 32$

So we see that $\left[r\right]+\left[5r\right]+\left[10r\right]+\left[20r\right]$ is never equal to $35$.

Hence, the given equation has no real solution.