Find all positive integers $x$ such that $[\frac{x}{5}]-[\frac{x}{7}]=1$, where, for any real number $t$, $[t]$ denotes the greatest integer $\le t$

Question

I have tried the following steps:

$[\dfrac{x}{5}]-[\dfrac{x}{7}]=1$

@Berci suggested this:

$[\dfrac{x}{5}]\le\dfrac{x}{5}$

$-[\dfrac{x}{7}]<-\dfrac{x}{7}+1$

$\implies [\dfrac{x}{5}]-[\dfrac{x}{7}]<\dfrac{x}{5}-\dfrac{x}{7}+1 $

$\implies x>0$

$\implies x> 0$

$\implies x \in [1,.... \infty]$

Next is a manual observation, we will see that numbers lets say $n\equiv1(mod 5)\implies n\equiv 5k+1 $ where $k\in \{1,2,3,4\}$

why $k\in \{1,2,3,4\}$? why not $k\in \mathbb{N}$?

Lets see:

Hence if we try numbers like $6,11,16,21$ we will see that $[\dfrac{x}{5}]-[\dfrac{x}{7}]=1$

But the moment one tries with $x=26,36$ we will see $[\dfrac{x}{5}]-[\dfrac{x}{7}]=2$

Again with $x=41,46,51,56$ we see $[\dfrac{x}{5}]-[\dfrac{x}{7}]=2$

And so on as will proceed and this $[\dfrac{x}{5}]-[\dfrac{x}{7}]>1$ and increases more and more.

Again if we try $x=12,13$,we get $[\dfrac{x}{5}]-[\dfrac{x}{7}]=1$

Also $x=5$ is a solution.

So is it the answer that $x \in {5,6,11,12,13,16,21}$?

Answer 1

To obtain a meaningful bound, you need the inequalities in the other way: $\left[\dfrac{x}{5}\right]\ge \dfrac{x}{5}-1$ and $\left[\dfrac{x}{7}\right]\le \dfrac{x}{7}$. Hence, $$\left[\dfrac{x}{5}\right]-\left[\dfrac{x}{7}\right]\ge \frac{x}{5}-1-\frac{x}7=\frac{2x}{35}-1$$ Now, for any $x\ge 36$, you have that $$x\ge 36\implies \frac{2x}{35}-1\ge \frac{2\cdot36}{35}-1>1$$ which implies that $$\left[\dfrac{x}{5}\right]-\left[\dfrac{x}{7}\right]>1$$ for any $x\ge 36$. So, your candidate solutions are the integers $1,\dots, 35$. Trial and error will work. A direct substitution shows that $35$ does not work. So, to do it more efficiently, observe that for $x\in \{1,\dots,34\}$, $\left[\dfrac{x}{5}\right]$ takes only $7$ values and $\left[\dfrac{x}{7}\right]$ only $5$. A systematic way to do it is the following:

• For $x=1,\dots,6$: $\left[\dfrac{x}{7}\right]=0$. Hence, $x=5,6$ will do, since for these values $\left[\dfrac{x}{5}\right]=1$.
• For $x=7,\dots,13$: $\left[\dfrac{x}{7}\right]=1$. Hence, $x=10,\dots,13$ will do, since for these values $\left[\dfrac{x}{5}\right]=2$.
• For $x=14,\dots,20$: $\left[\dfrac{x}{7}\right]=2$. Hence, $x=15,\dots,19$ will do, since for these values $\left[\dfrac{x}{5}\right]=3$.
• For $x=21,\dots,27$: $\left[\dfrac{x}{7}\right]=3$. Hence, $x=21,\dots,24$ will do, since for these values $\left[\dfrac{x}{5}\right]=4$.
• For $x=28,\dots,34$: $\left[\dfrac{x}{7}\right]=4$. Hence, $x=28,29$ will do, since for these values $\left[\dfrac{x}{5}\right]=5$.

This gives you the complete set of solutions, which is $$\{5,6,10,11,12,13,15,16,17,18,19,21,22,23,24,28,29\}$$