For any real number x: Prove that among the numbers x,2x,...,(n-1)x ,there is one that differs from an integer by at most $\frac{1}{n}$.
any hints for a pigeon solution.
Non-pigeon solution
Consider $|x|< 1\Rightarrow x=\pm \frac{1}{p}$, then
$\bullet $ if $p<n$, we get an integer.
$\bullet $ if $p\geq n$, we get for $x>0$ that $x-\frac{1}{n}=\frac{1}{p}-\frac{1}{n}\leq 0$ and for $x<0$ that $\frac{1}{n}+x=\frac{1}{n}-\frac{1}{p}\geq 0$.So in either case x is close to 0.
So if $x\in (a,b)$,where $a,b\in \mathbb{Z}$, we get $x-a$ is close to zero or an integer, thus by translating back x is close to a or an integer.
Consider the intervals $[0,1/n)$, $[1/n,2/n)$, $\ldots$, $[1-1/n,1)$. Consider each of the numbers $x$, $2x$, $\ldots$, $(n-1)x$ modulo $1$. If one of these numbers is contained in the first or the last of the intervals, we are done. If this is not the case, we have $n-1$ numbers to be distributed over $n-2$ intervals $[1/n,2/n)$, ..., $[1-2/n,1-1/n)$. By the pigeonhole principle, there are two numbers $ax$, $bx$ ($a>b$) in the same interval. Now $(a-b)x$ differs at most $1/n$ from an integer.