How to solve equation involving ceiling function I am trying to show the following holds when $n$ is not divisible by $1+\Delta$.
$⌈ \frac{n}{1+\Delta} ⌉ = n- \Delta \implies \Delta=n-2$
I have tried using the Quotient Reminder Theorem but I am unable to show this. Any help would be appreciated.
 A: First write n as:
$$n=m(1+\Delta)+r~~,~~ \lceil\frac{n}{1+\Delta}\rceil=m+1$$
Substituting and solving for $\Delta$ we obtain that:
$$
\Delta=(1-r)/(m-1)$$
Since $\Delta$ is a positive integer, the only choice is $r=1$. But this also sets $m=1$ and we obtain the desired result.
A: From
$$\left\lceil\frac{n}{1+\Delta}\right\rceil-1<\frac{n}{1+\Delta}<\left\lceil\frac{n}{1+\Delta}\right\rceil$$
and the given equation, we draw
$$n-\Delta-1<\frac{n}{1+\Delta}<n-\Delta$$
or after rewriting
$$n-2-\frac1\Delta<\Delta<n-1.$$
As soon as $\Delta>1$, we do have $\Delta=n-2$.

Note that $\Delta+1$ never divides $n$, except with $\Delta+1=n$, which is always a solution.
A: By defintion of ceiling function you get $n-\Delta -1 < \frac n {1+\Delta}$ and $n-\Delta \geq \frac n {1+\Delta}$. You can re-write this as $1+\Delta \leq N <\Delta+2+\frac 1 {\Delta}$. Since $\frac 1 {\Delta}$ is a fraction $N$ must be either $1+\Delta$ or $2+\Delta$. The hypothesis rules out the first option so $n=2+\Delta$ or $\Delta =n-2$. 
