a) Find the natural numbers ${\overline{ab}}$ such that
${\overline{ab}}= a+(a+1)+...+b$.
b) Exist natural numbers ${\overline{abc}}$ such that
${\overline{abc}} = a+(a+1)+...+{\overline{bc}}?$;
Generalization.
In case a), using arithmetic progression sum, obtain the equation ${\overline{ab}}= \frac{(a+b)(-a+b+1)}{2}$ solutions having: 15, 27.
$2.10^na+2x=2\overline{ax}=x(x+1)-a(a-1)$ where $x$ could be $\overline{b},\overline{bc}...$
$ x^2-x-a^2-(2.10^n-1)a=0$ has solutions $\displaystyle x=\frac{1\pm\sqrt{1+4a^2+4(2.10^n-1)a}}{2}$. There are no integer solutions for $n=2,3,4,6,7$. For $n=5$, $a=19048, x=47620$ is the only solution.