Let $\rm \overline{5^a5^b}=5^k$ and $\rm 10^{p-1}<5^b<10^p$.We have $\rm 5^k=\overline {5^a5^b}=5^a\cdot 10^p+5^b\Leftrightarrow 5^b(5^{k-b}-1)=5^{a+p}2^p$ $\rm u_5(LHS)=u_5(RHS)\Leftrightarrow b=a+r$ and $\rm 5^{k-b}-1=2^p$.Let $\rm k-b=q\geq 1$. We have that $\rm 5^q-1\equiv 2^p\pmod 3\Leftrightarrow (-1)^q-1\equiv (-1)^p\equiv \pm 1\pmod3 $ so $\rm q$ is odd and $\rm p $ is even.Let $\rm q=2r+1$.If $\rm p\geq 3$ then $\rm 5^{2r+1}-1\equiv 2^p\equiv 0\pmod 8\Leftrightarrow 25^r\cdot 5\equiv 1\pmod5\Leftrightarrow 5\equiv 1\pmod8 $,contradiction. So $\rm p=1,2\Rightarrow 5^q=3,5$ so $\rm p=2,q=1$. $\rm p=2\Rightarrow 10<5^b<10^2\Leftrightarrow b=2,k-b=q=1\Leftrightarrow k=3$. So $\rm 125=\overline{5^a25}\Leftrightarrow 5^a=1\Leftrightarrow a=0$,contradiction because $\rm a>0$. So there are no exist.