[tex]\displaystyle\it\\a=2^{n}\cdot5^{n+3}+1=2^n\cdot5^n\cdot125+1=10^n\cdot125+1=\\1\underbrace{\it 0000...00}_{\it de~n~ori~10}\cdot125+1=125\underbrace{\it 0000...00}_{\it de~n~ori~10}+1=\\125\underbrace{\it 000000...0000}_{\it de~n-1~ori~10}1,~suma~cifrelor~Este~1+2+5+1=9=\mathcal{M}_9 \implies\\\boxed{\it a\vdots9}.[/tex]
