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We will find a number $n$ in the form $10^k(10a+b)+c$ where $0\le c<10^k$ and $a,b$ are digits. Deleting the digit $b$ gives the number $n_1=10^{k}a+c$. Since $n-n_1=10^k(9a+b)$, to satisfy the conditions of the problem it is sufficient that $d$ divides $9a+b$ and $10^{k}a+c$.

The digit $b$ can be chosen so that $9$ divides $d-b$ and we set $a=\frac{d-b}{9}$. Then $d=9a+b$. Let $k$ be such that $10^{k-1}>d$. Then there are integers $q$ and $0\le r<d$ such that $10^{k}a+10^{k-1}=dq+r$. If we set $c=10^{k-1}-r>0$, then $10^{k}a+c=dq$, obviously a number divisible by $d$.