Find minimum of $\frac{n}{S(n)}$

For a mathematical proof of (b)

$$\frac{10a+b}{a+b}=2+\frac{8a-b}{a+b}$$

The only way this can be less than $2$ is if $a=1,b=9$. So the minimum is $2-\frac{1}{10}=1.9$.

Part (d) $$\frac{10^4a+10^3b+10^2c+10d+e}{a+b+c+d+e}-100=\frac{9900a+900b-90d-99e}{a+b+c+d+e}$$ The numerator of the RHS is clearly positive and so the minimum will occur for $c=9$. If instead of subtracting $100$ we subtracted $10$ we would obtain $d=9$ and subtracting $1$ gives $e=9$.

However subtracting higher powers of $10$ i.e. $1000$ and $10000$ produces a fraction where the numerator can be made negative and then it is best to make $a,b$ as small as possible i.e. $a=1,b=0$

The minimum is obtained for $10999$.