Find minimum of $\frac{n}{S(n)}$ For every Natural Number like $n$ consider:$\frac{n}{S(n)}$ so that $S(n)$ is sum 
of the digits of the number $n$ in base-10. find minimum of $\frac{n}{S(n)}$ when:
a)$9<n<100$
b) $99<n<1000$
c)$999<n<10000$
d)$9999<n<100000$
for $9<n<100$ I tried:
$n=10a+b$ and $Min(\frac{10a+b}{a+b})=Min(1+\frac{9a}{a+b})$ so It is obvious that $b$ should be 9. I put $a=1,2,3,...$ and realized that if $a=1$ it will be minimum so the answer of part (a) is 19 but I dont know How we can mathematically show that $a=1$ 
for part b,c ,d I cant find mathematically way to show when this fraction (for example for part b: $\frac{100a+10b+c}{a+b+c}$) is minimum
 A: 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 $10999$.
A: Let's try a sketch of a more general approach for large $d$:

*

*To minimise $\frac n{S(n)}$ you will want the initial digits to be as small as possible to reduce $n$ and the final digits to be as big as possible to increase $S(n)$

*So once you get past one digit, solutions will be of the form $100\ldots099\ldots999$.  If you have $d$ digits with $m$ of them $9$s then the number is $n=10^{d-1}+10^m-1$ and the digit sum will be $S(n)=9m+1$.

For given $d$ you are in effect trying to find the integer $m\lt d$ which minimises  $\frac{10^{d-1}+10^m-1}{9m+1}$.  This is the smallest $m$ for which $\frac{10^{d-1}+10^m-1}{9m+1}<\frac{10^{d-1}+10^{m+1}-1}{9m+10}$ i.e. for which $$ 9m 10^{m} > 10^{d-1} -1 $$
This does not seem to have a simple solution for $m$ given $d$ though it is easily dealt with for small $d$ giving the answers to your particular questions:
digits  m=number of 9s    n which minimises n/(S(n))  S(n)
   2         1                      19                  10
   3         2                     199                  19
   4         2                    1099                  19
   5         3                   10999                  28
   6         4                  109999                  37

The next time $m$ does not increase by $1$ when $d$ does is at $d=15$ when optimal $m=12$ rather than $13$.  You can expect a similar stutter every time $m$ is just over $\frac19$ of a power of $10$, so the next when $m=112$ and $d=116$ and the following stutter at $m=1112$ and $d=1117$.
