<s>[INST] 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
 [/INST] 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 $10999$.
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