Suppose a two digit whole number is divided by the sum of its digits, what are the largest and smallest possible values? So we can write a two digit whole number as $n = 10a+b$ where $1 \leq a,b \leq 10$ and we would have that we want to minimize/maximize the following functions:

$f(a,b) = 10a+b$

$g(a,b) = a+b$

I don't remember how one does this and I don't know if there is another approach that could work.

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    $\begingroup$ you say $1\le b,$ so multiples of $10$ are excluded? there aren't so many two-digit numbers, so trial and error is not prohibitive $\endgroup$ – J. W. Tanner Nov 19 '19 at 5:35
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    $\begingroup$ My mistake, should be $0 < a <10$ and $0\leq b < 10$. I know that trial and error is a possible way to do this but I was hoping to find a method that could generalize to n digits. $\endgroup$ – ElPerroBermudez Nov 19 '19 at 5:54

Write the ratio you are interested in $$\frac {10a+b}{a+b}=1+\frac {9a}{a+b}=10-\frac {9b}{a+b}$$ To make this large you want $a$ large and $b$ small. If you allow $b=0$ this becomes $10$ regardless of $a$. If you do not allow $b=0$ the maximum comes at $91$ with $\frac {91}{10}=9.1$. To make it small you want $b$ large and $a$ small, but we cannot have $a=0$, so the minimum is $19$ with $\frac {19}{10}=1.9$


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