1
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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 at 5:35
  • 1
    $\begingroup$ 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 at 5:54
3
$\begingroup$

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 $\frac {19}{10}=1.9$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.