0
$\begingroup$

I'm currently working in the following exercise:

Let $T$ be the function that sum the digits of an integer $n$. Let $a (n)$ be the number of times we must apply $T$ to an integer n so that it becomes a fixed point. For example $a(452) = 2$, since $T (452) = 11$, $T (11) = 2$, $T (2) = 2$. So we must apply twice T to get to the fixed point $2$. Find the smallest positive integer $n$ such that $a (n) = 3$. Find the smallest positive integer n such that $a (n) = 4$.

I've been trying descomposition in prime factors so that could help me to find the numbers required to find the answer but that option is not working. If anyone has a hint or any help will be really appreciated.

$\endgroup$
  • 2
    $\begingroup$ Prime factors has nothing to do with this. Find the smallest so that $a (n)=1$. That's clearly $10$. Find the smallest where $a(n)=2$ which is the smallest where the digits add to 10. That's clearly $19$ and so the smallest where $a (n)=3 is the smallest where the digits add to $19$ and that's 199. So the smallest where a (n)=4 will be the smallest where the digits add to 199. (It will have at least 22 digits so prime factors is definitely a useless idea). $\endgroup$ – fleablood Jan 30 at 2:50
3
$\begingroup$

Prime factorizations don't matter, digits do. So look at the smallest number such that $a(n)>0$.

Note that $T(d)=d$ for all single digits $d$, so try $T(10)=1$, so $a(10)=1$.

Going from here, we now know that if $T(n)=10$, then $a(n)=2$, since 10 is not a fixed point. To do that, simply make the digits sum to 10, and make the smallest number possible, which is $19$.

Next, if $T(n)=19$ then we know $a(n)=3$. I'll leave you to figure out what the smallest number that has a digit sum of 19 is.

$\endgroup$

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.