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.

  • 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

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.


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