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 to a fixed point.

Question

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.

Answer 1

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.