repeating number, when n is multiple of 3. If we take any integer  multiple of $3$, and add up the cubes of its decimal digits, then take 
the result and sum the cubes of its digits, and so on, we invariably end up with $153.$
Why the above result is true for only an integer multiple of $3?$ can we prove the reasons for it? Is there any such numbers will invariably end with any number and not multiple of $3?$ If yes, how to find them.
 A: A number is only divisible by three if the sum of its digits is a multiple of three. Since $a^3\equiv a\mod 3$, we see that after applying the 'step' (the function $s$) to a number divisible by three will still be divisible by three. Also, a number not divisible by $3$ will never be divisible by three after the step (because everything here works in two directions). Since $3|153$ (because $3|1+5+3=9$), only numbers divisible by $3$ can end in $153$.
In general, because $9^3<1000$, $s(n)<n$ when $n>10000$ and $s(n)<10000$ when $n\leq 10000$ (these are not a sharp bounds). This implies that the sequence $\left(s^k(n)\right)_{k\geq 0}$ will be decreasing until it falls below $10000$. Because of that, the function will eventually get in a finite loop. Thus, there is an $l$ such that, for a sufficient large $K$, $s^{k+l}(n)=s^k(n)$ for all $k>K$. It is possible to calculate all fixpoints $n$ of $s$ for which $s(n)=n$. I think this only applies to $153$, since it is given in this problem, but maybe there are some other numbers with this property.
EDIT As user44197 points out, the only other fixpoints (I assume) are $370$, $371$ and $407$. And of course $0$ and $1$, but they are trivial. As you can see, none of these numbers is divisible by $3$, so they can only end up in $153$ or in a cycle, but there aren't any cycles apparently. For other numbers then $3$, I don't know whether or not multiples will always end in the same number, but I think not. The only number that may be of interest is $11$, because a number is a multiple of $11$ when the alternating sum of its digits is a multiple of $11$. Problem is that $2^3=8\neq \pm 2\mod11$, so divisibility it not conserved. For any other interesting numbers, you can search for $n$ s.t. $i^3\equiv \pm i\mod n$ for all $1\leq i\leq n$. Thus, $2^3=8\equiv \pm 2\mod n$. This is true for $n=2$, $n=3$, $n=5$, $n=6$ and $n=10$. $n=2$ doesn't work, because $s(12)=1+8=9$ and $9$ is odd. Same for $n=10$: $s(10)=1+0=1$. For $n=6$, we get $s(12)=9$. For $n=5$, we get $s(10)=1$, so this won't work either. Thus, only multiples of $3$ have the property that they are still multiples of $3$ after applying $s$.
