Define $f(x)$ as the digit sum function in base $10$; Then, I am tasked with finding solutions to equation:
$$x = 2017f(f(x))\;.$$
My first idea was to use the congruence $x\equiv f(x) \bmod 9$ to obtain $x\equiv f(f(x)) \bmod 9$, hence $x-f(f(x))=9a$, for $a\in\mathbb Z$. Then, solving for $x$ yields from the first equation $x=2017a/224$, and this does yield solutions subject to the condition that $x$, $f(f(x))\ge0\in\mathbb Z$ when different values of $a$ are put in the general solution.
However, the solutions obtained actually satisfy the given equation for $a = 224n$, $n\in[0,9]$ only, and all solutions with $n\ge10$ do not satisfy it. Why is it so? Is there some better way to solve it?
The digit sum is like log, numbers have to get much larger for their digit sum to increase a little bit. Intuitively, this means that for a number to still be large after taking the digit sum twice, that number must be enormous- far larger than 2017 times the digit sum. This allows us to place an tight enough upper bound on how big $n=f(f(x))$ can be that we can just check everything below it.
For a number to have a digit sum $\ge n$ it must have at least $\frac{n}{9}$ digits, so in particular it must be $\ge 10^{n/9-1}$. So $$x \ge 10 \uparrow(\frac{1}{9}10\uparrow(\frac{f(f(x))}{9}-1)-1)$$ So by substituting $x = 2017n , f(f(x))=n$ $$2017n \ge 10 \uparrow(\frac{1}{9}10\uparrow(\frac{n}{9}-1)-1)$$ If we rearrange this to be about logs instead of exponentials we get: $$\log_{10}(\log_{10}2017+\log_{10} n + 1)+1 = \log_{10}\frac{1}{9} + \frac{n}{9}$$ Plot this in a grapher and you'll be able to see the RHS grows much more quickly, and any $n \ge 25$ is impossible.