if $ d$ is the number of digits in $ n$, then $ S(n)\leq 9d$, whereas $ n\geq 10^{d - 1}$. so we get the inequality \[ 10^{d - 1}\leq 1458d^3 + 8 \] which is false for $ d\geq 8$. in particular, this means that $ S(n)\leq 63$. now in light of the fact that $ S(n)\equiv n\pmod{9}$, we must have $ n\equiv 2n^3 + 8\pmod{9}$, whose only solution is $ n\equiv 1\pmod{9}$. so the only possible values of $ S(n)$ are 1, 10, 19, 28, 37, 46, 55. a quick check now shows that the only solutions are $ n = 10, 2008, 13726$.

pleurestique wrote: if $ d$ is the number of digits in $ n$, then $ S(n)\leq 9d$, whereas $ n\geq 10^{d - 1}$. so we get the inequality \[ 10^{d - 1}\leq 1458d^3 + 8 \] which is false for $ d\geq 8$. in particular, this means that $ S(n)\leq 63$. now in light of the fact that $ S(n)\equiv n\pmod{9}$, we must have $ n\equiv 2n^3 + 8\pmod{9}$, whose only solution is $ n\equiv 1\pmod{9}$. so the only possible values of $ S(n)$ are 1, 10, 19, 28, 37, 46, 55. a quick check now shows that the only solutions are $ n = 10, 2008, 13726$. In fact,you only need to check $ n=1,10,19,28,37,46$,the value $ 55>9\cdot 6$. My solution and official solutions is the same. Thank you.

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Superguy wrote: S(n)<=54 not 63 bto that bump though... btw how far did you have to scroll to get to this thread

please explain me why $S(n)$ = $1$ ( $mod 9$)?

inequality k<7