According to Stetson University, $27$ is the largest number $χ$ such that $\sum_{i \in δ(χ^3)} = χ$ where $δ(χ) =$ digits of $χ$.

I have tried proving that this is true, but do not know where to start. Could someone please help me with this question? $\because 27 = 3^3$, does this mean anything? Thanks.

PS: $i$ does not represent $\sqrt {-1}$ in this case.


closed as off-topic by Did, José Carlos Santos, Namaste, Shailesh, Xam Aug 15 '17 at 3:21

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  • 2
    $\begingroup$ $\delta(x)$ is the sum of the digits, maybe? The digits are just a collection. $\endgroup$ – Thomas Andrews Aug 5 '17 at 2:28
  • $\begingroup$ Yes. Sorry bout that. $\endgroup$ – George N. Missailidis Aug 5 '17 at 2:36
  • $\begingroup$ If $\delta(\chi)$ are the digits of $\chi$ then notationally the problem should be showing that $27$ is the largest $x$ such that $$\sum_{i\in \delta(x^3)} i = x$$ $\endgroup$ – adfriedman Aug 5 '17 at 2:41
  • $\begingroup$ Thanks for that. I will get to that right away $\endgroup$ – George N. Missailidis Aug 5 '17 at 2:43
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    $\begingroup$ @GeorgeN.Missailidis I notice that you use symbols such as $\because$ and $\therefore$ in your writing, and that you have been editing other user's questions and answers to replace their words with these symbols. You will find that in mathematics, it is often better to use words such as therefore or since, etc., rather than these symbols; using $\because$ does not make for better mathematical writing. The important thing is that the idea is clear. $\endgroup$ – Théophile Aug 5 '17 at 3:25

Intuitively, the sum of digits is like a logarithm, so $\delta (x^3)$, the sum of digits of $x^3$, is something like $3 \log x$ and clearly there will be a largest $x$ that satisfies your equation. We want to find an upper limit for $x$, then we can just search up to that and be done.

To make this precise, suppose $10^{n-1} \lt x \lt 10^n$, so $x$ has $n$ digits. Then $x^3 \lt 10^{3n}$ so $x=\delta(x^3) \lt 27n=27 \lceil \log_{10}x\rceil$ We can ask Alpha to find that for $x \ge 45$, there is no hope because $x \gt 27 \log_{10}x$. Then we can just search up to $45$ and discover that $27$ is the largest solution.


The number of digits of a positive integer $x$ is $\lfloor \log_{10} x\rfloor+1$. Let $S(x)$ be the sum of the digits of $x$. Then $$S(x)\le 9(\lfloor \log_{10} x\rfloor+1)$$

So $S(x^3)=x$ implies $$x\le 9(\lfloor 3\log_{10} x\rfloor+1)\le9(3\log_{10} x+1)$$

For a lazy solution, plot (with google, for example) the function $$f(x)=9(3\log_{10} x+1)-x$$ and you'll find that $f(x)\le 0$ for $x\ge 57$. For a more rigurous approach, you should differentiate $f$ and prove that $f'<0$ to guarantee that.

  • $\begingroup$ Thanks for the help guys. I really appreciate it. I would have given you a vote as well @ajotatxe but I have reached my limit for today. $\endgroup$ – George N. Missailidis Aug 5 '17 at 3:02

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