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Is there any number with $2017$ divisors whose sum of digits is $2017$?

We know that $2017$ is prime and any number satisfying the required condition is of the form $p^{2016}$, where $p$ is a prime number. From here I couldn't make any progress.

Any help or reference would be appreciated.

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  • $\begingroup$ From my observations from a computer program, the answer is negative. The least case $2^{2016}$ has sum of digits $2656$, and the sum of digits increases (most of the time, not always) so quickly that the number becomes larger than $10000$ when $p \ge 13$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 24 '18 at 13:58
  • $\begingroup$ I've written a program in Julia to see this. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 24 '18 at 14:04
  • $\begingroup$ I'm not sure if this is known to be impossible. However there are similar questions where it's easy to show non-existence: there are no numbers with $2011$ divisors whose sum of digits is $2011$. I wonder if this question was adapted from an earlier year :). $\endgroup$ – Erick Wong Mar 24 '18 at 16:53
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Heuristic argument

You have correctly identified that such a number must be of the form $p^{2016}$, where $p$ is a prime.

From Wolfram, $$\begin{array}{c|c}\text{number}&2^{2016}&3^{2016}&5^{2016}&7^{2016}&11^{2016}&13^{2016}&17^{2016}&19^{2016}\\\hline\text{sum of digits}&2656&4293&6211&7552&9559&10126&11539&11584\end{array}$$ Hence we would not expect any number to have such properties; otherwise at least $2746-2017=729$ digits have to be $0$ for the next possible number $23^{2016}$ (and every other digit must be $1$), which is extremely unlikely.

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