Let $D = \{ n\in\mathbb{N} | 128000 \vdots n\}$. Calculate $\sum_{n\in D} n$.

We can see that $n = 128000k$, and $n$ needs to be even. But I don't know what to do. Any ideas?

  • $\begingroup$ Your question does not make sense. As currently phrased the answer is trivially $\infty$ $\endgroup$ – KingJ May 17 at 14:15
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    $\begingroup$ What does $128000 \vdots n$ mean? Do you mean...$n$ is divisible by $128000$? But then the sum obviously diverges. $\endgroup$ – lulu May 17 at 14:15
  • $\begingroup$ I edited. Im sorry. $\endgroup$ – sticknycu May 17 at 14:18
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    $\begingroup$ Are you sure you didn't mean $n\,|\,128000$"? $\endgroup$ – lulu May 17 at 14:19
  • $\begingroup$ When using nonstandard notation you should define it. $\endgroup$ – Bill Dubuque May 17 at 18:15

I suppose that the sum of divisors of $128000$ was meant, i.e., $$ \sigma(128000)=\sum_{d\mid 128000} d=319332. $$ Here we can use an explicit formula for $\sigma(n)$, using the prime factorisation $128000=2^{10}\cdot 5^3$ see here:

Is there a formula to calculate the sum of all proper divisors of a number?

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  • $\begingroup$ Please close (abstract) dupes - don't answer them (with links) $\endgroup$ – Bill Dubuque May 17 at 18:16
  • $\begingroup$ @Gone I usually do this, but here I was not sure whether it is an exact duplicate. Actually, it was not even clear to me whether or not the question was about $\sigma(n)$. Nevertheless you are right, of course. $\endgroup$ – Dietrich Burde May 17 at 18:18
  • $\begingroup$ Then you can post a comment. It makes it harder to clean up now that it has an accepted answer, i.e. we need to waste more time doing so... $\endgroup$ – Bill Dubuque May 17 at 18:20
  • $\begingroup$ OK, next time I will post this as comment. $\endgroup$ – Dietrich Burde May 17 at 18:20
  • $\begingroup$ Great! $\phantom{.......................................}$ $\endgroup$ – Bill Dubuque May 17 at 18:21

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