# Find n with this condition [duplicate]

Problem:

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?

• Your question does not make sense. As currently phrased the answer is trivially $\infty$ – KingJ May 17 at 14:15
• What does $128000 \vdots n$ mean? Do you mean...$n$ is divisible by $128000$? But then the sum obviously diverges. – lulu May 17 at 14:15
• I edited. Im sorry. – sticknycu May 17 at 14:18
• Are you sure you didn't mean $n\,|\,128000$"? – lulu May 17 at 14:19
• When using nonstandard notation you should define it. – 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:
• @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. – Dietrich Burde May 17 at 18:18
• Great! $\phantom{.......................................}$ – Bill Dubuque May 17 at 18:21