# Finding $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+…+\frac{1}{d_k}$

If we assume that $d_1,d_2,d_3,...,d_k$ are the divisors for the positive integer $n$ except $1,n$ if $d_1+d_2+d_3+...+ d_k=72$ then how to find $$\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+...+\frac{1}{d_k}$$

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what are you trying ? – Maisam Hedyelloo Feb 14 '13 at 21:15
$1$ is not prime. I'm sure you meant all divisors – Norbert Feb 14 '13 at 21:16
Yeah, I'm confused - do you want all the divisors, or all the prime divisors? Excluding $1$ causes the confusion, but the problem is actually easier when you are using all the divisors other than $1,n$. – Thomas Andrews Feb 14 '13 at 21:18
– Norbert Feb 14 '13 at 21:20
I am surprised that this question has been closed as what used to be called a duplicate. True, a solution (if the reference to primes is a mistake) mostly uses similar ideas as the linked to post. But if one's criterion for duplicate is this broad, a large proportion of the questions should be closed. – André Nicolas Feb 14 '13 at 21:33

If you write: $$\sum_{d \mid n} \frac{1}{d} = \frac{1}{n} \sum_{d \mid n} \frac{n}{d} = \frac{1}{n} \sum_{d \mid n} d = \frac{72}{n}$$ The divisor sum $\sum_{d \mid n} d = \sigma(n)$ is easily seen to satisfy $\sigma(n) \ge n + 1$ (those two are divisors always; they are the only ones if $n$ is prime, otherwise there are more). So you'd have to check up to $n = 71$. To find out what $n$ is so that $\sigma(n) = 72$, a fast trip to http://oeis.org/A000203 shows that the full list is 30, 46, 51, 55, 71.
The divisor function satisfies the reflection formula, $$n^x\sigma_{-x}(n)=\sigma_{x}(n)$$
Sense $$\sum_{d\mid n} d^x=\sum_{d\mid n} (n/d)^x$$