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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
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$1$ is not prime. I'm sure you meant all divisors –  userNaN 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
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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

2 Answers 2

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.

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Great answer, +1. –  1015 Feb 15 '13 at 0:32

The divisor function satisfies the reflection formula, $$n^x\sigma_{-x}(n)=\sigma_{x}(n)$$

Sense $$\sum_{d\mid n} d^x=\sum_{d\mid n} (n/d)^x$$

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What does sense means? I haven't seen this before. Unless it is a typo then never mind. –  Arjang Feb 14 '13 at 23:55
    
@Arjang, typo for "since", I'll wager. –  Gerry Myerson Feb 15 '13 at 2:09

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