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Let $d_1, d_2, \dots , d_k$ be all the positive factors of a positive integer $n$ including $1$ and $n$. If $d_1+d_2+ \dots+d_k=72$, then find the value of $\frac{1}{d_1}+\frac{1}{d_2}+\dots+\frac{1}{d_k}$.

My attempt:

Let $a=p_1^{m_1}p_2^{m_2}\dots p_k^{m_k}$ be the prime factorization of $a$.

To calculate the sum of divisors function, $\sigma(a)$ we use the expression :

$(1+p_1+p_1^2+\dots+p_1^{m_1})(1+p_2+p_2^2+\dots+p_2^{m_2})\dots(1+p_k+p_k^2+\dots+p_k^{m_k})$

, which represents the sum of all possible positive factors of $a$ and is equal to:

$(\frac{p_1^{m_1+1}-1}{p_1-1})(\frac{p_2^{m_2+1}-1}{p_2-1})\dots(\frac{p_k^{m_k+1}-1}{p_k-1})$.

Now, if we replace each term in the each factor of the expression above the above expression by its reciprocal we shall obtain the sum of the reciprocals of all possible positive factors of $a$. This expression will be equal to:

$(1+1/p_1+1/p_1^2+\dots+1/p_1^{m_1})(1+1/p_2+1/p_2^2+\dots+1/p_2^{m_2})\dots(1+1/p_k+1/p_k^2+\dots+1/p_k^{m_k})$

which is equal to:

$(\frac{1-1/p_1^{m_1+1}}{1-1/p_1})(\frac{1-1/p_2^{m_2+1}}{1-1/p_2})\dots(\frac{1-1/p_k^{m_k+1}}{1-1/p_k})$ . This on simplication yields $\frac{\sigma(a)}{n}$.

Therefore for the information given in the question, the required value is $\frac{72}{n}$.

Is my reasoning and my answer correct? What are other methods to solve this question.

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    $\begingroup$ You are working much too hard. Just multiply that sum of reciprocals by $n$, and think about what you're adding up then. $\endgroup$ – Gerry Myerson Mar 29 at 5:38
  • $\begingroup$ Your answer is correct, and the comment of @GerryMyerson gives you the same value in one line of computation $\endgroup$ – user1952500 Mar 29 at 5:46
  • $\begingroup$ So $72$ is simply a dummy constant. That is, if the sum of the divisors is $m$, then the sum of the reciprocals is $\frac{m}{n}$. If weight is given to the choice of $72$, then $n=(30,51,55,71)$, and there is no single-valued answer. $\endgroup$ – Keith Backman Mar 29 at 18:58
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Your answer is correct, but there's a much simpler way as the comments show.

The key idea is that if $\{d_1,\ldots,d_k\}$ is the set of divisors of $n$ then $\{\frac n{d_1},\ldots,\frac n{d_k}\} = \{d_1,\ldots,d_k\}$ is also exactly the set of divisors of $n$, simply because each element of the first set is a divisor, and there is a bijection $d_k \to \frac n{d_k}$ for each $k$, so the number of elements in both sets are the same, so the first set covers all the divisors.

With this in mind, the sum of all elements of both sets are the same, hence we have : $$ d_1+\ldots+d_k = \frac{n}{d_1}+ \ldots + \frac n{d_k} \implies 72 = n\left(\frac 1{d_1}+ \ldots + \frac 1{d_k}\right) $$

which gives the result. Your proof required the full expression of the $\sigma$ function, which we did not use.

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