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

(a) $\frac{k^2}{72}$

(b) $\frac{72}{k}$

(c) $\frac{72}{n}$

(d) cannot be computed.

I can't express $n$ in terms of $d_1,d_2,\dots,d_k$ . Here I am stuck. Please help.

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@ThomasAndrews :I just able to do the step $\frac{1}{d_1}+\frac{1}{d_2}+\dots+\frac{1}{d_k}=\frac{d_2d_3\dots d_k+ d_1d_3\dots d_k+\dots +d_1d_2 \dots d_{k-1}}{d_1d_2 \dots d_k}$ – Argha Jan 22 '13 at 21:19
Does it help to write $1 = d_1 < d_2 < \cdots < d_k = n$? – Eric Jan 22 '13 at 21:21
up vote 4 down vote accepted

There are only finitely many cases where the factors add up to 72. And these are the numbers 30, 46, 51, 55 and 71. You may want to check what happens to the sum of the reciprocal of their factors.

Now, for the general case, let $1 = d_1 < d_2 < \cdots < d_k = n$ be the factors of $n$. Then $\dfrac{1}{d_1} = \dfrac{d_k}{n}$, and in general $\dfrac{1}{d_i} = \dfrac{d_{k-i+1}}{n}$. Thus,

$\dfrac{1}{d_1} + \cdots \dfrac{1}{d_k} = \dfrac{d_k + d_{k-1} + \cdots + d_1}{n} = \dfrac{\sigma(n)}{n}$,

where $\sigma(n)$ is the sum of factors of $n$.

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+1 for nice detailed way. ;-) – Babak S. Feb 19 '13 at 15:25

There's no need to find $n$. Note that for every $d$ dividing $n$, $\frac{n}{d}$ is also a divisor of $n$. Therefore the sum of all $d_i$ is equal to the sum of all $\frac{n}{d_i}$, which means...

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The sum of divisors function is usually written with a letter $\sigma,$ so they are asking about $\sigma(n) = 72.$ Now, for a prime number $p,$ we do get $\sigma(p) = 1 + p,$ so in particular $\sigma(71) = 72.$ And that leads to at least one value for the sum of the reciprocals of the divisors.

We know that $\sigma(n) \geq n+1.$ So, the target 72 demands that $n \leq 71.$ Find out what other numbers solve $\sigma(n) = 72.$

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