Let $d_1, d_2, \cdots d_k$ be all the factors of a positive integer '$n$' including $1$ and $n$. Suppose $d_1 + d_2 + d_3+\cdots+d_k = 72$. Then find the value of $\frac{1}{d_1}+\frac{1}{d_2}+\cdots + \frac{1}{d_k}$.
Attempt:
Consider, $d_1, d_2, \cdots d_k$ are the factors of $n$ ascending order. Then
$d_1d_k=n$, $d_2d_{k-1}=n$, $\cdots$, $d_kd_{1}=n$ …………… (1)
Assume, $\frac{1}{d_1}+\frac{1}{d_2}+\cdots + \frac{1}{d_k}=P$
Then $\frac{n}{d_1}+\frac{n}{d_2}+\cdots + \frac{n}{d_k}=nP$
i.e
$d_k+d_{k-1}+\cdots +d_2+d_1=nP$ (using 1)
i.e $72=nP$, or $P=72/n$.
Problem: I have considered that number of divisors of $n$ are even as every divisor $d$ of $n$ has a twin divisor $n/d$. But it is not true if $n$ is a perfect square i.e $n=d^2$ as in this case $n$ has odd number of divisors (for example, $4=2^2$, $4$ has three divisors $1,2 ~\&~ 4$).
Best Answer
Your argument is fine and squares are not a problem. You are not pairing up the divisors, you are just reordering them. Let's walk through your calculation with $n=9$, where the sum of divisors is $1+3+9=13$. We then are asking what the value of $\frac 11 + \frac 13 + \frac 19=S$ is. We multiply by $9$ and get $9S=9+3+1=13, S=\frac{13}9$. The term $3$ does not cause a problem.