Indian Statistical Institute B.Math & B.Stat : Number Theory

Let $d_1,d_2,d_3, \dots ,d_k$ be all the factors of a positive integer $n$ including $1$ and $n$. Suppose $d_1+d_2+d_3+ \dots +d_k = 72$. Then find the value of $$\frac{1}{d_1}+ \frac{1}{d_2}+\dots +\frac{1}{d_k}$$ Since $n$ is positive and $d_i$ is a factor of $n$ for each $i \in \{1,2,3,\dots,k\} \quad \exists \quad \lambda_i > 0 \quad \lambda_i \in \mathbb{N}$ such that $n=d_i \lambda_i$. This also shows that $\lambda_i$ is a factor of $n$. We now show that $\lambda_i \neq \lambda_j$ for $i \neq j$. If possible let $\lambda_i = \lambda_j$ for some $i,j$ where $i \neq j$. $\implies d_i \lambda_i = d_j \lambda_i \implies d_i = d_j$, a contradiction since $d_i$ and $d_j$ are distinct factors of $n$.Thus $\lambda_1,\lambda_2,\lambda_3, \dots ,\lambda_k$ are also the possible factors of $n$ $\implies \{ d_1,d_2,d_3, \dots ,d_k\}=\{\lambda_1,\lambda_2,\lambda_3, \dots ,\lambda_k \}$ in some order. $d_1+d_2+d_3+ \dots +d_k = 72 \implies \frac{1}{\lambda_1}+ \frac{1}{\lambda_2}+\dots +\frac{1}{\lambda_k} = \frac{72}{n} \implies \frac{1}{d_1}+ \frac{1}{d_2}+\dots +\frac{1}{d_k} = \frac{72}{n}$