Permutations and Combinations
13.0 Sum of all Possible Divisors of a Natural Number
13.0 Sum of all Possible Divisors of a Natural Number
The sum of all possible divisors, $$ = (p_1^0 + p_1^1 + p_1^2 + .... + p_1^{{\alpha _1}})(p_2^0 + p_2^1 + p_2^2 + .... + p_2^{{\alpha _2}})(p_3^0 + p_3^1 + p_3^2 + .... + p_3^{{\alpha _3}})......$$
Question 27. Find the number of divisors of $5040$. How many proper factors are there for the same? Also find the sum of the divisors.
Solution: $5040$ can be factorised as, $$\begin{equation} \begin{aligned} = 5040 \\ = {2^4} \times {3^2} \times 5 \times 7 \\\end{aligned} \end{equation} $$
Total number of divisors,
$$\begin{equation} \begin{aligned} = (4 + 1)(2 + 1)(1 + 1)(1 + 1) \\ = (5)(3)(2)(2) \\ = 60 \\\end{aligned} \end{equation} $$
The total number of proper factors,
$$\begin{equation} \begin{aligned} = [(4 + 1)(2 + 1)(1 + 1)(1 + 1)] - 2 \\ = [(5)(3)(2)(2)] - 2 \\ = 58 \\\end{aligned} \end{equation} $$
Sum of all divisors,
$$ = ({2^0} + {2^1} + {2^2} + {2^3} + {2^4})({3^0} + {3^1} + {3^2})({5^0} + {5^1})({7^0} + {7^1})$$
$$\begin{equation} \begin{aligned} = (1 + 2 + 4 + 8 + 16)(1 + 3 + 9)(1 + 5)(1 + 7) \\ = (31)(13)(6)(8) \\ = 19,344 \\\end{aligned} \end{equation} $$ $$\begin{equation} \begin{aligned} 2\left| {\underline {\,5040\,} } \right. \\ 2\left| {\underline {\,2520\,} } \right. \\ 2\left| {\underline {\,1260\,} } \right. \\ 2\left| {\underline {\,630\,} } \right. \\ 3\left| {\underline {\,315\,} } \right. \\ 3\left| {\underline {\,105\,} } \right. \\ 5\left| {\underline {\,35\,} } \right. \\ 7\left| {\underline {\,7\,} } \right. \\ 1 \\\end{aligned} \end{equation} $$
