Question

The sum of all positive divisors of $960$ is?

• A. $3048$
• B. $3087$
• C. $3047$
• D. $2180$

Answer 1

The correct option is A

$3048$

Explanation for correct option

The prime factors of $960$ are

$960=2^6\times 3\times 5$

We know that if the prime factorization of $n$ is

$n=\prod _k{p}_k^{a_k}$

Where $p_k$ are the prime factors and $a_k$ are the positive integer exponents, the sum of all the positive integer factors is

$\prod _k\left(\sum _{i=0}^{a_k}{p}_k^i\right)$

Therefor the sum of the positive divisor of $960$ is

$$\begin{gathered} =\left(1+2+2^2+2^3+2^4+2^5+2^6\right)·\left(1+3\right)·\left(1+5\right) \\ =\left(2^7-1\right)·4·6 \\ =127·4·6 \\ =3048 \end{gathered}$$

Hence the correct option is $\mathrm{Option}\left(\mathrm{A}\right)$