Question

The sum of all positive divisors of $960$ is
1) $3048$
2) $3087$
3) $3047$
4) $2180$

Answer 1

Hint: First we need to find the prime factorisation of the number. Then to find the sum of the positive divisors, if the prime factorisation is of the form $n_1^{{l_{_1}}} \times n_2^{{l_2}} \times ..... \times n_m^{{l_k}}$, we will use the formula,
$S = \left( {1 + {n_1} + n_1^2 + n_1^3 + ... + n_1^{{l_1}}} \right) \times \left( {1 + {n_2} + n_2^2 + n_2^3 + ... + n_2^{{l_2}}} \right) \times ... \times \left( {1 + {n_m} + n_m^2 + n_m^3 + ... + n_m^{{l_k}}} \right)$
Where, $n,l,m,k$ are positive integers.
Using this formula we can find the sum of divisors of the given number.

Complete step-by-step answer:
 The number is $960$.
Therefore, the prime factorisation of $960$ is ${2^6} \times 3 \times 5$.
Now, to find the sum of divisors of $960$, we will use the formula,
$S = \left( {1 + {n_1} + n_1^2 + n_1^3 + ... + n_1^{{l_1}}} \right) \times \left( {1 + {n_2} + n_2^2 + n_2^3 + ... + n_2^{{l_2}}} \right) \times ... \times \left( {1 + {n_m} + n_m^2 + n_m^3 + ... + n_m^{{l_k}}} \right)$.
Therefore, substituting the values of the prime factorisation of $960$, we get,
$S = \left( {1 + 2 + {2^2} + ... + {2^6}} \right) \times \left( {1 + 3} \right) \times \left( {1 + 5} \right)$
Simplifying the calculations, we get,
$ \Rightarrow S = \left( {1 + 2 + {2^2} + ... + {2^6}} \right) \times \left( 4 \right) \times \left( 6 \right)$
Now, we can see that the series present in the bracket $\left( {1 + 2 + {2^2} + ... + {2^6}} \right)$ forms a GP.
Therefore, we can find the sum of n terms of a GP using the formula $\dfrac{{a\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}$. So, we get,
$\left( {1 + 2 + {2^2} + ... + {2^6}} \right) = \dfrac{{1\left( {{2^7} - 1} \right)}}{{2 - 1}}$
Simplifying the expression,
$ \Rightarrow \left( {1 + 2 + {2^2} + ... + {2^6}} \right) = \dfrac{{\left( {{2^7} - 1} \right)}}{1}$
$\left( {1 + 2 + {2^2} + ... + {2^6}} \right) = {2^7} - 1$
Therefore, we can write $S$ as,
$ \Rightarrow S = \left( {{2^7} - 1} \right) \times \left( 4 \right) \times \left( 6 \right)$
Now, putting in the value of ${2^7}$ as $128$, we get,
$ \Rightarrow S = \left( {128 - 1} \right) \times \left( 4 \right) \times \left( 6 \right)$
Carrying out the calculations, we get the expression as,
$ \Rightarrow S = \left( {127} \right) \times 24$
$ \Rightarrow S = 3048$
Therefore, the sum of all positive divisors of $960$ gives us $3048$. Hence, Option (A) is the correct answer.
So, the correct answer is “Option B”.

Note: The formula we used basically in this problem is just a shortened and simpler form of the fact that we are adding all the divisors of $960$ one by one to get the required answer. We must know the formula to find the sum of n terms of a GP in order to solve such a problem. One must be accurate while doing calculations in order to be sure of the final answer.