Question

The sum of all positive divisors of 960 is

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

Answer 1

Answer: (A)

3048

Answer 2

Given number is 960, we know that 960 = 2⁶ x 3¹ x 5¹. Therefore bases are p₁ = 2, p₂ = 3 and p₃ = 5, and powers a₁ = 6, a₂ = 1 & a₃ = 1. Thus sum of all the positive divisors of 960

= $\left( \frac{{p_{1}}^{a_{1 + 1}} - 1}{p_{1} - 1} \right)\left( \frac{{p_{2}}^{a_{2 + 1}} - 1}{p_{2} - 1} \right)\left( \frac{{p_{3}}^{a_{3 + 1}} - 1}{p_{3} - 1} \right)$

= $\left( \frac{2^{6 + 1} - 1}{2 - 1} \right)\left( \frac{3^{1 + 1} - 1}{3 - 1} \right)\left( \frac{5^{1 + 1} - 1}{5 - 1} \right)$ = (127) (4) (6) = 3048.