The sum of all positive divisors of 960 is | vedclass

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Question

(A) The prime factorization of $960$ is $960 = 2^6 	imes 3^1 	imes 5^1.$The sum of all positive divisors of a number $N = p_1^{a_1} 	imes p_2^{a_2}

Vedclass · Accepted Answer

$3048$

Vedclass · Answer

(A) The prime factorization of $960$ is $960 = 2^6 	imes 3^1 	imes 5^1.$The sum of all positive divisors of a number $N = p_1^{a_1} 	imes p_2^{a_2} 	imes p_3^{a_3}$ is given by the formula $\left( \frac{p_1^{a_1+1}-1}{p_1-1} ight) \left( \frac{p_2^{a_2+1}-1}{p_2-1} ight) \left( \frac{p_3^{a_3+1}-1}{p_3-1} ight).$Substituting the values $p_1=2, a_1=6, p_2=3, a_2=1, p_3=5, a_3=1$:Sum $= \left( \frac{2^{6+1}-1}{2-1} ight) 	imes \left( \frac{3^{1+1}-1}{3-1} ight) 	imes \left( \frac{5^{1+1}-1}{5-1} ight)$Sum $= \left( \frac{128-1}{1} ight) 	imes \left( \frac{9-1}{2} ight) 	imes \left( \frac{25-1}{4} ight)$Sum $= 127 	imes \frac{8}{2} 	imes \frac{24}{4}$Sum $= 127 	imes 4 	imes 6 = 3048$.

The sum of all positive divisors of $960$ is

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