A) 3048
B) 3087
C) 3047
D) 2180
Correct Answer: A
Solution :
Given number is 960, we know that \[960={{2}^{6}}\times {{3}^{1}}\times {{5}^{1}}.\] Therefore bases are \[{{p}_{1}}=2,\,{{p}_{2}}=3\] and \[{{p}_{3}}=5.\] and powers \[{{a}_{1}}=6,\,\,{{a}_{2}}=1\] and \[{{a}_{3}}=1.\]Thus sum of all the positive divisors of 960 \[=\left( \frac{{{p}_{1}}^{{{a}_{1}}+1}-1}{{{p}_{1}}-1} \right)\,\left( \frac{{{p}_{2}}^{{{a}_{2}}+1}-1}{{{p}_{2}}-1} \right)\,\left( \frac{{{p}_{3}}^{{{a}_{3}}+1}-1}{{{p}_{3}}-1} \right)\] \[=\left( \frac{{{2}^{6+1}}-1}{2-1} \right)\,\left( \frac{{{3}^{1+1}}-1}{3-1} \right)\,\left( \frac{{{5}^{1+1}}-1}{5-1} \right)\,=(127)(4)(6)=3048\].
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