A) 3048
B) 3087
C) 3047
D) 2180
Correct Answer: A
Solution :
\[960={{2}^{6}}\times {{3}^{1}}\times {{5}^{1}}.\]Therefore, bases are \[{{P}_{1}}=2,{{P}_{2}}=3\]and \[{{P}_{3}}=5\]and powers are \[{{a}_{1}}=6,{{a}_{2}}=1,{{a}_{3}}=1.\]Thus sum of all the positive divisors of \[960=\left( \frac{{{P}_{1}}^{{{a}_{1}}+1}-1}{{{P}_{1}}-1} \right)\left( \frac{{{P}_{2}}^{{{a}_{2}}+1}-1}{{{P}_{2}}-1} \right)\]\[\left( \frac{{{P}_{3}}^{{{a}_{3}}+1}-1}{{{P}_{3}}-1} \right)\] \[=\left( \frac{{{2}^{6+1}}-1}{2-1} \right)\left( \frac{{{3}^{1+1}}-1}{3-1} \right)\left( \frac{{{5}^{1+1}}-1}{5-1} \right)\] \[=127\times 4\times 6\] \[=3048\]
You need to login to perform this action.
You will be redirected in
3 sec
