Question

The number of divisors of the number $\frac{\displaystyle\sum_{ r =1}^{20} 5^{ r -1}{ }^{19} C _{ r -1}}{\displaystyle\sum_{ r =0}^{17}{ }^{17} C _{ r } 4^{17- r } 2^{ r }}$ is equal to

• A. 4
• B. 6
• C. 9
• D. 18

Answer 1

Answer: (C)

$\frac{\displaystyle\sum_{ r =1}^{20} 5^{ r -1}{ }^{19} C _{ r -1}}{\displaystyle\sum_{ r =0}^{17}{ }^{17} C _{ r } 4^{17- r } 2^{ r }}=\frac{\displaystyle\sum_{ r =1}^{20} 5^{ r -1} \cdot 1^{20- r } \cdot{ }^{19} C _{ r -1}}{\displaystyle\sum_{ r =0}^{17}{ }^{17} C _{ r } 4^{17- r } 2^{ r }}=\frac{6^{19}}{6^{17}}=36$
Total number of factor of 36 is 9