NTA Abhyas JEE Main2020MathematicsPermutation and CombinationPractice
The total number of divisors of the number N = 2 5 ⋅ 3 4 ⋅ 5 10 ⋅ 7 6 that are of the form 4 K + 2 ,   ∀ K ∈ N is equal to
Options
- A385
- B384
- C96
- D77
Correct answer
B. 384
Step-by-step solution
The required number is of the form 2 2 K + 1 , i.e. it will contain exactly one power of 2 . Also, we have to exclude 2 as K ∈ N . Number of required divisors = 1 ⋅ 5 ⋅ 11 ⋅ 7 - 1 = 385 - 1 = 384