No. of divisors

How many different divisors have number
134⋅2413^{ 4} \cdot 2^{ 4}
?

Result

n =  25

Solution:

n1=4 n2=4  n=(n1+1)⋅ (n2+1)=(4+1)⋅ (4+1)=25n_{ 1 } = 4 \ \\ n_{ 2 } = 4 \ \\ \ \\ n = (n_{ 1 }+1) \cdot \ (n_{ 2 }+1) = (4+1) \cdot \ (4+1) = 25

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