Let n 2635 a How many positive divisors are there of n b Ho

Let n = 2⁶3⁵. (a) How many positive divisors are there of n? (b) How many positive divisors of n are there whose prime factorization contains a different number of 2’s and 3’s?

It is under combinatorics.

Solution

prime factorization os 2635 = 5*17*31

no of divisiors = (1+1)*(1+1)*(1+1) = 2*2*2 = 8
note : here the bold numbers are the powers of the prime factors

Since there are no prime factors containing 2 and 3 hence all the prime factors will have the same no of 2\'s and 3\' i.e zero hence zero positive divisors of n are there whose prime factorization contains a different number of 2’s and 3’s

Please check if your n is correct other