Question

how many consecutive zeroes will be there in a number N, whose prime factorisation is given as N = 2² × 3² × 5² × 7 × 13?​

Answer 1

Answer:

Given that

The prime factorization of the given natural number N is 2² × 3² × 5² × 7 × 13

Prime factorization: by using prime factorization we can find out which prime numbers multiple together to make the original number.

To find the consecutive zeroes in a number, split out the 2’s and 5’s which can finally sum up together to give 10.

N = 2² × 3² × 5² × 7 × 13

N = 3² × 7 × 13 × 2² × 5²

N = 3² × 7 × 13 × (2 × 5)²

N = 3² × 7 × 13 × (10)²

N = 3² × 7 × 13 × 100

Thus, in the given natural number 'N' there are 2 consecutive zeroes.

Hope it helps u...