If the prime factorization of a natural number n is 23 × 32 × 52 × 7, write the number of consecutive zeros in n.

Question

Philoid · Accepted Answer

If any number ends with the digit 0, it should be divisible by 10,

i.e. it will be divisible by 2 and 5.

Prime factorization of n is given as 2³ × 3² × 5² × 7.

It can be observed that there is (2 × 5) × (2 × 5)

⇒ 10 × 10 = 100

Thus, there are 2 zeros in n.

The number of consecutive zeros in n is 2.

If the prime factorization of a natural number n is 2³ × 3² × 5² × 7, write the number of consecutive zeros in n.