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