If n = 23 × 34 × 54 × 7, then the number of consecutive zeros in n, where n is a natural number, is
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 × 34 × 54 × 7.
It can be observed that there is (2 × 5) × (2 × 5) × (2 × 5)
⇒ 10 × 10 × 10 = 1000
Thus, there are 3 zeros in n.