If n = 23 × 34 × 54 × 7, then the number of consecutive zeros in n,

1.	If n = 23 × 34 × 54 × 7, then the number of consecutive zeros in n, where n is a natural number, isA. 2B. 3C. 4D. 7
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 × 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.

If n = 23 × 34 × 54 × 7, then the number of consecutive zeros in n, where n is a natural number, isA. 2B. 3C. 4D. 7

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 × 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.