Q. 34.5( 13 Votes )

# If n = 2^{3} × 3^{4} × 5^{4} × 7, then the number of consecutive zeros in n, where n is a natural number, is

A. 2

B. 3

C. 4

D. 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 2^{3} × 3^{4} × 5^{4} × 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.

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