Q. 34.5( 13 Votes )

If n = 23 × 34 × 54 × 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 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.

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