If the prime factorization of a natural number n is $2^{3} \times 3^{2} \times 5^{2} \times 7$ , find the number of consecutive zeros in n.

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answer is D.

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Detailed Solution

Given natural number is

n = 2^{3} \times 3^{2} \times 5^{2} \times 7

{n = 2}^{3} \times 3^{2} \times 5^{2} \times 7

\Rightarrow n = (2 \times 2 \times 2) \times (3 \times 3) \times (5 \times 5) \times 7

\Rightarrow n = 8 \times 9 \times 25 \times 7

\Rightarrow n = 72 \times 175

\Rightarrow n = 12600

Thus, it can be clearly concluded that the consecutive number of zeros in n, which is 12600 is 2.
Hence, option 4 is correct.

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