To get zeroes it must have (2×5 ) as factors so (2×5)^3 will give zeroes on multiplication.

Prime Factorization of N
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The prime factorization of a natural number N represents the expression of N as a product of its prime factors. In this case, the prime factorization of N is given as:

N = 2⁴ × 3⁴ × 5³ × 7

To determine the number of consecutive zeroes in N, we need to analyze the prime factors and their exponents.

Prime Factors and Exponents
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The given prime factorization shows that N is composed of four prime factors: 2, 3, 5, and 7. Each of these prime factors has a corresponding exponent indicating the number of times it appears in the factorization.

- The prime factor 2 has an exponent of 4, which means it appears 4 times in the factorization.
- The prime factor 3 has an exponent of 4, indicating it appears 4 times as well.
- The prime factor 5 has an exponent of 3, meaning it appears 3 times.
- The prime factor 7 has an exponent of 1, indicating it appears only once.

Consecutive Zeroes
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To determine the number of consecutive zeroes in N, we need to consider the prime factorization and identify the lowest exponent among the prime factors. In this case, the prime factor 7 has the lowest exponent of 1.

Since the number of consecutive zeroes in a natural number is determined by the power of 10, which is composed of 2 and 5 as its prime factors, we need to consider the exponents of 2 and 5 in the prime factorization.

- The exponent of 2 is 4, indicating there are four 2's in the factorization.
- The exponent of 5 is 3, meaning there are three 5's in the factorization.

Since the exponent of 5 (3) is lower than the exponent of 2 (4), we can conclude that the number of consecutive zeroes in N is equal to the exponent of 5, which is 3.

Therefore, the number of consecutive zeroes in N is 3.