To determine the number of consecutive zeroes at the end of the number $n$, we need to find the number of factors of 10 in $n$. Each factor of 10 is composed of a factor of 2 and a factor of 5.

Given:

$n = 2^3 \times 3^4 \times 5^4 \times 7$

The number of factors of 2 in $n$ is 3 (from $2^3$), and the number of factors of 5 in $n$ is 4 (from $5^4$). Since each pair of a factor of 2 and a factor of 5 forms a factor of 10, the number of trailing zeros is determined by the minimum of the exponents of 2 and 5.

Here, the number of factors of 2 is 3, and the number of factors of 5 is 4. Therefore, the number of factors of 10 is:

$\min(3, 4) = 3$

Thus, the number of consecutive zeroes at the end of $n$ is 3.

Would you like further clarification, or do you have any other questions?

Here are 5 related questions:

1. What would be the number of trailing zeroes if $n = 2^5 \times 3^3 \times 5^2 \times 11$?
2. How do the exponents of 2 and 5 affect the number of trailing zeroes in a product?
3. If you had the number $n = 2^6 \times 5^3 \times 7^2$, how many trailing zeroes would there be?
4. What if the problem included additional prime factors other than 2 and 5, how would that affect the number of trailing zeros?
5. How do you calculate the number of trailing zeroes in a factorial, such as $100!$?

Tip: Remember that for trailing zeroes, you only need to count the pairs of factors of 2 and 5. The limiting factor is the lesser exponent between them.