If n = 2 ^ 3 * 3 ^ 4 * 5 ^ 4 * 7 , then the number of consecutive zeroes in n, where n is a natural number, is​

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.

Learn how to determine the number of consecutive zeroes at the end of the number n = 2^3 * 3^4 * 5^4 * 7 using prime factorization and exponent analysis. Understand the relationship between factors of 2 and 5 in determining trailing zeros. Suitable for students in Grades 9-12.

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