Consider the number n = 16^4 x 81^4 x 6^6. A) find the prime factorization of n. B) provide two other factorization of n. C) without multiplying, briefly explain how you know that the two factorizations you provided in part b are factorizations of n.

Let's break down the problem step by step:

### A) Prime Factorization of \( n = 16^4 \times 81^4 \times 6^6 \)

1. **Factorizing each base number into primes**:
   - \( 16 = 2^4 \)
   - \( 81 = 3^4 \)
   - \( 6 = 2 \times 3 \)

2. **Substitute these factorizations into the expression for \( n \)**:
   \[
   16^4 = (2^4)^4 = 2^{16}
   \]
   \[
   81^4 = (3^4)^4 = 3^{16}
   \]
   \[
   6^6 = (2 \times 3)^6 = 2^6 \times 3^6
   \]

3. **Now combine the powers of 2 and 3**:
   \[
   n = 2^{16} \times 3^{16} \times 2^6 \times 3^6
   \]
   Group the powers of 2 and 3:
   \[
   n = 2^{16 + 6} \times 3^{16 + 6} = 2^{22} \times 3^{22}
   \]

Thus, the **prime factorization** of \( n \) is:
\[
n = 2^{22} \times 3^{22}
\]

---

### B) Two Other Factorizations of \( n \)

1. First Factorization:
   \[
   n = 4^{11} \times 9^{11}
   \]
   Explanation:
   - \( 4 = 2^2 \) and \( 9 = 3^2 \), so \( 4^{11} = (2^2)^{11} = 2^{22} \) and \( 9^{11} = (3^2)^{11} = 3^{22} \), matching the prime factorization.

2. Second Factorization:
   \[
   n = 16^5 \times 81^5 \div 2^2 \times 3^2
   \]
   Explanation:
   - \( 16^5 \times 81^5 \) leads to powers \( 2^{20} \times 3^{20} \), and dividing by \( 2^2 \times 3^2 \) gives \( 2^{22} \times 3^{22} \), which matches the prime factorization.

---

### C) How We Know These Are Factorizations of \( n \)

1. **First factorization**: 
   - \( 4^{11} \times 9^{11} \) simplifies back to \( 2^{22} \times 3^{22} \), which directly matches the prime factorization found in part A.

2. **Second factorization**: 
   - Starting with \( 16^5 \times 81^5 \), we know that \( 16 = 2^4 \) and \( 81 = 3^4 \), so the total powers add up as \( 2^{20} \times 3^{20} \). Dividing by \( 2^2 \times 3^2 \) adds the required exponents to match \( 2^{22} \times 3^{22} \).

### Do you want more details or have any questions?

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**Five Related Questions**:
1. How would the prime factorization change if one of the bases in \( n \) was a different number?
2. Can you find the prime factorization of \( n = 25^3 \times 18^2 \times 7^5 \)?
3. How does prime factorization help in simplifying complex multiplications?
4. Can \( n \) be factorized using different bases other than powers of 2 and 3?
5. How can prime factorization be used to find the greatest common divisor of two numbers?

**Tip**: When factoring large expressions, always start by breaking down each base into its prime components. This simplifies the process and helps in managing exponents effectively.

Learn how to find the prime factorization of n = 16^4 x 81^4 x 6^6. This problem covers exponents and prime factorization techniques, involving the fundamental theorem of arithmetic and breaking down numbers into their prime components. Suitable for Grades 9-12.

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