Consider the number M = 7^2 x 23 x 31, where 7, 23, and 31 are prime numbers. List all of the factors of M without computing the value of M first. Clearly show or explain how you were able to find all of the factors.

The number \( M = 7^2 \times 23 \times 31 \) consists of the prime factorization of \( M \), where the prime numbers are 7, 23, and 31, and 7 is raised to the power of 2. Let's find all the factors of \( M \) by using the prime factorization directly without computing the product.

### Steps to find all factors:

1. **Prime factorization**: 
   The prime factorization of \( M \) is \( 7^2 \times 23^1 \times 31^1 \).

2. **Number of factors**:
   The total number of factors of a number is determined by adding 1 to each of the exponents in its prime factorization and multiplying these results.
   - For \( 7^2 \), there are \( 2 + 1 = 3 \) possible factors: \( 7^0, 7^1, 7^2 \).
   - For \( 23^1 \), there are \( 1 + 1 = 2 \) possible factors: \( 23^0, 23^1 \).
   - For \( 31^1 \), there are \( 1 + 1 = 2 \) possible factors: \( 31^0, 31^1 \).

   So, the total number of factors is:
   \[
   3 \times 2 \times 2 = 12
   \]
   There are 12 factors in total.

3. **Generating factors**:
   To find all the factors, we take all combinations of powers of 7, 23, and 31:
   - Factors of the form \( 7^a \times 23^b \times 31^c \), where \( a = 0, 1, 2 \), \( b = 0, 1 \), and \( c = 0, 1 \).

   Here are all the possible combinations:
   - \( a = 0, b = 0, c = 0 \): \( 7^0 \times 23^0 \times 31^0 = 1 \)
   - \( a = 1, b = 0, c = 0 \): \( 7^1 \times 23^0 \times 31^0 = 7 \)
   - \( a = 2, b = 0, c = 0 \): \( 7^2 \times 23^0 \times 31^0 = 49 \)
   - \( a = 0, b = 1, c = 0 \): \( 7^0 \times 23^1 \times 31^0 = 23 \)
   - \( a = 1, b = 1, c = 0 \): \( 7^1 \times 23^1 \times 31^0 = 7 \times 23 = 161 \)
   - \( a = 2, b = 1, c = 0 \): \( 7^2 \times 23^1 \times 31^0 = 49 \times 23 = 1127 \)
   - \( a = 0, b = 0, c = 1 \): \( 7^0 \times 23^0 \times 31^1 = 31 \)
   - \( a = 1, b = 0, c = 1 \): \( 7^1 \times 23^0 \times 31^1 = 7 \times 31 = 217 \)
   - \( a = 2, b = 0, c = 1 \): \( 7^2 \times 23^0 \times 31^1 = 49 \times 31 = 1519 \)
   - \( a = 0, b = 1, c = 1 \): \( 7^0 \times 23^1 \times 31^1 = 23 \times 31 = 713 \)
   - \( a = 1, b = 1, c = 1 \): \( 7^1 \times 23^1 \times 31^1 = 7 \times 23 \times 31 = 4973 \)
   - \( a = 2, b = 1, c = 1 \): \( 7^2 \times 23^1 \times 31^1 = 49 \times 23 \times 31 = 34693 \)

### List of all factors:
\[
1, 7, 23, 31, 49, 161, 217, 713, 1127, 1519, 4973, 34693
\]

These are all the factors of \( M \), found by systematically combining the powers of the primes.

### Would you like more details on how these combinations were formed, or do you have any other questions?

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Here are 5 related questions to deepen understanding:

1. How do you compute the number of divisors from a prime factorization?
2. What is the difference between finding prime factors and finding all factors of a number?
3. Can you explain how to find the greatest common divisor using prime factorization?
4. How would the process change if \( 7 \), \( 23 \), and \( 31 \) were not prime numbers?
5. What are the applications of factorization in number theory or algebra?

**Tip**: For large numbers, prime factorization makes finding divisors and solving equations much easier.

Learn how to find all factors of the number M = 7^2 x 23 x 31 without computing M directly. This problem explains prime factorization, combinatorics, and provides a step-by-step approach to listing all factors of M. Suitable for students in Grades 9-12.

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