If p, m and n are prime numbers, none of which is equal to the other two, what is the greatest common factor of `24p m^2n^2, 9p mn^2 `and `36p(mn)^3`?

Options

A. `3p mn `
B. `3p^2m^2n^2`
C. `3p mn^2`
D. `3p mn^3`

Answer

To find the greatest common factor (GCF) of the expressions \( 24pm^2n^2 \), \( 9pmn^2 \), and \( 36p(mn)^3 \), we will follow these steps:

### Step 1: Factor each expression
1. **First expression: \( 24pm^2n^2 \)**
- Factor \( 24 \): \( 24 = 2^3 \times 3^1 \)
- The expression becomes: \( 24pm^2n^2 = 2^3 \times 3^1 \times p^1 \times m^2 \times n^2 \)

2. **Second expression: \( 9pmn^2 \)**
- Factor \( 9 \): \( 9 = 3^2 \)
- The expression becomes: \( 9pmn^2 = 3^2 \times p^1 \times m^1 \times n^2 \)

3. **Third expression: \( 36p(mn)^3 \)**
- Factor \( 36 \): \( 36 = 2^2 \times 3^2 \)
- The expression becomes: \( 36p(mn)^3 = 2^2 \times 3^2 \times p^1 \times m^3 \times n^3 \)

### Step 2: Identify the common factors
Now we need to find the GCF by taking the minimum power of each prime factor present in all three expressions.

- **For \( 2 \)**:
- In \( 24pm^2n^2 \): \( 2^3 \)
- In \( 9pmn^2 \): \( 2^0 \) (not present)
- In \( 36p(mn)^3 \): \( 2^2 \)
- **Minimum power**: \( 2^0 \)

- **For \( 3 \)**:
- In \( 24pm^2n^2 \): \( 3^1 \)
- In \( 9pmn^2 \): \( 3^2 \)
- In \( 36p(mn)^3 \): \( 3^2 \)
- **Minimum power**: \( 3^1 \)

- **For \( p \)**:
- In all expressions: \( p^1 \)
- **Minimum power**: \( p^1 \)

- **For \( m \)**:
- In \( 24pm^2n^2 \): \( m^2 \)
- In \( 9pmn^2 \): \( m^1 \)
- In \( 36p(mn)^3 \): \( m^3 \)
- **Minimum power**: \( m^1 \)

- **For \( n \)**:
- In \( 24pm^2n^2 \): \( n^2 \)
- In \( 9pmn^2 \): \( n^2 \)
- In \( 36p(mn)^3 \): \( n^3 \)
- **Minimum power**: \( n^2 \)

### Step 3: Combine the common factors
Now we combine the common factors with their minimum powers:
\[ GCF = 3^1 \times p^1 \times m^1 \times n^2 = 3pmn^2 \]

### Final Answer
The greatest common factor of \( 24pm^2n^2 \), \( 9pmn^2 \), and \( 36p(mn)^3 \) is:
\[ \boxed{3pmn^2} \]