If 24 is a factor of h and 28 is a factor of k, must 21 be a factor of hk?

Answer

To determine whether 21 is a factor of the product \( hk \) given that 24 is a factor of \( h \) and 28 is a factor of \( k \), we can follow these steps:

### Step 1: Express \( h \) and \( k \) in terms of their factors
Since 24 is a factor of \( h \), we can express \( h \) as:
\[ h = 24n \quad \text{(for some integer } n\text{)} \]
Similarly, since 28 is a factor of \( k \), we can express \( k \) as:
\[ k = 28m \quad \text{(for some integer } m\text{)} \]

### Step 2: Calculate \( hk \)
Now, we can find the product \( hk \):
\[ hk = (24n)(28m) = 24 \times 28 \times n \times m \]

### Step 3: Simplify \( hk \)
Next, we can simplify \( 24 \times 28 \):
\[ 24 = 3 \times 8 \quad \text{and} \quad 28 = 4 \times 7 \]
Thus,
\[ hk = (3 \times 8)(4 \times 7) \times n \times m = 3 \times 8 \times 4 \times 7 \times n \times m \]

### Step 4: Factor \( hk \)
Now, let's calculate \( 24 \times 28 \):
\[ 24 \times 28 = 672 \]
Next, we can factor \( 672 \):
\[ 672 = 3 \times 8 \times 4 \times 7 \]
This can be rearranged as:
\[ 672 = 3 \times 4 \times 7 \times 8 \]

### Step 5: Check divisibility by 21
Now, we need to check if \( hk \) is divisible by 21. The prime factorization of 21 is:
\[ 21 = 3 \times 7 \]
Since \( hk \) contains both 3 and 7 as factors, we can conclude that:
\[ hk = 3 \times 7 \times \text{(other factors)} \times n \times m \]
This indicates that \( hk \) is divisible by 21.

### Conclusion
Thus, we can conclude that 21 is indeed a factor of \( hk \).