For any positive integer `m, g(m)= 24m- 2m (7m+1)` is perfectly divisible by

Options

A. 2
B. 4
C. 7
D. All of them

Answer

To solve the problem, we need to analyze the expression given for \( g(m) \) and simplify it to determine what it is perfectly divisible by.

### Step 1: Write down the expression for \( g(m) \).
The expression given is:
\[ g(m) = 24m - 2m(7m + 1) \]

### Step 2: Distribute the \( -2m \) in the expression.
We can distribute \( -2m \) across the terms inside the parentheses:
\[ g(m) = 24m - 2m \cdot 7m - 2m \cdot 1 \]
This simplifies to:
\[ g(m) = 24m - 14m^2 - 2m \]

### Step 3: Combine like terms.
Now, we combine the like terms \( 24m \) and \( -2m \):
\[ g(m) = (24m - 2m) - 14m^2 \]
This results in:
\[ g(m) = 22m - 14m^2 \]

### Step 4: Factor out common terms.
Next, we can factor out the common term from the expression:
\[ g(m) = 2m(11 - 7m) \]

### Step 5: Analyze the factors.
The expression \( g(m) = 2m(11 - 7m) \) shows that \( g(m) \) is divisible by \( 2m \). Since \( m \) is a positive integer, \( 2m \) is always a positive integer.

### Conclusion:
Thus, for any positive integer \( m \), \( g(m) \) is perfectly divisible by \( 2m \).