The number of divisors of the form `4K + 2 , K ge 0` of the integers 240 is

Options

A. 4
B. 8
C. 10
D. 3

Answer

To find the number of divisors of the integer 240 that are of the form \(4k + 2\) (where \(k \geq 0\)), we can follow these steps:

### Step 1: Factorize the number 240
First, we need to factorize 240 into its prime factors.

\[ 240 = 2^4 \times 3^1 \times 5^1 \]

### Step 2: Identify the form of divisors
Divisors of the form \(4k + 2\) can be expressed as \(2 \times \text{(odd number)}\). This means that any divisor of this form must include exactly one factor of 2 (to ensure it is even) and the remaining part must be odd.

### Step 3: Determine the odd factors
To find the odd factors of 240, we can ignore the factor of \(2^4\) and only consider the remaining factors, which are \(3^1\) and \(5^1\).

The odd part of 240 is:

\[ 3^1 \times 5^1 \]

### Step 4: Count the odd factors
The number of odd factors can be calculated using the formula for the number of factors. If a number is expressed as \(p_1^{e_1} \times p_2^{e_2} \times \ldots \times p_n^{e_n}\), the number of factors is given by \((e_1 + 1)(e_2 + 1) \ldots (e_n + 1)\).

For \(3^1 \times 5^1\):

\[ (1 + 1)(1 + 1) = 2 \times 2 = 4 \]

The odd factors of 240 are \(1, 3, 5, 15\).

### Step 5: Form the divisors of the form \(4k + 2\)
Now, we multiply each of the odd factors by 2 to find the divisors of the form \(4k + 2\):

- \(1 \times 2 = 2\)
- \(3 \times 2 = 6\)
- \(5 \times 2 = 10\)
- \(15 \times 2 = 30\)

So, the divisors of 240 that are of the form \(4k + 2\) are \(2, 6, 10, 30\).

### Step 6: Count the divisors
We have found four divisors that fit the criteria:

- \(2\)
- \(6\)
- \(10\)
- \(30\)

Thus, the total number of divisors of the form \(4k + 2\) of the integer 240 is **4**.

### Final Answer
The number of divisors of the form \(4k + 2\) of the integer 240 is **4**.

---