The number of divisors of 240 in the form 4n + 2, (n ≥ 0) is equal to:

• A) 4
• B) 8
• C) 10
• D) 3

To find the number of divisors of 240 that are in the form of 4n + 2, we first need to determine the prime factorization of 240.

Prime Factorization of 240

The prime factorization of 240 is:

• 240 = 2^4 × 3^1 × 5^1

Identifying Divisors

Divisors of 240 can be expressed in the form:

• 2^a × 3^b × 5^c

where:

• 0 ≤ a ≤ 4
• 0 ≤ b ≤ 1
• 0 ≤ c ≤ 1

Form of 4n + 2

Divisors in the form of 4n + 2 must be even and not divisible by 4. This means:

• The exponent of 2 (a) must be exactly 1 (since 2^1 = 2, which is of the form 4n + 2).

Calculating Valid Combinations

Now, we can choose values for b and c:

• For b (the exponent of 3), we can have 0 or 1 (2 choices).
• For c (the exponent of 5), we can also have 0 or 1 (2 choices).

Total Combinations

The total number of divisors of the form 4n + 2 is:

• 1 (for a = 1) × 2 (for b) × 2 (for c) = 4

Thus, the number of divisors of 240 in the form 4n + 2 is 4.

The correct answer is A) 4.