Question

Number of divisors of the form 4n+2(n≥0) of the integer 240 is

• A. 4
• B. 8
• C. 10
• D. 3

Answer 1

Answer: (A)

4

Answer 2

240 = 2⁴ x 3 x 5

Since, 4n + 2= (2n+1), required divisor must be of the form 2 x an odd integer.

Here number of ways in which we can choose odd positive integer.

= (1+1)(1+1) = 4 (including I)

∴ Required number = 4 ($\because$we can take only one two out of 4 twos)