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

The correct option is D $3$

$240=(2{)}^4.3.5$ $,$ $4n+2$$=2(2n+1)$ $,$ $n>0$

$\therefore$ Divisors must be of the for $2\times \left(oddnumber\right)$.

Possible divisors are $2\times 3,2\times 3\times 5,2\times 5$

$\therefore$ $3$ such divisors.

Hence, the answer is $3.$