Question

The total number of divisors of the form $4n+2(n\ge 0)$ of integer $240$ is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is B $4$

$240=2^4\times 3\times 5$
$4n+2=2(2n+1)=2\times$odd
$\therefore$ the required number of divisors
$=$ the number of selections of one $2$ from four ${}^{\prime }{2}^{\prime }s$, any number of ${}^{\prime }{3}^{\prime }s$ from one $3$ and any number of ${}^{\prime }{5}^{\prime }s$ from one $5$.
$=1\times (1+1)\times (1+1)=1\times 2\times 2=4$