Question

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

• A. $4$
• B. $8$
• C. $10$
• D. $3$

Answer 1

The correct option is A $4$

We can write $240$ as $2^4{.3}^1{.5}^1$
But we want divisors of the form $4n+2$

That is, we want even divisors of the form $2(2n+1)$ and $n⩾0$

The divisor is $2$ (when $n=0$) or divisors are odd multiples of $2$

So the divisors are $2$, $6$, $10$ and $30$
Hence the answer is $4$