Question

The number of divisors of $240$ in the form $4n + 2$, $\left( {n \geqslant 0} \right)$ is equal to:
A) $4$
B) $8$
C) $10$
D) $3$

Answer 1

Hint:
We will factorize $240$ for simplifying it to be divisible by $4n+2$. Then, we will calculate all the possible divisors of 240 and after that we can write down the terms (factors of 240) for $n \geqslant 0$ in the form of $4n+2$. Hence, we can count the possible number of divisors which are of the form $4n + 2$.

Complete step by step solution:
We are required to calculate the total number of divisors of $240$ in the form of $4n + 2$ for $n \geqslant 0$.
On factoring $240$, we get
$ \Rightarrow 240 = 2 \times 2 \times 2 \times 2 \times 3 \times 5$
Or, we can write them as
$ \Rightarrow 240 = {2^4} \times 3 \times 5$
Now, looking at the divisors of $240$, we have
Divisors: $1,2,3,4,5,6,8,10,12,15,16,20,24,30,40,48,60,80,120{\text{ and }}240$
So, there are a total 20 divisors of $240$.
In the form of $4n + 2$, $n \geqslant 0$ , we can have terms as:
For $n = 0$, $4n + 2 = 2$
For $n = 1$, $4n + 2 = 6$
For $n = 2$, $4n + 2 = 10$
For $n = 3$, $4n + 2 = 14$
For $n = 4$, $4n + 2 = 18$
For $n = 5$, $4n + 2 = 22$
For $n = 6$, $4n + 2 = 26$
For $n = 7$, $4n + 2 = 30$
For $n = 8$, $4n + 2 = 34$ and so on.
We will have divisors of $240$: $2,6,10{\text{ and 30}}$ only in the form of $4n + 2$ for $n = 0,1,2{\text{ and 7}}$ respectively.
Therefore, a total of 4 divisors only of $240$ are of the form $4n + 2,n \geqslant 0$.

Hence, option (A) is correct.

Note:
In this question, you may go wrong in calculation of the divisors of $240$ in order to check if they are of the form $4n + 2$ for non – negative value of n. You can also see that if we put the value of n in $4n + 2$, we obtain an Arithmetic Progression with common difference $d = 4$ i.e., $2,6,10,14,18...$ is the obtained A.P. After this, we can write $4n + 2$ as $2\left( {2n + 1} \right)$ , therefore, only those divisors will be of this form which are both multiple of 2 and even as well since $2n + 1$ is representation of any odd number (2 included since $n = 0$ as well). Hence, only possible divisors of $240$ satisfying this condition are $2,6,10{\text{ and 30}}$.