Divisors of $240$ of the form $2m+1$

Question

Just read a question where it was asked , ‘What is the number of divisors of $240$ that are of the form $4m+2$ ?’ According to the answer , the divisor must be of the from $2(2m+1)$ Hence the number of divisors is given by $1(1+1)(1+1) = 4$ But I was wondering , the $(1+1)$ also corresponds to the case where the odd divisors of $240$ i.e. $3$ and $5$ are not chosen at all simultaneously. How will I then get the odd divisor of the form $2m+1$ ?

score 3 · Answer 1 · answered Oct 15 '17 at 15:30

3

$240=2^3\cdot3\cdot5$. If you want only odd (positive) divisors, you can just take the divisors of $3\cdot5=15$ so you get the four divisors $1,3,5$ and $15$.

answered Oct 15 '17 at 15:30

Piquito

29,594

Thank you , I was just wondering whether the factor $1$ corresponds to that way of choosing the divisors of $15$ when we leave $3$ and $5$ completely ? – Aditi Oct 15 '17 at 15:42
You are welcome. – Piquito Oct 15 '17 at 15:44

score 1 · Answer 2 · answered Oct 15 '17 at 15:16

1

I assume you are looking for positive divsors. You are looking for odd divisors of $15=3\cdot 5$. There are $(1+1)(1+1)$ of them because you can either take the $3$ or not for the first $(1+1)$ and you can take the $5$ or not for the other one. If we list them there are $1,3,5,15$

answered Oct 15 '17 at 15:16

Ross Millikan

374,822

1

To add to the specific question asked: If we leave out 3 and 5, then we have the divisor 2, which is of the form $4m + 2$ where $m=0$. Hence, we can also leave out 3 and 5 completely, yielding $2m + 1 = 1$. – DWe1 Oct 15 '17 at 15:22
Ohh okay thank you I understand it now ! – Aditi Oct 15 '17 at 15:36

score 1 · Answer 3 · answered Oct 15 '17 at 15:22

Let $n\in\mathbb{N}$. We wish to find it's odd divisors. First, let $m$ be the smallest number such that $m|n$ and $\frac{n}{m}$ is a power of $2$. This number exists by Well-Ordering Principle because the set of numbers with this property is bounded below by $1$ and non-empty (since $n$ itself falls in this set). Put another way, $m$ is the number $n$ with all the factors of $2$ taken out. Notice that $m$ and $n$ have exactly the same odd factors.

Since $m$ is odd, all of the factors of $m$ are odd, so we just need to find all the factors of $m$. This is straightforward and is detailed here if you don't know how.

In your case, $240=2^4\cdot 15$ and so $m=15$. The factors of $15$ are $\{1,3,5,15\}$.

Divisors of $240$ of the form $2m+1$

3 Answers3