Divisors And Their Properties

I. Counting The Number Of Divisors

If we have to answer the question $\unicode{0x201C}How\ many\ factors\ are\ there\ for\ the\ number\ 56?\unicode{0x201D}$, how are we going find out the answer?

Many of us would go about trying to divide $56$ starting with $1$, $2$ and likewise go upto $56$ to find out all the factors,

and will eventually find out that $56$ is divisible by $1$, $2$, $4$, $7$, $8$, $14$, $28$ and of course the number $56$ itself.

Therefore there are $8$ possible factors of $56$, including $1$ and the number itself. Try doing it for a slightly larger number like $108$. It will take a lot of time, and you will find it quite tiring.

Luckily for us, there is an easy way to find the number of factors for any given number, without actually finding out all the factors one by one. Here are the steps:

$\rightarrow First\ find\ all\ the\ prime\ factors\ of\ the\ given\ number$

$\rightarrow Count\ the\ occurrences\ of\ each\ of\ these\ factors$

$\rightarrow Add\ 1\ to\ each\ of\ these\ occurrence\ values$

$\rightarrow Multiply\ the\ numbers\ obtained\ in\ the\ previous\ step$

$\rightarrow The\ resulting\ product\ is\ the\ number\ of\ different\ factors\ that\ the\ given\ number\ has$

We will understand this method better with a few examples.

-----------book page break-----------

II. Divisor Counting Examples

We will illustrate this method using our example of $56$.

Let's first factorise it into prime factors.

$56 = 2 \times 2 \times 2 \times 7$

Now we will count the number of times each factor occurs:

$2$ occurs $3$ times.

$7$ occurs $1$ time.

We will add $1$ to each of the occurrences and multiply them.

So, we get $(3 + 1) \times (1 + 1) = 4 \times 2 = 8$.

We already saw before that there are $8$ factors of $56$.

Let's take a little larger number and repeat the exercise. We will use the number $150$.

The prime factors of $150$ are:

$150 = 5 \times 5 \times 3 \times 2$

Count the occurrences of each factor:

$5$ occurs $2$ times

$3$ occurs $1$ time

$2$ occurs $1$ time.

Add $1$ to each occurrence and multiply them:

We get:

$(2 + 1) \times (1 + 1) \times (1 + 1)$

$= 3 \times 2 \times 2$

$= 12$.

So we have $12$ different factor for $150$. Let's check. The factors of $150$ are:

$1$, $2$, $3$, $5$, $6$, $10$, $15$, $25$, $30$, $50$, $75$, and $150$.

That's $12$ different factors.

-----------book page break-----------

III. Parity Of Count Of Divisors

Now that we know how to find the total number of factors without actually counting them, what else can we tell about them? For example, what will be the parity of the count of divisors of a given integer? When we say parity, we mean whether the number is odd or even. Let's see what we can find out about this.

Let's consider any number, let's say $28$. Let us try to write this number using pair of factors.

Like:

$28 = 1 \times 28$

$28 = 2 \times 14$

$28 = 4 \times 7$

We can see that in this case every divisor has a distinct pair. Therefore, the parity of the count of divisors will be even.

Now, let us consider the number $36$ which is a perfect square and carry out the same exercise.

$36 = 1 \times 36$

$36 = 2 \times 18$

$36 = 3 \times 12$

$36 = 4 \times 9$

$36 = 6 \times 6$

Here we can see that every divisor has a distinct pair, except for the divisor $6$, which, being the square root of $36$, has no distinct pair, but itself.

It is easy to see that this will be true for every perfect square and its square root.

Therefore, we can conclude that the count of distinct divisors of a perfect square is always odd and for all other integers, the count of divisors is even.