This lesson will cover a few examples relating to number of divisors of a number.

**Example 1 **Let N = 10800. Find the number of divisors of N. How many of them are

(i) even? (ii) odd? (iii) multiples of 3? (iv) multiples of 15? (v) of the form 4k + 2?

**Solution **Here N = 2^{4}3^{3}5^{2}. To find the number of divisors, we’ll simply use the formula (4 + 1)(3 + 1)(2 + 1) =** 60**.

Now let’s move on to the complicated stuff.

(i) Remember that to form factors, we’re doing nothing but selections from a lot of four 2s, three 3s and two 5s, and multiplying them together.

To form an even factor, we must select atleast one ‘2’ from the lot, which will ensure that whatever be the remaining selection, their multiplication will always result in an even factor.

The number of ways to select atleast one ‘2’ from a lot of four identical ‘2’s will be 4 (i.e. select 1 or select 2 or select 3 or select 4. Refer to the previous examples for details.)

And, we’ll select any number of ‘3’s and ‘5’s, in 4 and 3 ways respectively.

The required number of ways will therefore be 4 x 4 x 3 = **48**.

Now the next parts should be easy.

(ii) To count the odd factors, we’ll get rid of the ‘2’s. We’ll make the selection from the ‘3’s and the ‘5’s only. The number of selections (or factors) will therefore be (3 + 1)(2 + 1) = **12**.

Note that this could also have been obtained by subtracting the even factors from the total, i.e. 60 – 48, which will give the same answer.

(iii) This one is similar to (i). For factors to be multiples of 3, we must select at least one ‘3’ (3 ways), and any number of ‘2’s and ‘5’s (5 and 3 ways respectively). The required number will be 5 x 3 x 3 = **45**.

(iv) What about this one?

For factors to be multiples of 15, we must select atleast one ‘3’ (3 ways) and one ‘5’ (2 ways), and any number of ‘2’s (5 ways). The required number is 5 x 3 x 2 = **30**.

(v) Notice that 4k + 2 = 2(2k + 1), which looks like a ‘2’ multiplied by an odd number. That means, we need to select **exactly one ‘2’**, and any number of 3s and 5s (as they’ll give an odd number on multiplication).

To select exactly one ‘2’, there is only 1 way. And the ‘3’s and the ‘5’s can be selected in 4 and 3 ways respectively. Therefore the required number of factors = 1 x 4 x 3 = **12**.

That’ll be all for this lesson. As an exercise, try finding out the number of factors of N which are (a) perfect squares, and (b) of the form 4k + 1. (not so easy)

The next lesson covers a new topic – division of objects into groups.