A milestone on a tortuous road..

It is funny how one problem gives rise to another. Of course this happens because of the irritating habit that mathematicians have of asking, “if it was something else and not something like this, what would be the solution?” And then you start all over again.

Here is a good problem I was discussing with a colleague-

In how many ways can you write the number 210 as a product of three integers?

210 = 2 × 3 × 5 × 7. Since the problem asks for integers, we will first calculate for positive numbers and then assign signs to these numbers.

Your first instinct is to write down 2, 3, 5 and 7 in a row and place two partitions between them. The partitions will give you three groups of numbers. Something like this:

But you should quickly realize that in this way no matter where you place your partitions, the number 2 and 7, 3 and 7, 2 and 5 will never be together. The partition method won’t work.

In fact, the situation is similar to placing 4 similar balls in three similar boxes. How would you do that? The hard way of course:

So we have to place 4 prime numbers in 3 places. If a place remains empty after distribution that means we will assume number 1 over there. Let’s find the number of ways of distributing these prime numbers. Since the box are all same only the different grouping of number matters.

As we are talking about integers, the numbers of possible cases are:

· All positive

· One positive and two negative (we will have to find different ways of assigning positive and negative signs also in this case)

There! 55 is the answer.

But do you stop at that? No, you ask another question:

So how do we do it?

The hard way again.

· Case I (positive, positive, positive)

· Case II (positive, negative, negative) e.g. (2, - 1, - 210)

The number of cases will be equal to the number of cases when all of them are positive. All we have to do is to write the integer triplets (x, y, z) of case I and then assign negative signs to y and z.

Therefore, total number of cases = 81.

· Case III (negative, positive, positive) e.g. ( - 2, 1, 210)

· Case IV (negative, negative, negative)

The number of ways in this case will again be equal to the number of ways in case III as we can assign negative sign to y and z to all the cases of case III. Therefore, the number of ways = 54

Therefore, the total number of ways = 81 + 81 + 54 + 54 = 270.

There. We have found answer to one more question. But what if it was…? The road is endless..