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A milestone on a tortuous road..
by Total Gadha - Saturday, 17 March 2007, 10:35 AM

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 Ã— 3 Ã— 5 Ã— 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:

partition method

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)

distribution of prime factors

There! 55 is the answer.

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

number theory question

So how do we do it?

The hard way again.

· Case I (positive, positive, positive)

distribution of numbers

· 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)

number theory

distribution of numbers

· 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..

Re: A milestone on a tortuous road..
by Sanju P - Friday, 22 June 2007, 12:43 PM


Awesome article! I just loved reading that! big grin


S P smile

Re: A milestone on a tortuous road..
by Vimalesh Kobla - Thursday, 16 August 2007, 09:26 PM
  Hi TG,
   what is that prime factor? can you explain me again?
Re: A milestone on a tortuous road..
by Sodium Hydro Phosphate - Tuesday, 21 August 2007, 10:22 PM
  Badhiya hai TG. Tortuous road pe chalne se kuch reluctancy aa gayi hai ab to. I would have to overhaul my systems again. Is baar 'A' nikaalna hai smile
Re: A milestone on a tortuous road..
by vikas choudhary - Wednesday, 19 September 2007, 01:44 PM
  Awesome article TG. Keep it up.....
Re: A milestone on a tortuous road..
by ramya n - Thursday, 20 September 2007, 10:49 AM

hi... it may sound stupid bt i really don understand this

if xyz = 210 and x = 2^a then, 2^(ayz) = 210.


Then how wil ayz = 210 be possible?

pls clarify

Re: A milestone on a tortuous road..
by RIDHIMA CHOPRA - Saturday, 27 October 2007, 11:09 PM

tg sir plz can u explain this wid some simpler method

i have not understood anythng

Re: A milestone on a tortuous road..
by Ashish Sharma - Monday, 19 October 2009, 04:10 PM
  hmmm... suppose we want to find no. of ways we can write 210 as a product of three natural no.s...
according to your method it should be 14...but m getting 15 as the answer...
four prime factors: 2,3,5,7

now let x has "a" prime factors of 2,3,5,7 ; similarly y has "b" and z has "c"....hence
the non-negative solutions of these equation will be 6C2=15....

please tell where m mistaken.... which extra case i m counting sad
Re: A milestone on a tortuous road..
by Total Gadha - Wednesday, 21 October 2009, 09:50 PM

Before applying a concept learn a concept completely. Half-cooked learning is dangerous. In this equation (0, 0, 4), (4, 0, 0) and (0, 4, 0) are different solution but they are actually same for the problem- 210 × 1 × 1. Similarly, (1, 1, 2), (2, 1, 1) etc.

Total Gadha
Re: A milestone on a tortuous road..
by Ashish Sharma - Tuesday, 3 November 2009, 10:30 PM
  thanks TG. smile specially for this post and the other one on grouping and arrangments...cuz i always try to use the balls and walls problem in arrangements... Thanks again 
Re: A milestone on a tortuous road..
by spunky arora - Wednesday, 22 June 2011, 01:37 PM

y in case of no. of cases wen all positive u multiply it by 3!

i dint get it

pls explain sir