Revision Notes

In the entire following, \(f\) refers to the number factors of \(n\).

Case 1: When \(n\) is not a perfect square

Because \(n\) is not a perfect square, the case of the 2 numbers being same does not arise and the possible ways are only those where the two numbers are distinct. And since \(n\) is not a perfect square, \(f\) will be even.

i) Writing \(n\) as product of two unordered natural numbers = \(\frac{f}{2}\)

ii) Writing \(n\) as product of two ordered natural numbers = \(f\)

iii) Writing \(n\) as product of two un-ordered integers = \(f\)

iv) Writing \(n\) as product of two ordered integers = \(2 \times f\)

The above are the most common situations you can prepare for. But then the question can add some different conditions and you need to be able to manage them on a case-to-case basis. One such example is:

(v) If \(|a \times b |= n\) where \(a, b \in \mathbb{Z}\), find the number of different solutions to \(\left(a, b\right)\).

If \(|a| \times |b| = n\) where \(a, b \in \mathbb{Z}\), find the number of different solutions to \(\left(a, b\right)\).

Both the above questions are pretty much the same. And the answer to either of the question = \(4 \times f\)

And the case \(a, b \in \mathbb{N}\) in the above questions doesn’t really make any logic, because then the mods in the question have no role and the question will be exactly same as (ii)

Case 2: When \(n\) is a perfect square

Because \(n\) is a perfect square, now we will have 1 way of writing \(n\) as product of identical numbers. Also because \(n\) is a perfect square, \(f\) will be odd.

i) Writing \(n\) as product of two unordered natural numbers = \(\frac{f+1}{2}\)

1 way of these will be where both the numbers are identical. And hence number of ways of writing n as product of two distinct unordered natural numbers = \(\frac{f+1}{2}-1\) which is same as \(\frac{f-1}{2}\)

ii) Writing \(n\) as product of two ordered natural numbers = \(f\)

1 way of these will be where both the numbers are identical. And hence number of ways of writing n as product of two distinct ordered natural numbers = \(f-1\)

iii) Writing \(n\) as product of two un-ordered integers = \(f+1\)

2 ways of these will be where both the numbers are identical e.g. if \(n\) = 64, then 8 × 8 and (−8) × (−8). And hence number of ways of writing n as product of two distinct ordered natural numbers = \(f-1\)

iv) Writing \(n\) as product of two ordered integers = \(2 \times f\)

2 ways of these will be where both the numbers are identical (same example given above). And hence number of ways of writing n as product of two distinct ordered natural numbers = \(2 \times f\;-\;2\)

(v) While all variations cannot be exhausted, the one variation which we did earlier … If \(|a \times b |= n\) OR \(|a| \times |b| = n\) where \(a, b \in \mathbb{Z}\), find the number of different solutions to \(\left(a, b\right)\).

The answer is \(4 \times f\).

2 of these will be where the two numbers are identical and hence the answer, if \(a\) and \(b\) are distinct integers, will be \(4 \times f \;-\; 2\)

If you are wondering what other variations are possible, it could be writing \(n\) as a product of two even numbers; or one even and one odd number; or as product of two numbers, each greater than/less than a specific value. For all these cases, you will need to get into details and cannot just rely on the above formulae.