Reasons for Post:

- To introduce factors and primes through a geometric approach.
- To introduce an algorithm to quickly find how many factors a number has.
- To discuss why the algorithm works based on combinations of factors.

Recently I gave my Year 8 (Grade 7) students a mini-investigation based on factors and primes – without saying it was about factors and primes.

*“For this investigation, consider a square to be a special type of rectangle. If the number 10 represents the area of a rectangle, how many different rectangles can you make with a whole number length and width? Investigate this for different numbers.”*

Before discussing two examples as a class, I told them that if anyone figured out what this investigation was actually about, then they should keep it to themselves. Of course some realised quite quickly that the width and length of a rectangle or square represented a factor pair of the area. Those students generally went on to look at investigating how many factors each number has and whether there is a way to figure this out. I didn’t expect anyone to get far with this.

I was genuinely surprised when two of the students working together came up with the standard algorithm for finding the number of factors using the indices within the product of prime factors.

For example, to find how many factors the number 30 has.

Step 1: Break 30 into a product of primes: 30 = 2¹ x 3¹ x 5¹

Step 2: Add one to each of the indices and then multiply these numbers together.

(1 + 1) × (1 + 1) × (1 + 1) = 2 × 2 × 2 = 8 factors

Of course, by finding the product of prime factors, you’re essentially finding the fundamental building block of each factor. Still, quite impressed that they’d found an algorithm. I personally don’t know any other way of doing it. Anyone else know a different way?

I told them that as a mathematician – our aim is to understand why an algorithm works but that might be something they’d be able to investigate in a few years once they’d learnt more mathematics. In this case that boils down to why we add one to each of the indicies and then multiply? I spent part of break with a coffee making sure I understood why the algorithm works (on a similar, but less rigorous line to Erdos; “A mathematician is a machine for changing coffee into theorems”). Not hugely difficult to make sense of/justify for numbers with only one base prime factor (e.g 8 = 2³; 2 is the base prime factor) or two base prime factors (e.g 15 = 5 x 3; 5 and 3 are the base prime factors) and then intuitively extend from there.

With one base prime factor, e.g. 8 = 2³. The factors are 2, 2², 2³ and then 1 (you could think of this as 2° if you fancied it). Hence why there are 3 + 1 = 4 factors. So for N=a^x, there will be x + 1 factors in total.

With two base prime factors, take the number 100 = 2² × 5²

By the algorithm, we know that this number has (2 + 1)(2 + 1) = 9 factors.

If you expand out the brackets (2 + 1)(2 + 1) = 2×2+1×2+2×1+1×1 = 4 + 2 + 2 + 1 (*) then we can see that these 9 factors are split up into groups. i.e. 4 factors + 2 factors + 2 factors + 1 factor.

Expanding out any bracket of the form (a + 1)(b + 1)(c + 1)… will always give you a last term which is just a load of one’s multiplied together. So the last 1 in the calculation above represents the fact that 1 is a factor of every single number.

So where do the 4 factors + 2 factors + 2 factors come from in equation (*) above. I think the geometrical representation below sheds the most light on this.

It’s clear to see that the 4 factors are made up of all of the combinations of the two base prime factors. The (2 factors + 2 factors) can be seen in the picture as factors which are only made up from one of the base prime factors.

So for any number with 2 base prime factors, we have:

Of course if each exponent is equal (i.e. x = y), then we have the second binomial expansion (x + 1)². With three base prime factors all being equal (x = y = z), then you have the third binomial expansion (x + 1)³, and so on. There’s the connection to Pascal’s triangle and hence the lead into the different combinations of all of the factors.

So an interesting change in my approach to factors and primes. I’m still not sure whether I’d teach the algorithm above – wouldn’t be too hard to add in some understanding of this but then usually I’d teach factors and primes in Year 7 (Grade 6) which would be before looking at breaking numbers into a product of prime factors. It’s therefore difficult to know where this would all be placed in the curriculum. Possibly worthwhile as a closed investigation in Year 9 as a follow on from expanding brackets.

Note: Bear in mind that for the above example, we couldn’t use the calculation 4C2 (4 things choose 2 of them) for the 4 factors part because you can’t put 2 and 2² together or 5 and 5² together. You have to see the base prime factors as separate groups (2, 2²) and (5, 5²) and pick an element from one group to pair with an element from another group.

Hi Dan,

Can’t you sum the powers of the prime factorization and the raise 2 to that power? For each factor of the number, it either contains each prime factor or doesn’t. 1 is the factor that does not include any of the prime factors.

Sorry bout that. Home sick catching up on blog reading with a fever and neglected to consider that one must consider that some of the prime factors are identical and that leads to overcounting of total some of the prime factors. Yep, that makes it a bit nastier.