I was trying to figure out the number of positive divisors of large numbers and came across something interesting.

I'm not sure whether this is a theorem that already exists, it probably does and I've just never come across it.

So if you consider the prime factor $1 000 000$,

That's $10^6= (5\times2)^6 = 5^6 \times 2^6$

Now this is as simplified product of prime factors you can get for $1 000 000$.

I manually calculated the number of positive divisors there are for $1 000 000$ and I figured out that there were $49$ of them.

What I noticed however, was $49$ was the $(6+1) \times (6+1)$, 6 being the powers of the simplified product of prime factors.

Now I let this be to coincidence but I tried it for another number.

The prime factors of $315 000$

$315 \times 10^3 = 63 \times 5 \times 5^3 \times 2^3$ = $9 \times 7 \times 5^4 \times 2^3$

When you manually calculate the number of prime divisors, you obtain $120$, which is also the product of $(3+1)(2+1)(4+1)(1+1) = 120, 3,2,4$ and $1$ being the powers of the product of prime factors.

Now, does this apply to all prime factorisations and number of divisors or am I looking silly and just stating a well-known theorem or is this is just a massive coincidence?

This possibly involves a hint of combinatorics. Any explanation is appreciated!

  • $\begingroup$ Yes, this is a known result - multiply together one more than each distinct prime exponent for the number of factors. Good discovery work. $\endgroup$ – Joffan Jan 2 '17 at 21:18
  • $\begingroup$ Thanks! Is there a reason to why it works or does it just simply work? $\endgroup$ – TripleA Jan 2 '17 at 21:19

In order to generate each factor of a number $n = \prod p_i^{k_i}$, you need to select a power from each dividing prime to multiply in to that factor. For each dividing prime $p_i$ with exponent $k_i$, you have $k_i+1$ choices. These are all independent choices so the total number of factors is as you have found, $\prod (k_i+1)$.

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    $\begingroup$ Mmmm i see where the combinatorics comes in now, thanks a lot, but I seem like an idiot for asking ;) $\endgroup$ – TripleA Jan 2 '17 at 21:26
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    $\begingroup$ Not at all, it's great to find these things out independently. Bonus fact - this shows why only square numbers have an odd number of factors. $\endgroup$ – Joffan Jan 2 '17 at 21:27

This is known as one of the divisor functions, the one that sums the zeroth powers of the divisors. Taking your example of $315000$, the divisor can have anywhere from $0$ to $3$ powers of $2$, which is four choices, and so on for the other factors. It is a well known and useful result in number theory. If you search the site for divisor function you will find many questions that use it.


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