Product of Factors of an Integer

The product of (positive integer) factors of a positive integer $n$ is denoted by $\pi(n)$. It is closely connected to the number of divisors function $d(n)$.

(Note: this $\pi(n)$ is not to be confused with the prime counting function.)

There is a very simple formula for $\pi(n)$, which can be derived by grouping the divisors elegantly. Try to figure out how. If you have no idea, an example, like $n=36$ may help.

The trick is to notice that the divisors of $n$ can be arranged into pairs which multiply to $n$. If the divisors of $n$ are $1=a_1<a_2<...<a_{d(n)}=n$, and the pair of $a_i$ is $b_i=\frac{n}{a_i}$, then since $n=b_1>b_2>...>b_{d(n)}=1$ also contains each divisor of $n$ exactly once, we may write

$$a_1a_2\dots a_{d(n)}=b_1b_2\dots b_{d(n)}=\pi(n).$$

However, as $a_ib_i=n$ for each $i$, we conclude that

$$a_1b_1\cdot a_2b_2\cdot \dots \cdot a_{d(n)}b_{d(n)}=n^{d(n)}.$$

Recognizing $\pi(n)^2$ on the left-hand side of this last equation and taking square roots, we arrive at

$$\pi(n)=\sqrt{n}^{d(n)}.$$

Indeed, this is not multiplicative. For example, $\pi(4)=8$ and $\pi(3)=3$, but $\pi(12)=\sqrt{12}^6\neq 24=\pi(4)\pi(3)$.

One interesting result when looking at the factors of an integer is to find the product of the factors. Lets consider the product of the factors of 30. First, lets look at the factor pairs of 30: $$1\cdot 30=30$$ $$2\cdot15=30$$ $$3\cdot10=30$$ $$5\cdot6=30$$ By definition, each of these factor pairs has a product of 30 as shown. So the product of the factors is $$(1\cdot30)(2\cdot15)(3\cdot10)(5\cdot6)$$ However, as shown above, each of these factor pairs has a product of 30 so we can substitute 30 for each product. $$(30)(30)(30)(30) = 30^4=810000$$ Since the results of these types of problems is usually so large then we often leave the answer in exponential form. Try a couple on your own.

However, listing out factor pairs is no efficient for big numbers. So instead, we utilize another important skill when working with numbers, finding the number of factors . Thus we look at a big number like 648. The prime factorization of 648 is $648 = 2^3\cdot3^4$. Using the algorithm about number of factors we know that 648 has $(3+1)\cdot(4+1)=20$ factors. As we have shown above, the factors can be arranged into pairs that have a product of 648. So 20 factors are arranged into 10 factor pairs. Each of these are multiplied together to get $648^{10}$.

Special Case: Perfect Squares

Consider the easy case of 9. The factors pairs of 9 are $1\cdot9$ and $3\cdot3$. However, the second pair has 3 twice, but we only count it as a factor once. Thus the answer is $9\cdot3$. To further illustrate this, we look at 36. The factor pairs of 36 are $(1\cdot36), (2\cdot18), (3\cdot12), (4\cdot9),$ and $(6\cdot6)$. So, the product of the first four factor pairs is $36^4$ as we illustrated above. But we must include one of the 6's, thus the actual answer is the product of the factors of 36 is $(36^4)\cdot6$. There are actually two other ways to write this. We can recognize that $6=\sqrt{36}=36^\frac{1}{2}$. Thus, $(36^4)\cdot6=(36^4)\cdot(36^\frac{1}{2})=36^{4\frac{1}{2}}$. The other way to write the answer is to recognize that 36 is a perfect square so $(36)^4=(6^2)^4=6^8$. Thus our answer would be $(36^4)\cdot6=(6^8)\cdot6=6^9$.