Multiplying prime numbers

If I multiply $13$ and $17$ to get $221$ I can only get $221$ by multiplying $13$ and $17$ (excluding $1$ and $221$) does the same rule apply to multiplying $3$ numbers? (excluding the use of $1$)

Solution 1:

If you mean multiplying 3 prime numbers, then yes there is only one way (this is known as the fundamental theorem of arithmetic).

For instance $715=5\cdot 11\cdot 13$, and only these three primes will produce it (if you exclude $715=(-5)\cdot (-11)\cdot 13$ and so on)

Solution 2:

Take any composite number (non-prime number) $n$ where $1 < n < k$ such that $$n = \prod_{i=1}^k \alpha_i$$ for which $\alpha_i$ is a prime number. Since $\alpha_i$ is prime, it follows that $$\exists!\{1, \alpha_i\}\mid \alpha_i$$ Therefore, by letting $N^* = \{$smallest numbers $> 1$ that divide $n$$\}$, the only elements of $N^*$ will be $\alpha_1 ,\alpha_2, \ldots, \alpha_k$. This means that if you multiply any $3$ numbers to produce a number $n$, these three numbers will be the only numbers to divide $n$ on the condition that these three numbers are prime, since there will exist no smaller number than each of these three numbers that divides $n$ apart from $1$. This statement yields the fact that every composite number $n$ has a unique prime factorisation (or decomposition) as opposed to every other composite number, which is described to be The Fundamental Theorem of Arithmetic, along with how $0$ and $1$ are neither prime nor composite. This is also derived from propositions made by Euclid in his Elements, Book VII-IX.

Let $f(X)$ be a function that describes the cardinal number of a given set $X$, then $f(N^*) \in (0, \infty)$ such that $f(N^*) = 1$ implies that $n$ is a prime number and $\alpha_1 = n$. The cardinal function is typically denoted as $n(X)$, but since we already have a value $n$, I let this cardinal function be $f$ instead to avoid confusion. There are also similar notations to express $f(N^*)$ like $\text{card}(N^*)$ and etc, but that is irrelevant to the subject.

Solution 3:

So you have two distinct positive primes, let's call them $p$ and $q$. Then their product has precisely four divisors among the positive integers: 1, $p$, $q$ and $pq$ itself, and we verify that $1 \times pq = pq$.

In your example with $p = 13$ and $q = 17$ (or $p = 17$ and $q = 13$, if you prefer), we verify that $1 \times 221 = 13 \times 17$.

What if we throw a third distinct positive prime, $r$, into the mix? Then $pqr$ has eight divisors: 1, $p$, $q$, $r$, $pq$, $pr$, $qr$ and $pqr$ itself.

To expand on the previous example, let's say $r = 29$. Then we see that 6409 can be expressed as a product of two positive integers four different ways: $$1 \times 6409 = 13 \times 493 = 17 \times 377 = 29 \times 221$$ But none of those ways consist of two primes, because the number is the product of three primes.