Fractions of the form $\frac{a}{k}\cdot\frac{b}{k}\cdot\frac{c}{k}\cdots=\frac{n}{k}$

Question

For any $n,k\in\mathbb{Z}^+$, $n\gt k$. I found that the majority of the time, either there are no positive integers, $a,b,c\dots,$ such that

$$\frac{a}{k}\cdot\frac{b}{k}\cdot\frac{c}{k}\cdots=\frac{n}{k}$$

and $k \lt a, b, c,... \lt n$

(such $n/k$ are prime in a sense)

or if we can find positive integers that satisfy the above constraints, then it is possible to express $n/k$ as the product of at least two fractions of the form shown below.

$$\frac{a}{k}\cdot\frac{b}{k}=\frac{n}{k}$$

An example is $$\frac{5}{2}\cdot\frac{6}{2}=\frac{15}{2}$$

It appears that most "composite" $n/k$ can be expressed in this form.

However I found that there are rare cases of "composite" $n/k$ that cannot be expressed as the product of two such fractions. The only examples I've found are $$\frac{44}{42}\cdot\frac{45}{42}\cdot\frac{49}{42}=\frac{55}{42}$$ and $$\frac{45}{42}\cdot\frac{49}{42}\cdot\frac{52}{42}=\frac{65}{42}$$

What makes these type of fractions so much rarer? Can we find more examples that are the product of at least three such fractions?

Or can we find examples that are the product of at least four such fractions.

Answer 1

It appears that you are using the concept of numbers being relatively prime without actually saying it, so I'll introduce the terminology here.

For any $a,b \in \mathbb{Z}$, let $\gcd(a,b)$ denote the greatest common divisor between $a$ and $b$. We say that $a$ and $b$ are relatively prime if $\gcd(a,b) = 1$; that is, $a$ and $b$ are relatively prime if they have no factors in common other than $1$, which is equivalent to saying that the fraction $\frac{a}{b}$ cannot be reduced.

Let's consider $$\frac{a_1}{k} \cdot \frac{a_2}{k} \cdot \dots \cdot \frac{a_r}{k} = \frac{n}{k} $$ where $k < a_1, a_2, \dots, a_r < n$ and $\gcd(n,k) = 1$. This is equivalent to $$a_1 \cdot a_2 \cdot \dots \cdot a_r = nk^{r - 1} $$

Essentially, then, the $a_1, a_2, \dots, a_r$ can each be thought of as a regrouping of the prime factorization of $nk^{r - 1}$.

For example, you had $n = 15, k = 2,$ and $r = 2$. Then $$nk^{r - 1} = (15)(2^{2 - 1}) = 30 = 2 \cdot 3 \cdot 5$$ We have 3 prime factors, but notice that one of them is $k = 2$. Thus, one of the numbers will need to consist of $2$ and another prime factor so that it satisfies $a > k$, and the other will need to be the prime factor that's left out. That is, the possible choices are: $$\frac{6}{2} \cdot \frac{5}{2} = \frac{10}{2} \cdot \frac{3}{2} = \frac{15}{2} $$

Notice that the key to being able to find the right number of combinations here is that $nk^{r - 1}$ needs to have many prime factors. Given enough prime factors that multiply to a number greater than $k$, you'll be able to get the desired form.

Your issue with cases for which $n/k$ is reducible is no different. Suppose that you want a string of $r = 4$ fractions, and you have $n = 60$ and $k = 12$. Then $nk^{r - 1} = (60)(12^{4 - 1}) = 103680 = 2^8 \cdot 3^4 \cdot 5$. We can group these prime factors as $(2^4), (2^3)(3), (2)(3^2), (3)(5) > 12$. Then, $$\frac{16}{12} \cdot \frac{24}{12} \cdot \frac{18}{12} \cdot \frac{15}{12} = \frac{60}{12} $$ As you can see, the larger that we want $r$ to be, the greater that $n$ must be relative to $k$, and the more prime factors $nk^{r - 1}$ needs to have.

Answer 2

When $n$ is not much more than $k$ you have less flexibility in choosing the fractions to multiply, so it becomes harder to find a solution. You need to cancel out all the extra factors of $k$ in the denominator on the left, spreading those factors over the numerators on the left. Suppose we give $k$ four prime factors that aren't so different, like $k=23\cdot 29 \cdot 31 \cdot 37=765049$ Now we will give $a,b,c,d$ each three of the factors and round up the fourth a bit, getting $$\frac {24\cdot 29 \cdot 31 \cdot 37}{23\cdot 29 \cdot 31 \cdot 37}\cdot\frac {23\cdot 30 \cdot 31 \cdot 37}{23\cdot 29 \cdot 31 \cdot 37}\cdot\frac {23\cdot 29 \cdot 32 \cdot 37}{23\cdot 29 \cdot 31 \cdot 37}\cdot\frac {23\cdot 29 \cdot 31 \cdot 40}{23\cdot 29 \cdot 31 \cdot 37}=\frac {24\cdot 30 \cdot 32 \cdot 40}{23\cdot 29 \cdot 31 \cdot 37}=\frac {921600}{765049}$$ You can clearly find solutions like this which require as many fractions as you like, though the numbers will get large. You can also just make $k$ a prime power, so $$\frac {3^5\cdot 23}{3^6}\cdot\frac {3^5\cdot 24}{3^6}\cdot\frac {3^5\cdot 25}{3^6}\cdot\frac {3^5\cdot 26}{3^6}\cdot\frac {3^5\cdot 27}{3^6}\cdot\frac {3^5\cdot 28}{3^6}=\frac {\frac{28!}{22!}}{3^6}=\frac{271252800}{729}$$ but you may be able to shuffle the factors in the numerators here.