I was playing around with numbers the other day, and I found an interesting symmetry, that I would like to know if it has any specific name assigned to it.
Let's assume the notation $n:a$ to refer to any number whose modulus $n = a$. That is, which if written with an $n$-radixed numeral system, would have a least significant digit of $a$. (If anyone knows the standard notation for expressing this, feel free to add that info.)
We can then write $10:3 \times 10:7 = 10:1$, because if we multiply any two integers with least significant decimals of $3$ and $7$ respectively, then the result will always have a least significant decimal of $1$, for example:
$$\begin{align} 33 \times 77 &= 2541\\ 103 \times 97 &= 9991\\ 73 \times 37 &= 2701 \end{align}$$
We can also write $\dfrac{10:3}{10:7} = 10:9$, because if we divide a number with least significant decimal of $3$, by a number with least significant decimal of $7$, then (given that it is evenly divisible), the result will always have a last decimal of $9$. For example:
$$\begin{align} \frac{7743}{87} &= 89\\ \frac{1083}{57} &= 19\\ \frac{3353}{7} &= 479 \end{align}$$
So far, nothing weird is going on. However, the strange thing starts to happen if we use certain radixes, for example 24 instead of 10, then
$$(24:17 \times 24:23) = \left(\frac{24:17}{24:23}\right) = \left(\frac{24:23}{24:17}\right) = 24:7$$
Let us simplify the above a bit, and write it as:
$$24:\left((17 \times 23) = \frac{17}{23} = \frac{23}{17} = 7\right)$$
Let us try to test the truthness of this. By creating the numbers $24:17$ and $24:23$, and multiplying them, we can test if the product $= 24:7$. If $24:\left(\frac{23}{17} = 7\right)$, then $24:((17 \times 7) = 23)$, and if $24:(\frac{17}{23} = 7)$, then $24:((7 \times 23) = 17)$. In order to test the above expression, let us therefore multiply the numbers:
- $24:17$ and $24:23$ (the result should be $24:7$)
- $24:23$ and $24:7$ (the result should be $24:17$)
- $24:17$ and $24:7$ (the result should be $24:23$)
For this demonstration I will only do the multiplication once for each of the three types, above, but of course one can try as many combinations of numbers with $24$-moduluses specified above that one feels is sufficient to convince oneself that it is always true.
A number that equals $24:17$, that is, which has a $24$-modulus of $17$, must have the form $17 + 24 \times x$, where $x$ is any arbitrary integer. We can use the same method to create arbitrary numbers with any specific modulus. For this test, let's let $x = 0$. Let us use these arbitrary numbers with moduluses $17$, $7$ and $23$:
$$\begin{align} 17 + 0 \times 24 &= 17\\ 7 + 0 \times 24 &= 7\\ 23 + 0 \times 24 &= 23 \end{align}$$
Then let us examine if there is a symmetry between the result of them being multiplied and divided by one another.
$$\begin{align} 17 \times 7 &= 119\\ \frac{119}{24} &= 4 + \frac{23}{24} \end{align}$$
The above shows that $24:\left(\frac{23}{17} = 7\right)$
$$\begin{align} 23 \times 7 &= 161\\ \frac{161}{24} &= 6 + \frac{17}{24} \end{align}$$
The above shows that $24:\left(\frac{17}{23} = 7\right)$
$$\begin{align} 17 \times 23 = 391\\ \frac{391}{24} = 16 + \frac7{24} \end{align}$$
The above shows that $24:((17 \times 23) = 7)$
You are just noticing some basic relations in the rings $\mathbb{Z}/n\mathbb{Z}$. You may not be familiar with it : it's just the formalization of your idea that for any numbers $a$ and $b$, the last digit in the $n$-adic expansion of $a+b$ and $ab$ depends only on the last digits of $a$ and $b$.
So you can think of $\mathbb{Z}/n\mathbb{Z}$ as the set $\{0,1,\dots,n-1\}$, and for any $a\in \mathbb{Z}$ you can write $[a]\in \mathbb{Z}/n\mathbb{Z}$ its residue modulo $n$. Then you get $[a+b]=[a]+[b]$ and $[ab]=[a][b]$.
Now in your example modulo $24$, you take $a=23$ and $b=17$. But $[a]=[-1]$, so quite naturally you get $[ab]=[-b]=[b/a]$ because $(-1)b=b/(-1)$. The fact that $[-b]=[a/b]=[-1/b]$ means that $[-b^2]=[-1]$, so $[b^2]=1$. And indeed, $[17^2]=[289]=[12\cdot 24+1]=[1]$.
So for any $n$, if you find $b\in \mathbb{Z}$ such that $[b^2]=[1]$ (which is always possible) and $a$ such that $[a]=[-1]$ (for instance, $a=n-1$), then you will always have $[ab]=[b/a]=[a/b]$.