A dance between numbers

Given the following equation:

$$\Delta = \Biggl(\frac{t \mod 6}{(t \mod r) + 1}\Biggr)^2$$

Find the relationship between $$t$$ and $$r$$, along with the properties of each, that ensures $$\Delta = 9$$. For clarity, the correct answer is not looking for values that ensure $$\Delta = 9$$, but rather an explanation on how one could determine if a given value for $$r$$ is valid with respect to a given value for $$t$$ and vice-versa.

This puzzle focuses on the relationship used to build my previous puzzle. I felt as though it was nice enough to be shared since it wasn't exposed with the original puzzle.

• Wow, thanks for the bounty - there was no need to give that! My answer wasn't that comprehensive... Thank you anyway though
– Anon
Oct 19 at 2:05

$$t$$ is any number of the form $$3(2k+1)$$ for integer $$k$$, i.e. $$t\in\{3,9,15,...\}$$, and $$r$$ is any number that divides $$t$$

See this as follows (assuming $$t$$ and $$r$$ are positive integers):

$$\Delta=9$$ implies: $$\Biggl(\frac{t \mod 6}{(t \mod r) + 1}\Biggr)^2=9$$ $$\therefore \Biggl(\frac{t \mod 6}{(t \mod r) + 1}\Biggr)=3$$ $$\therefore t \mod 6=3(t \mod r)+3$$ Now write $$t=6k+j$$, for $$j\in\{0,1,2,3,4,5\}$$ and integer $$k$$, and similarly $$t=rm+n$$ for $$n\in\{0,1,2,3,...,m-1\}$$ and integer $$m$$

Therefore we have: $$j=3n+3$$ Now the only possible solution of this is $$n=0$$ ($$n=1$$ or greater would imply $$j>5$$ which is impossible by construction), therefore we have: $$j=3$$ Therefore: $$t=6k+3=3(2k+1)$$ and $$t=rm$$ for integer $$m$$ which is what we wanted to show.

Thus the first few options for $$t$$ and $$r$$ are as follows:

$$t=3$$ and $$r\in\{1,3\}$$

$$t=9$$ and $$r\in\{1,3,9\}$$

$$t=15$$ and $$r\in\{1,3,5,15\}$$

$$t=21$$ and $$r\in\{1,3,7,21\}$$

$$t=27$$ and $$r\in\{1,3,9,27\}$$

$$t=33$$ and $$r\in\{1,3,11,33\}$$

$$t=39$$ and $$r\in\{1,3,13,39\}$$

$$t=45$$ and $$r\in\{1,3,5,9,15,45\}$$
etc.

So, finally to explicitly answer the question:

The relationship between $$t$$ and $$r$$ is that $$t$$ is a multiple of $$r$$.

If given any $$t$$, if and only if it can be written as $$t=6k+3$$ can it be a part of a $$(t,r)$$ pair, in which case any $$r$$ that divides it will work.

If given any $$r$$, it can be part of a $$(t,r)$$ pair if and only if it is odd
- If it is even it cannot be a divisor of the odd number $$3(2k+1)$$
- If it is odd, then a possible $$(t,r)$$ pair is $$(t=3r,r)$$.

• I love this answer overall; but I especially love the fact that you demonstrated the relationship using the first few values of $t$. It clearly shows that ROT13(gur bayl inyvq inyhrf sbe e, tvira n inyhr sbe g ner gur snpgbef bs g). Based on timing, I must give the check to TCooper's answer, but I intend to award a bounty to yours due to the effort you've given and details you've included. Oct 14 at 0:08
• @Tacoタコス I think it more correct enough to deserve the check > a few minutes difference in answer time - and mine has technical inaccuracies in the wording Oct 14 at 14:05

Not sure how to show my work/explain this, it's mostly intuitive, but

delta = 9 where t mod 6 = 3 and t mod r = 0

So we know t must be a multiple of 3 but not 6 and a multiple of r

so I guess I can best simplify it as t=nr where t mod 6 = 3 and n >= 1

In this answer I'm implying the requirements of r based on its relation to t - Anon's answer is a far superior explanation

• @Anon correct, I misread it the first go around, apologies TCooper! Oct 13 at 23:34
• I didn't see @TCooper's answer pop up while I was writing mine. Nevertheless I'll leave mine up (for the time being at least) as it goes into more detail on the derivation, and how to determine which pairs $(t,r)$ work
– Anon
Oct 13 at 23:37
• @Anon please do leave yours up - it's a much better explanation. I feel like half (or more of) this puzzle was breaking it down clearly rather than the "challenge" of the equation Oct 14 at 13:58
• @Tacoタコス No worries, just shows why we need Anon's answer to clarify ;) Oct 14 at 13:59