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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.

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

$t$ 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 $(t=3r,r)$.

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    $\begingroup$ 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. $\endgroup$ Oct 14 at 0:08
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    $\begingroup$ @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 $\endgroup$
    – TCooper
    Oct 14 at 14:05
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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

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  • $\begingroup$ @Anon correct, I misread it the first go around, apologies TCooper! $\endgroup$ Oct 13 at 23:34
  • $\begingroup$ 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 $\endgroup$
    – Anon
    Oct 13 at 23:37
  • $\begingroup$ @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 $\endgroup$
    – TCooper
    Oct 14 at 13:58
  • $\begingroup$ @Tacoタコス No worries, just shows why we need Anon's answer to clarify ;) $\endgroup$
    – TCooper
    Oct 14 at 13:59

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