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title: Digit Sum Patterns and Modular Arithmetic title: Digit Sum Patterns and Modular Arithmetic
date: 2021-12-30T15:42:40-08:00 date: 2021-12-30T15:42:40-08:00
tags: ["Ruby", "Mathematics"] tags: ["Ruby", "Mathematics"]
draft: true description: "In this article, we explore the patterns created by remainders from division."
--- ---
When I was in elementary school, our class was briefly visited by our school's headmaster. When I was in elementary school, our class was briefly visited by our school's headmaster.
@ -79,39 +79,39 @@ divisor (what a mouthful). In mathematics, we typically use
\\(a|b\\) to say \\(a\\) divides \\(b\\), or, as we have seen above, that the remainder of dividing \\(a|b\\) to say \\(a\\) divides \\(b\\), or, as we have seen above, that the remainder of dividing
\\(b\\) by \\(a\\) is zero. \\(b\\) by \\(a\\) is zero.
Working with remainders actually comes up pretty commonly in discrete math. A well-known Working with remainders actually comes up pretty frequently in discrete math. A well-known
example I'm aware of is the [RSA algorithm](https://en.wikipedia.org/wiki/RSA_(cryptosystem)), example I'm aware of is the [RSA algorithm](https://en.wikipedia.org/wiki/RSA_(cryptosystem)),
which works with remainders resulting from dividing by a product of two large prime numbers. which works with remainders resulting from dividing by a product of two large prime numbers.
But what's a good way to write, in numbers and symbols, the claim that "\\(a\\) divides \\(b\\) But what's a good way to write, in numbers and symbols, the claim that "\\(a\\) divides \\(b\\)
with remainder \\(r\\)"? Well, we know that dividing yields a quotient (possibly zero) and a remainder with remainder \\(r\\)"? Well, we know that dividing yields a quotient (possibly zero) and a remainder
(also possibly zero). Let's call the quotient \\(k\\). (also possibly zero). Let's call the quotient \\(q\\).
{{< sidenote "right" "r-less-note" "Then, we know that when dividing \(b\) by \(a\) we have:" >}} {{< sidenote "right" "r-less-note" "Then, we know that when dividing \(b\) by \(a\) we have:" >}}
It's important to point out that for the equation in question to represent division It's important to point out that for the equation in question to represent division
with quotient \(k\) and remainder \(r\), it must be that \(r\) is less than \(a\). with quotient \(q\) and remainder \(r\), it must be that \(r\) is less than \(a\).
Otherwise, you could write \(r = s + a\) for some \(s\), and end up with Otherwise, you could write \(r = s + a\) for some \(s\), and end up with
{{< latex >}} {{< latex >}}
\begin{aligned} \begin{aligned}
& b = ka + r \\ & b = qa + r \\
\Rightarrow\ & b = ka + (s + a) \\ \Rightarrow\ & b = qa + (s + a) \\
\Rightarrow\ & b = (k+1)a + s \Rightarrow\ & b = (q+1)a + s
\end{aligned} \end{aligned}
{{< /latex >}} {{< /latex >}}
In plain English, if \(r\) is bigger than \(a\) after you've divided, you haven't In plain English, if \(r\) is bigger than \(a\) after you've divided, you haven't
taken out "as much \(a\) from your dividend as you could", and the actual quotient is taken out "as much \(a\) from your dividend as you could", and the actual quotient is
larger than \(k\). larger than \(q\).
{{< /sidenote >}} {{< /sidenote >}}
{{< latex >}} {{< latex >}}
\begin{aligned} \begin{aligned}
& b = ka + r \\ & b = qa + r \\
\Rightarrow\ & b-r = ka \\ \Rightarrow\ & b-r = qa \\
\end{aligned} \end{aligned}
{{< /latex >}} {{< /latex >}}
We only really care about the remainder here, not the quotient, since it's the remainder We only really care about the remainder here, not the quotient, since it's the remainder
that determines if something is divisible or not. From the form of the second equation, we can that determines if something is divisible or not. From the form of the second equation, we can
deduce that \\(b-r\\) is divisible by \\(a\\) (it's literally equal to \\(a\\) times \\(k\\), deduce that \\(b-r\\) is divisible by \\(a\\) (it's literally equal to \\(a\\) times \\(q\\),
so it must be divisible). Thus, we can write: so it must be divisible). Thus, we can write:
{{< latex >}} {{< latex >}}
@ -332,7 +332,7 @@ Let's explore this some more with a different divisor, say 12. We fill find that
of 3, 7 has a cycle length of 12, 9 has a cycle length of 4. What's of 3, 7 has a cycle length of 12, 9 has a cycle length of 4. What's
happening here? To see, let's divide 12 __by these cycle lengths__. For 8, we get (12/3) = 4. happening here? To see, let's divide 12 __by these cycle lengths__. For 8, we get (12/3) = 4.
For 7, this works out to 1. For 9, it works out to 3. These new numbers, 4, 1, and 3, are actually For 7, this works out to 1. For 9, it works out to 3. These new numbers, 4, 1, and 3, are actually
the __greatest common factors__ of 8, 3, and 7 with 12, respectively. The greatest common factor the __greatest common factors__ of 8, 7, and 3 with 12, respectively. The greatest common factor
of two numbers is the largest number that divides them both. We thus write down our guess of two numbers is the largest number that divides them both. We thus write down our guess
for the length of a cycle: for the length of a cycle:
@ -410,7 +410,18 @@ Lastly, recall that our patterns would spiral away from the center whenever a \\
\\(k\\) is, we can restate this as "\\(d/\\text{gcd}(d,n)\\) is divisible by 4". But if we pick \\(k\\) is, we can restate this as "\\(d/\\text{gcd}(d,n)\\) is divisible by 4". But if we pick
\\(n=d-1\\), the greatest common factor has to be \\(1\\) (see [this section below](#divisors-of-n-and-n-1)), so we can even further simplify this "\\(d\\) is divisible by 4". \\(n=d-1\\), the greatest common factor has to be \\(1\\) (see [this section below](#divisors-of-n-and-n-1)), so we can even further simplify this "\\(d\\) is divisible by 4".
Thus, we can state simply that any divisor divisible by 4 is off-limits, as it will induce loops. Thus, we can state simply that any divisor divisible by 4 is off-limits, as it will induce loops.
For example, pick \\(d=4\\). Running our algorithm for \\(n=d-1=3\\), we indeed find an infinite For example, pick \\(d=4\\). Running our algorithm
{{< sidenote "right" "constructive-note" "for \(n=d-1=3\)," >}}
Did you catch that? From our work above, we didn't just find a condition that would prevent spirals;
we also found the precise number that would result in a spiral if this condition were violated!
This is because our proof is <em>constructive</em>: instead of just claiming the existence
of a thing, it also shows how to get that thing. Our proof in the earlier section (which
claimed that the divisor 9 would never create spirals) went by contradiction, which was
<em>not</em> constructive. Repeating that proof for a general \(d\) wouldn't have told us
the specific numbers that would spiral.<br>
<br>
This is the reason that direct proofs tend to be preferred over proofs by contradiction.
{{< /sidenote >}} we indeed find an infinite
spiral: spiral:
{{< figure src="pattern_3_4.svg" caption="Spiral generated by the number 3 with divisor 4." class="tiny" alt="Spiral generated by the number 3 by summing digits." >}} {{< figure src="pattern_3_4.svg" caption="Spiral generated by the number 3 with divisor 4." class="tiny" alt="Spiral generated by the number 3 by summing digits." >}}
@ -420,7 +431,7 @@ Let's try again. Pick \\(d=8\\); then, for \\(n=d-1=7\\), we also get a spiral:
{{< figure src="pattern_7_8.svg" caption="Spiral generated by the number 7 with divisor 8." class="tiny" alt="Spiral generated by the number 7 by summing digits." >}} {{< figure src="pattern_7_8.svg" caption="Spiral generated by the number 7 with divisor 8." class="tiny" alt="Spiral generated by the number 7 by summing digits." >}}
A poem comes to mind: A poem comes to mind:
> Turning and turning in the wydening gyre > Turning and turning in the widening gyre
> >
> The falcon cannot hear the falconner; > The falcon cannot hear the falconner;
@ -450,7 +461,7 @@ First, let's update our condition. Before, we had "\\(d\\) cannot be divisible b
we aren't constraining ourselves to only 4, but rather using a generic variable \\(c\\). we aren't constraining ourselves to only 4, but rather using a generic variable \\(c\\).
We then end up with "\\(d\\) cannot be divisible by \\(c\\)". For instance, suppose we kept We then end up with "\\(d\\) cannot be divisible by \\(c\\)". For instance, suppose we kept
our divisor as 9 for the time being, but started turning 3 times instead of 4. This our divisor as 9 for the time being, but started turning 3 times instead of 4. This
violates are divisibility condtion, and we once again end up with a spiral: violates our divisibility condtion, and we once again end up with a spiral:
{{< figure src="pattern_8_9_t3.svg" caption="Pattern generated by the number 8 in base 10 while turning 3 times." class="tiny" alt="Pattern generated by the number 3 by summing digits and turning 120 degrees." >}} {{< figure src="pattern_8_9_t3.svg" caption="Pattern generated by the number 8 in base 10 while turning 3 times." class="tiny" alt="Pattern generated by the number 3 by summing digits and turning 120 degrees." >}}
@ -483,8 +494,8 @@ a triangle:
As long as you rotate by the same amount each time (and you will, since the cycle length determines As long as you rotate by the same amount each time (and you will, since the cycle length determines
how many times you turn, and the cycle length never changes), you can do so for any number how many times you turn, and the cycle length never changes), you can do so for any number
of directions. For instance, here's a similar visualization for a five-turn system, where of directions. For instance, here's a similar visualization in which
each turn is 72 degrees: there are 5 possible directions, and where each turn is consequently 72 degrees:
{{< figure src="turn_5_anim.gif" caption="Pentagon formed by five 72-degree turns." class="small" alt="Pentagon formed by five 72-degree turns." >}} {{< figure src="turn_5_anim.gif" caption="Pentagon formed by five 72-degree turns." class="small" alt="Pentagon formed by five 72-degree turns." >}}
@ -537,14 +548,14 @@ after \\(360/\\text{gcd}(360, t)\\) turns, we'll end up facing the way that we s
So far, so good. What I don't like about this, though, is that we have all of these So far, so good. What I don't like about this, though, is that we have all of these
numbers of degrees all over our equations: 72 degrees, 144 degrees, and so forth. However, numbers of degrees all over our equations: 72 degrees, 144 degrees, and so forth. However,
something like 73 degrees (in a 5-turn system) is just not a valid bearing, something like 73 degrees (if there are five possible directions) is just not a valid bearing,
and nor is 71. We have so many possible degrees (360 of them, to be exact), but we're only and nor is 71. We have so many possible degrees (360 of them, to be exact), but we're only
using a handful! That's wasteful. Instead, observe that for a system with \\(c\\) turns, using a handful! That's wasteful. Instead, observe that for \\(c\\) possible turns,
the smallest possible turn is \\(360/c\\). Let's call this turn \\(\\theta\\) (theta). the smallest possible turn angle is \\(360/c\\). Let's call this angle \\(\\theta\\) (theta).
Now, notice that we always turn in multiples of \\(\\theta\\): a single turn moves us \\(\\theta\\) Now, notice that we always turn in multiples of \\(\\theta\\): a single turn moves us \\(\\theta\\)
degrees, two turns move us \\(2\\theta\\) degrees, and so on. If we define \\(r\\) to be degrees, two turns move us \\(2\\theta\\) degrees, and so on. If we define \\(r\\) to be
the number of turns after a single cycle, we have \\(t=r\\theta\\), and our turning the number of turns that we find ourselves rotated by after a single cycle,
equation can be written as to: we have \\(t=r\\theta\\), and our turning equation can be written as:
{{< latex >}} {{< latex >}}
kr\theta \equiv 0\ (\text{mod}\ c\theta) kr\theta \equiv 0\ (\text{mod}\ c\theta)
@ -567,7 +578,7 @@ And finally, observing that \\(kr=kr-0\\), we have:
This equivalence says the same thing as our earlier one; however, instead of being in terms This equivalence says the same thing as our earlier one; however, instead of being in terms
of degrees, it's in terms of the number of turns \\(c\\) and the turns-per-cycle \\(r\\). of degrees, it's in terms of the number of turns \\(c\\) and the turns-per-cycle \\(r\\).
Now, recall once again that for this equivalence to hold in \\(k>0\\) steps, we need Now, recall once again that the smallest number of steps \\(k>0\\) for which this equivalence holds is
\\(k = c/\\text{gcd}(c,r)\\). \\(k = c/\\text{gcd}(c,r)\\).
We're close now: we have a sequence of \\(k\\) steps that will lead us back to the beginning. We're close now: we have a sequence of \\(k\\) steps that will lead us back to the beginning.
@ -623,6 +634,18 @@ is one more than the _divisor_ (we started with numbers in the usual base 10, bu
{{< figure src="pattern_2_11_t7.svg" caption="Pattern generated by the number 2 in base 12 while turning 7 times." class="tiny" alt="Pattern generated by the number 2 by summing digits in base 12 and turning 51 degrees." >}} {{< figure src="pattern_2_11_t7.svg" caption="Pattern generated by the number 2 in base 12 while turning 7 times." class="tiny" alt="Pattern generated by the number 2 by summing digits in base 12 and turning 51 degrees." >}}
### Conclusion
Today we peeked under the hood of a neat mathematical trick that was shown to me by my headmaster
over 10 years ago now. Studying what it was that made this trick work led us to play with
the underlying mathematics some more, and extend the trick to more situations (and prettier
patterns). I hope you found this as interesting as I did!
By the way, the kind of math that we did in this article is most closely categorized as
_number theory_. Check it out if you're interested!
All that remains are some proofs I omitted from the original article since they were taking
up a lot of space (and were interrupting the flow of the explanation). They are listed below.
### Referenced Proofs ### Referenced Proofs
#### Adding Two Congruences #### Adding Two Congruences
@ -676,7 +699,7 @@ if \\(a=fc\\) and \\(b=fd\\), then \\(c\\) and \\(d\\) have no common factors ot
\\(\\text{gcd}(c,d)=1\\)). \\(\\text{gcd}(c,d)=1\\)).
__Proof__: Suppose that \\(c\\) and \\(d\\) do have sommon factor, \\(e\\neq1\\). In that case, we have __Proof__: Suppose that \\(c\\) and \\(d\\) do have sommon factor, \\(e\\neq1\\). In that case, we have
\\(c=ei\\) and \\(d=ej\\) for some \\(i\\) and \\(j\\). then, we have \\(a=fei\\), and \\(b=fej\\). \\(c=ei\\) and \\(d=ej\\) for some \\(i\\) and \\(j\\). Then, we have \\(a=fei\\), and \\(b=fej\\).
From this, it's clear that both \\(a\\) and \\(b\\) are divisible by \\(fe\\). Since \\(e\\) From this, it's clear that both \\(a\\) and \\(b\\) are divisible by \\(fe\\). Since \\(e\\)
is greater than \\(1\\), \\(fe\\) is greater than \\(f\\). But our assumptions state that is greater than \\(1\\), \\(fe\\) is greater than \\(f\\). But our assumptions state that
\\(f\\) is the greatest common divisor of \\(a\\) and \\(b\\)! We have arrived at a contradiction. \\(f\\) is the greatest common divisor of \\(a\\) and \\(b\\)! We have arrived at a contradiction.