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The Sensorimotor Underpinnings of Intelligence
Posted Dec 5 2023 by Mariven
Human mathematicians generally aren't thinking in terms of formal proofs when they're solving problems; what, then, goes on in their heads that allows them to correctly solve these problems? It's remained largely mysterious.
!tab Here, I conduct a first-person investigation of the mental structures that underlie mathematical problem solving, paying particular attention to the way in which my mind uses concepts as a way to turn abstract mathematical structures into virtual spatiotemporal representations which it can productively manipulate with the sensorimotor schemas it's learned through experience.
!tab Note: the core of this was written over the course of a couple hours one night in July 2022. I was a bit slow to clean it up.
How do we do mathematics?
In general, attempts to formalize cognition have always irritated me—not because I'm against the idea, but because everyone who does it pretty much immediately gets it blatantly wrong. For instance, if an approach posits some particular set of symbols, or procedure for learning symbols, and some particular rules for inference, and says "this is how cognition works"Think GOFAI, not psychology. Saying "this is a useful frame in which to analyze cognition" is fine and good, but saying that some simplistic breakdown of the mental world actually gets you cognition is a mistake., I immediately look inwards and observe that it doesn't fit. This pattern baffled me for a long time—why do they not notice how obviously ridiculous their models are? Is this some sort of massive in-joke, or a way to launder reputation, like a zero-effort modern art piece that sells for six digits? No—that ability to 'look inwards' is just not something that most people have—or use, in any case—and the people that do have it tend to only weakly have it. In other words, they're doing the best they can; it's only my abnormal inclination towards metacognition that gives me a clear view of their errors.
!mar{why is cognition of formal structures so difficult to formalize?} Despite the subject matter, mathematical cognition is one of the least formalizable kinds of cognition. You can ask a mathematician a question, and they'll just stare blankly for a solid minute before giving you an answer which they then have to do additional work to explain or formally prove
For instance, what is the fundamental group of the real projective plane, $\pi_1({\mathbb R\mathbb P}^2)$?If you don't know what a fundamental group is, then this may help you..
The real projective plane ${\mathbb R\mathbb P}^2$ is the set of lines in 3-dimensional Euclidean space ${\mathbb R}^3$ that pass through the origin. Not rays or signed lines (in which case the space would be isomorphic to a sphere), but just lines as one-dimensional subspaces, making ${\mathbb R}{\mathbb P}^2$ isomorphic to a sphere whose antipodes are identified. This is hard to visualize, since this identification makes the space no longer embeddable in ${\mathbb R}^3$, but the relevant consequence is that you can spin a line $180^\circ$ and it won't be changed. Such a $180^\circ$ spin specifies a path in ${\mathbb R\mathbb P}^2$ that begins and ends at the same place (i.e., a loop $S^1 \to {\mathbb R}^3$), and it's intuitively clear that you can't continuously deform this loop into a constant map, since at some point you'll have to unhook one of the endpoints to undo the spin, and then it's no longer a loop. So $G = \pi_1({\mathbb R\mathbb P}^2)$ is a nontrivial group. It feels obvious (though it's not at all obvious how I'd actually prove it) that this is the 'gimmick' that the space has here, the phenomenon that produces a generator $g$ of the fundamental group $G$.
!tab But what are the integer multiples of the generator? Is, for instance, a $360^\circ$ spin ($g^2$) continuously deformable to a constant loop? If you're not thinking about it enough, it's easy to say no and conclude that a $540^\circ$, $720^\circ$, etc. loop won't be deformable into a loop of any other degree, making $G$ isomorphic to ${\mathbb Z}$. But in fact you can deform a $360^\circ$ loop: if you imagine the loop as starting at a line passing through Earth's equator which spins around the entire equator, then you can deform it into a loop that in its spin goes a bit north until the 180 degree point, at which point it starts going south; make it go further and further north, bring it around the north pole, and then contract it to a constant map. So, while $G$ isn't trivial, its one generator becomes the identity when squared, whence $G \cong {\mathbb Z}_2$.
!tab I don't expect this explanation to be easy to understand(this kind of proves my point), and I can't immediately see how to make it rigorous, but many algebraic topologists would probably use the same intuition to get this answer. Granted, they'd know how to prove it formally as well—I've forgotten a lot of basic algebraic topology, but a significant part of it is about tools that let you formalize intuitive answers. But you often have to have the intuitive answer before you formalize it—while in lower-level math you use the tools to figure out the answer, abstract math reverses this formula: the answer is often intuited first, and the manner of intuition is used to select the tools that can formally yield that answer.
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There's generally a "spark of creativity" that leads to any given mathematical proof—a special subjective factor noticed 'spontaneously'—but reading the proof itself oftenNot always: the proof that there are infinitely many prime numbers is just a restatement of the intuition that we could multiply finitely many primes $p_i$ and add one to get a number divisible by no $p_i$. But often. fails to reveal this spark. Reports of what the mind is actually doing during mathematical cognition seem to be really rare—more common are sequences of idealizations arranged into narratable stories, which shed no insight onto how the idealizations were initially generated. So I decided to create such a report myself, by keeping very close track of my mind and writing down exactly what I noticed going on while solving a math problem. What follows isn't so much a phenomenological report so much as a metacognitive report that's very phenomenological in character.
!mar{a choice of Schelling problem} !tab For this experiment, I took problem A1 from the 2020 Putnam exam. Being the first problem on the most recent editionAt time of writing (07/2022), it was the most recent problem I could source. I was unable to find the 2021 exam for some reason, and I guess I wrongly assumed that it had been cancelled due to COVID. of the most famous math problem challenge on which I could actually reliably solve a question
The problems on the Putnam exam are arranged roughly from easiest to hardest. I'm not that bad at competitive mathematics, having scored in the ~95th percentile on an earlier administration of the exam, and figured before looking at it (which would've given my brain a head-start on parsing it) that I could likely solve the first problem without much difficulty. This is important, since more challenging problems not only leave less cognitive resources available for introspection, but make the cognitions more complicated such that introspection doesn't go as far.
, this was as close to a Schelling problem as I could come up with. I gave this problem to myself to solve, with the constraint that I would write down everything I could identify as contributing to actual cognition of and along the problem; normally, such a simple problem would take me perhaps twenty minutes to solve, but this introspective task expanded it to several hours. This undoubtedly distorted my cognitive approach to the problem, but I think it doesn't really change the nature of the construction of this approach. If you go to pizza school and the professor demonstrates pizza construction by spending an hour on each step—flatten the dough for an hour, spread the sauce for an hour, sprinkle the cheese in slow motion, etc.—the final result will taste unusually stale and stratified, but the process will still have been faithfully demonstrated; I'm hoping something like that might also happen here.
Cognitive Theses
First, I'll outline a couple of fundamental theses about the structure of cognition that underlie the way I look at my own mind. These came about through lots of time spent investigating my own cognitive processes, and had to be fully internalized in order to continue progressing with this investigation—if you don't already know what I'm pointing to in each case, consider investigating it further in your own mind. (You can skip this).
Automaticity Thesis: !mar{pre-conscious mental actions} Agency comes "after the fact" of most mental actions. There is an intuition, an image, a connection, a decision, and then there is its attribution to the concept of the self. This concept is a useful groove that the brain is prone to carving out and staying within, but it doesn't quite get at the actual causal factors powering conscious cognition, the "why" we think this or that given thought, have this or that given intuition. So when I say that something happens "pre-consciously", I mean that it happens in a way that it would be misleading to attribute to "me"
There is the simian "me" (the biological organism) and the constructed "me" (the concept-borne convenient fiction); I mean the latter here. .
If you look at any object in front of you, everything that it is to you will have been constructed by your brain—it's what sketches the lines, fills them in, conceptually labels the object, attributes materiality and spatial extension to it—so can you say that it's you, your conceptualized self, that constructs the object? No... it kind of just happens to happen, right? It's in this same way that intuitive models spin up as percepts; there's a labeling of the intuition as a cognition, and my mind consequently tags it as being 'by me', but in reality there was no agency behind it, no self-oriented train of thought—just mind acting on its own prior to notions of "mine" or "not mine", "conscious" or "unconscious". This is what I mean when I say it's pre-conscious.
Multitemporality Thesis: !mar{phenomenal nonlinearity} The cognitive process does not proceed in a single timeline except at the macro-scale. Mental events [or their underlying neural causes; no need to make a distinction] all have causal power by virtue of their happening throughout physical time, but the unfolding processes of interaction by which they influence and create one another occur along a second dimension of 'processing' time. These mental events are linearized to create a single consistent history, a third dimension of 'phenomenal' time, but consciousness tends to be "spread out" along phenomenal time at any particular phenomenal time (i.e., it's more like a wave packet), which creates a fourth dimension of 'reflective' time. To say "I thought this, then this, then this" is only a large-scale approximation; in reality, there's a sort of quantum foam at the base of consciousness in which all phenomena just appear without reference to time, and they're only linearized into an internal history after the fact of their appearing.
!mar{2-time approximation} !tab While there are many ways to split temporality in the brain, I tend to just use two dimensions, by identifying phenomenal time with physical time as $t_p$ and discounting processing time, leaving reflective time $t_r$ as the second dimension. The two dimensions can be phrased like "what you experience having experienced at time $t_r$, at time $t_p$". After all, any traces at time $t_p$ of your experience at a prior time $t_p'$ can only exist as causal influences of neural states at time $t_p'$ on neural states at time $t_p$: the past isn't preserved for us, but only exists insofar as it affects the present. Consequently, the preservation of what was experienced at a given time $t_r$ doesn't come for free; it requires the brain to keep some record of the experience at $t_r$, but there's no reason that this recording ought to have perfect fidelity—it systematically changes experience in order to bring it in line with one's internal narrative and "make it make sense".
Indeterminacy Thesis: !mar{indeterminate properties} The feeling of a mental image as being given, like a developed photograph, is fake; the image never really exists as a concrete object, but is only approximated through virtualized percepts determined by conditions of mutual coherence and conceptual congruence. Insofar as it doesn't possess any particular value of a property, that property is indeterminate. For instance, if one visualizes a balloon without simultaneously visualizing a particular color of that balloon, the color is an indeterminate property. I distinguish between two ways that mental images can have indeterminate properties, and two corresponding ways that properties can be determined on-demand when the brain looks for their values.
!mar{mental superposition} !tab First, possible values may be superposed, or in a state similar to quantum mechanical superposition (metaphorically NOT literally). For instance, if I'm prompted to imagine a traffic light, the mentally imaged light might seem to me to have a color, but there will often be no particular color that it was. It could just as easily be red as it could be green or yellow. If I'm asked which color it was, and I'm not being sufficiently attentive to note that there was no which color, my mind might just collapse (metaphorically) the superposition and overwrite my past experience of having imagined it, so that I think I originally experienced imagining it to be red
Using 2-time, we can say: at time $0_p$, I experience having experienced a color at time $0_r$; at time $1_p$, I experience having experienced red at time $0_r$ and time $1_r$. This is, for instance, how chronostasis works.
!mar{mental deferral; detail-on-demand} !tab Second, possible values may be deferred. For instance, if I ask you to visualize a cat eating a grape, and then ask you if you imagined the cat's eyes to be a certain color, you likely won't have visualized a particular eye color at first, and (conditional on that) will suddenly be aware of an eye color when you check your visualization for that detailIt's possible that you did visualize it, e.g. if your visualization had you feeding the grape to the cat, so that its face was especially detailed; in that case, find yourself some other question to ask, or some other thing to visualize. Another example: I visualized a dog eating a grape, and it was instantly and clearly a basset hound, but there was no information on whether the dog's tail was visible until my mind scrambled to add it when I looked.. There wasn't a prior superposition of colors that's being collapsed—it's more like there was nothing there until you investigated, at which point the detail was conjured by the brain's detail-on-demand system.
!mar{collapsing and filling in} !tab Usually, properties that are indeterminate aren't either deferred or superposed; often they're in some complex combination of the two that we may only resolve to one or the other. But insofar as a property is in superposition, we can say that it collapses to a determinate value; insofar it is deferred, we can say that it is filled in with a determinate value.
Problem Statement
Problem A1 reads as follows:
[Line 1] How many positive integers $N$ satisfy all of the following three conditions?
[Line 2] (i) $N$ is divisible by 2020.
[Line 3] (ii) $N$ has at most 2020 decimal digits.
[Line 4] (iii) The decimal digits of $N$ are a string of consecutive ones followed by a string of consecutive zeros.
Here's the play-by-play:
Reading the Problem
(Line 1) !mar{number grid} As I read the first line, my mind pre-consciously begins to spin up a general intuition it has for "counting-like problems": here, the natural numbers are arranged into a grid, each number capable of being "on" or "off" like an LED light. It's neither the case that any given one of them is specifically on or off—rather, my mind's putting them in superposition, such that I can just as easily take the default to be either on or off. Nor do the axes of the grid count off anything in particular yet: I just imagine that there'll be some repeated pattern that allows me to calculate the count with some formula, and my mind wants to stuff individual iterations of the pattern in the x-axis.
!tab Note that I'm not explicitly trying to create or shape this sketch, and I probably wouldn't even notice it if I weren't trying to introspect (in which case it's unlikely there'd've been any mental object that could've been called an "LED light" in the first place; such detail likely comes from filling-in of a deferred visual texture). It's a transparency faintly projected onto my attention by the cognitive process. There's no intentionality behind my mind's choice to spin this intuition up; it just arrives, and the image it brings appears only very faintly in my imagination as I continue.
(Line 2) !mar{number sieve} Upon reading the next line"$N$ is divisible by 2020"., my mind adds a 'sieve' to the scene, which I picture as a horizontal row patterned with some number (between maybe four and seven
However, it is fundamentally not a kind of mental object characterizable by such numbers. It's just that when I look at it closely, the superposed system tends to end up like this.
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of filled squares, then an empty square, then repeat. The sieve carries the general form of the pattern, which is why it's horizontal (each pattern spatially extends across the x-axis). The sieve moves upwards in a perceptible start and then an instant flash, and everything on the same column as an empty part of the sieve is marked on (lit), while everything on columns where the sieve has a filled square is marked off (unlit); somehow, this pattern has extended below the sieve as well.
!tab But this all remains faint, and I'm mainly verbalizing (via my internal dialogue) something like "ok, $0 \bmod 2020$"—I see the text $\bmod 2020$ written out on a white background in a visualization disjoint from the above, though the zero isn't really present as an ellipse but as a feeling, a nullity or a default which is taken to exist implicitly. This sketch isn't developed further, and in fact seems to drop away from my visual imagination, but remains "in play" in the sense that I have the ability to call upon it without having to reconstruct it.
(Line 3) !mar{bounding the grid} As I move on to the next line, marked (ii)"$N$ has at most 2020 decimal digits"., I think to myself "bounded", not verbalizing that word internally but merely activating an intuition corresponding to it. The grid of LED lights becomes bounded, a rectangular boundary having appeared around a certain set of lights without my noticing it; the lights in this boundary are unaltered, and a column very close to the rightmost one is lit up (representative of all the lit up columns in the rectangle, for I 'sense' that there are more but have not visualized them), but the lights in this boundary are now especially dim, near black (where they were once in superpositions of green, yellow-orange, or red; they didn't have any particular color, but they did have some sort of 'LED color'), the import of this clearly being that they're deactivated, incapable of being turned on.
All of this happens more or less instantaneously in the visual imagination, and again a secondary visual imagination forms, this one containing the upper bound which I explicitly take to be a one with 2020 zeroes after it (though I'm unsure of this, and there's a certainty in my mind of the possibility of an off-by-one error here, though my mind dismisses it as irrelevant at this point); the actual content of the visual imagination is a character $1$ on the left-hand side, a string $0000000$ on the right (between one and roughly twenty zeroes are actually imagined, though the more are drawn the less each one actually exists, and the more they form a sort of blur), and a giant blur in the middle where all the rest of the zeroes lie.
!mar{there has to be a trick} !tab I have a quick idea to combine the divisibility filter and the upper bound to get a general sense of a new upper bound on the number of integers we want, which I figure has the form $10^{2020}/2020 = 10^{(2020 - \log_{10} 2020)}$; the logarithm is quickly taken as being something like three, and my mind notices that $2020$ is ridiculously large compared to $3$, dismissing this idea. No way is the actual number gonna be that large; it's gonna be something like 7 or 2991 or it's going to have some cutesy form like $3^{2020} - 2020^3$ or whatever—it's going to be something that's nontrivial and can be written down exactly. This all happens as I'm reading line (ii), and perhaps as I semiconsciously pause for a second after reading (ii) to give these thoughts space to settle down.
!mar{the interesting part} (Line 4) As I read (iii)"The decimal digits of $N$ are a string of consecutive ones followed by a string of consecutive zeros"., there's a flash of awakeness and vigor in my mind—this is the key point, this is what makes the problem go brrr—as I imagine something like
$$1111\ \text{[blur of ones]}\ 11\ \text{[invisible divider]}\ 00\ \text{[blur of zeroes]}\ 0000$$
with my moving around of the divider changing the point at which the consecutive ones become consecutive zeroes; the flash is due to my immediate understanding that this is the essential key to solving the problem. I move the divider around in my mind, and figure that the approach to the problem should be to:
Get all the possible places the divider can be—of which there should be basically $2020$, but again my mind is certain of the possibility of an off-by-one, or in this case an off-by-two error, since the invisible dividers at the very ends of each number seem potentially problematic,
Find out which of these are divisible by $2020$ (there's a recollection of the rule for checking if a number is divisible by three, but this recollection is tamped down by an expectation that nothing of the sort will work for $2020$, since it's got a stranger factorization),
Count them.
!mar{how the possibilities look} Importantly, I'm halted for a fraction of a second by the realization that the length isn't fixed at $2020$ digits, but is instead at most 2020 digits; in response, my mind places a second flexible bar at the left-hand side of the number, which I can move around to change the length of the number up to a maximum of $2020$ digits. So it's clear that the possibilities lie within a sort of triangle. To be explicit about how this triangle looks after the fact:
One leg of the triangle is bounded by numbers of the form 1, 10, 100, ..., 1000[...]0000;
A second leg is bounded by numbers of the form 1000[...]000, 1100[...]000, 1110[...]000, ..., 1111[...]111;
The hypotenuse is bounded by numbers of the form 1, 11, ..., 1111[...]1111.
(Clearly 0 doesn't count, even though it is divisible by 2020, since only positive integers are allowed).
Solving the Problem
Now, my mind perceptibly switches gears, going from a mode of understanding the problem to a mode of working on the problem. During the gear shift, I sort of withdraw mentally, and find myself just sitting and not seemingly thinking anything for maybe a couple seconds, before redeveloping conscious attention and sitting up straight, working through the problem at full speed.
Building Conceptualizations
!mar{conceptualizing panda numbers} My first thought is to investigate what's going on with divisibility by $2020$: how is it that one of these $111\ldots10\ldots000$ numbersThe shape of the form is much more "compact" in my thought than in text, so let me just call them panda numbers
My mind made a rather strange series of leaps in the creation of the name "panda number", and I got lucky enough to observe the play by play: it moved from the general format of ones followed by zeroes to one particular scene in the Animatrix (?!) where the numeric sequence "zero one" is represented by a black rectangle next to a white rectangle (it's actually the other way around, see this picture) to the black and white coloration of a panda bear, stopping there because "panda" was nice and compact.
!tab I seriously doubt that my mind pulled a picture of B-roll footage from a movie I haven't watched in years as a necessary component of this sequence, though — more likely there was an underlying tangle of conceptual relations being activated which was only relayed to my conscious mind in that particular form for a variety of second-order reasons. I know that panda bears aren't strictly black on one side and white on the other — that would look really weird — but somehow my mind interprets that property as being reminiscent of a panda bear.
here while noting that I just think of them as $111\ldots10\ldots000$ numbers. could be divisible by $2020$? I transpose this question into "$2020$ times what is a panda?", and it occurs to me that $2020 \times 5 = 10100$. This calculation does not appear immediately to me like $6 \times 7 = 42$ does, but it appears without any conscious work on my part as and only when when I impel the answer to form.
!mar{conceptualizing monochrome numbers} !tab Now, $10100$ isn't a panda, but it is what I'll call a monochrome number, composed of just ones and zeroesAgain, the word 'monochrome' is just for the text; I don't think it, but the concept is nevertheless as compact as the one word.. I realize that multiplying 10100 by ten and adding it to itself will "fill the hole", giving us a proper panda $2020 \times (5\times 11) = 111100$. I've shifted to a mental model whereby I have a number in decimal notation stopping at the decimal point and extending to the left, which contains an infinite sea of "virtual zeroes", for as many digits as it has. Multiplication by ten just shifts each digit one position to the left by adding a zero to the right, and addition occurs placewise; I feel there ought to be no carrying—no addition of lots of monochromatic numbers that can't just be expressed as a multiplication into another monochromatic number, and therefore (I guess) as a combination of multiplication by ten and "hole-filling" addition.
!mar{shift-and-drop multiplication} !tab It's clear to me that, by shifting $111100$ to the left four times and adding it to $111100$, I can extend the ones part of the initial panda number by four digits; repeating this, I can obtain any panda number which looks like a multiple of four ones followed by two zeroes. This shift-and-add operation is visualized as lifting a copy of $111100$ up from its place on the mental register, shifting that copy to the left enough that its ones don't overlap with existing ones — this happening to be four times — and putting it down, $111100(0000) + 111100 = 1111111100$. At this point, I'll start to use regex notation for "ease" of the reader (I don't have to use it in my head, as I don't have to word the concepts at all). So what I've deduced so far is that all numbers like $(1111)^+00$ are pandas.
!mar{bounding the zeroes below} The urge spontaneously comes to me at this point to check that no panda number can have below two zeroes, and this is trivial: if a panda has one zero, it ends in $10$, which can't be produced by multiplying $20$ by an integer; if a panda has no zeroes, it ends in 1, and you can't get an odd number by multiplying an even number. Therefore, two zeroes is the minimum. This train of thought leads me to ask about three zeroes, and I ascertain that we may get such numbers by multiplying two-zero pandas by ten — it occurs to me that any number like $(1111)^+(00)(0)^*$ is a panda. In fact, this ability to add zeroes leads me to intuit that figuring out the pandanesses of numbers with ones totaling a number that's not zero mod four will lead me to solve the problem.
!mar{bounding the zeroes above 2 mod 4} !tab I feel I can do this by studying divisibility, so I decide to factor $2020$. I immediately see the $5$ and draw out a $404$, immediately see the $4$ and draw out a $101$, and do a series of very quick tests to determine that $101$ is prime
Obviously any non-square composite number $N$ is gonna be divisible by some $2 \le n < \sqrt N$, and therefore some prime number in that range. So, as a general heuristic, you can check if $N$ is prime by checking for divisibility by each prime up to $\sqrt N$. For 101, these are 2, 3, 5, and 7; the first three tests fail near-instantly, the fourth passes in a fraction of a second after seeing that 28+70=98 is a multiple of 7.
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So $2020 = 2^2 \times 5 \times 101$. In particular, it isn't divisible by 11, and I'm led to think that no number whose ones are $2\bmod4$ will be divisible by 2020. After all, if one were, we could scale $111100$ high enough to match the highest four digits of such a number, subtract it to leave another number divisible by $2020$, and repeat to get $11 \times 10^{\text{something}}$, which is not divisible by $2020$ (since it is not a multiple of $101$).
!mar{bounding the zeroes above 1, 3 mod 4} !tab And what of pandas with a total number of ones either one mod four or three mod four? Could they be divisible by $2020$? Well, they better be over half the digit limit, I think, otherwise scaling and adding them to themselves will give us a number of $1+1=3+3=2\bmod 4$ ones which is divisible by $2020$, which can't happen. This turns out to be a confusion — even if they are over half the digit limit, they still can't be divisible by $2020$, since the contradiction here is purely arithmetical, having nothing to do with the digit limit.
Endgame
!mar{how to calculate $N$} With this settled, we're pretty much done. All that's left is to count the number of digits of the form $(1111)^+(00)(0)^*$ with at most $2020$ digits. My first thought is to set up a recurrence relation and use memoization to calculate a solution, but I realize that this is supposed to be a math competition, not a programming competition. My second thought is that this isn't necessary: I can just go through numbers of the form $(1111)^+(00)$ and, for each one, consider how many times we can add zero to it. This is simpler: the numbers of the form $(1111)^+(00)$ have $2$ zeroes and can have $4, 8, \ldots, 2012$, up to $2016$ many ones before we break the barrier.
!mar{running the calculation} !tab Dividing $4, 8, \ldots, 2016$ by four, I get that there are $1, 2, \ldots, 504$ numbers of the form $(1111)^+(00)$. The first one, with $4$ ones, can be scaled—let me think this through to avoid an off-by-one-error—scaling once makes it seven digits long, so scaling $n$ times should make it $n+6$ digits long, and we should be able to scale it $2014$ times. The number with eight ones can be scaled $2010$ times, the number with twelve $2006$ times, and so on, up until the final one, which has $2018$ digits and can be scaled two times. Adding one to each of $2014, 2010, \ldots, 2$ to make sure we count the unscaled panda numbers, I find that
$$N = 3 + 7 + 11 + ... + 2011 + 2015$$
I have the idea of calculating this sum by converting it to a sum of the form $1+2+\ldots+n$, the form of which everyone knows by heart. So, let's get them lined up for a division by adding one to each and subtracting $504$ from the total:
$$N = (4+8+\ldots+2016)-504$$
I'm happy it worked out so quickly, and factor the $4$ out of the parenthesized component to obtain
$$N = 4 \times (1+2+\ldots+504)-504$$
Gauss simplifies this to $4\frac{504(505)}{2}-504 = 2(504)(505)-504=504(1010)-504=504(1009)$; the algebraic manipulations happen very quickly. The final multiplication is also computed very quickly, since it's simple: in my mind, the $1000$ leaves the $9$ to extend a copy of $504$ into $504000$, while the $9$ takes another copy of $504$ and splits it into $500$ and $4$, converting the former into $4500$ and the latter into $36$; $504000$ and $4500$ add to $508500$, and the $36$ is tacked on to the end to give us the final answer, $$N = 508536$$
!mar{formalization} All that's left to do now is write my solution down in a formal manner. I essentially rehearse in my mind the sequence of pivotal steps that I made:
Figuring out the general form of the numbers I was being asked to find ('panda' numbers divisible by $2020$ and with $2020$ or fewer digits),
Demonstration of a basic class of cases ($(1111)^+(00)(0)^*$ numbers with length $2020$ or fewer digits being valid),
Refutation of analogous classes which form the rest of the totality (where the number of ones is $1, 2,$ or $3\bmod 4$ rather than 0 mod 4),
Counting of the numbers in the specific class I want (the actual computation leading to $508536$).
There aren't really any complications in my generation of this proof, so I'll skip this step. You can check the 2020 solutions to see that this is the correct answer.