These Simple Equations Are Levels of an Infinite Pattern

[@ComboClass]

Thanks for watching! Make sure you're also subscribed to my other channel www.youtube.com/@Domotro for shorts, livestreams, and bonus videos. And check out my patreon at www.patreon.com/comboclass if you want to get some cool bonus content (like behind-the-scenes clips, my music, etc) and get your name on a whiteboard in the Grade -1 finale!

[@Bibibosh]

"Eureka! The craziest of all experiments hath begun in my lunatic laboratory! Where madness and brilliance collide, creating an explosion of wild ideas. The beakers boil, the test tubes fizz, the equations scribbled upon the chalkboard, a chaotic symphony of the insane! But, ah ha! Lo and behold, the numbers, the glorious numbers! They tell the truth, revealing the beauty hidden within the madness. And so, I say unto thee, the results of my chaotic experiments are irrefutable, a triumph of truth over the absurdity of the human mind!"

[@Guil118]

I just silently assumed all my life, even though I was intrigued by the 2x2=2+2 case, that nothing more than it existed. Being shown a couple examples surprised me, but being shown an infinite amount of answers blew my mind!

[@lnx0007]

pick any collection of integers, find the difference between their sum and their product, add that many 1s to the set, and there you have a set of numbers whose sum equals it's product. its rather trivial because for any collection (of positive whole numbers) the sum will always be less than the product so you just add as many ones as it takes balance it out.

[@Guil118]

@@lnx0007 Well yeah it's trivial when you see it this way. But my intuition was wrong lol

[@lnx0007]

@@Guil118 i hadn't really put much thought into this until watching the video too. His explanations are very verbose and slow to the point so i always like to try racing him to see if i can figure out where hes taking something before he gets there. I figured out that trick above before he got to the 1,1,1...1,2,n pattern. I always bring my own whiteboard and marker to combo class lol

[@Guil118]

@@lnx0007 Same, I found the 11112n pattern while watching too. But assumed it was the only way. I should stop assuming hahaha

[@JoshuaWillis89]

Wow. You really demystified this for me. Now I’m just trying to figure out how I want to incorporate this into my classes. Obviously, rational functions is a perfect fit.

[@quryshna]

I am not great at math. Dyscalculia isn't fun. Regardless, I'm doing my best to learn more. I love, love, love Combo Class! Your enthusiasm is contagious. 3Blue1Brown is a relaxing journey. Standup Maths is funny. Your "LOOK AT THIS COOL MATH STUFF" vibe is what I need to get hyped to learn. I'm so glad I found your channel!

[@SunroseStudios]

dyscalculia gang! we love math but hate numbers lol

[@kbin7042]

Dude you have dyscalculia and you're watching 3blue1brown.. you're really doing your best huh

[@Fire_Axus]

your feelings are irrational

[@SoberCake]

​​@@kbin7042 his voice and animations are relaxing, even if I don't understand a thing of what I'm seeing or hearing

[@joshuasalem5022]

The KZbin algorithm will pick you up and bring your channel to great heights. You’re doing amazing work.

[@mathguy37]

fun fact: a solution similar to 2,2 and sqrt(3),sqrt(3),sqrt(3) is in general for n numbers the n-1th root of n

[@TheMagicFellow]

This sort of pattern also shows up in triangles with an incircle of r =1 and cutting the lengths of the triangle where the incircle is tangent with the triangle.

[@YonDivi]

Dude 3sqrt(3) = sqrt(3)^3 is mind-blowing. As a kid I always found it really really interesting how 2+2=2×2 and always wondered if there were other versions of this equations. You scratched an itch I forgot was there lol. Good job man you earned a subscribe

[@harriehausenman8623]

Yeah, the sqrt(3) thing got me giggling for quite a while 😊

[@YonDivi]

I know right? Made me come up with a generalised formula for numbers like that X^n = nX will give you numbers like that I'm pretty sure

[@chester_m]

@@YonDiviYes this works in general. Nice find! This can be proven. As a hint, divide on both sides by X (for nonzero x). I'll put a proof below. To read it, extend my answer. X^n = nX. For X = 0 it holds for all nonzero n. For nonzero X, divide by X. We arrive at X^(n-1) = n. Raise both sides to the 1/(n-1)-th power. We get X = n^(1/(n-1)). In the case of n = 3 this becomes 3^(1/2) = sqrt(3) as desired.

[@YonDivi]

Yeah I was on that same track too. I'm gonna try see what happens if I use fractional powers and irrational powers too just like Domotro would want

[@HansLemurson]

4 * cube_root(4) works too for 4 equal numbers. I think it should generalize to n^(1/(n-1). 4th root of 5, 5th root of 6, etc. You just need the first n-1 elements to multiply together to get n.

[@Alan_Clark]

Fun fact: in any triangle ABC, tanA + tanB + tanC = tanA x tanB x tanC.

[@yunoewig3095]

are all solutions contained in this formula?

[@Alan_Clark]

@@yunoewig3095 Yes,, it is true for any triangle.

[@matematicke_morce]

Fun fact: You can phrase the (2,2) example as "twice in a row twice in a row" = "twice in a row" twice in a row

[@davidrogers8030]

I prefer two twice = twice two.

[@DavidSartor0]

@@davidrogers8030 Three thrice = thrice three.

[@azai.mp4]

Sounds like you could formalize that using Church numerals. In Church numerals, "2 f" means "f applied twice," so (2 f)(x) = f(f(x)) 2+2 in Church numerals, then, is "apply f twice, and then apply f twice again (for a total of 4 applications)" In math equations, that's: ((2 + 2) f)(x) = (2 f)((2 f)(x)) = f(f(f(f(x)))) = (4 f)(x) 2*2 in Church numerals is "apply f twice to get g, and then apply g twice (for a total of 2 * 2 = 4 applications)" In math equations: ((2 \* 2) f)(x) = (2 (2 f))(x) = (2 f)((2 f)(x)), which we already saw equals (4 f)(x) Finally, 2^2 in Church numerals is "apply 'apply twice' twice to f" Which in equations is: ((2 ^ 2) f)(x) = ((2 2) f)(x) = (2 (2 f))(x) which again equals (4 f)(x) I think in your phrasing, "'twice in a row' twice in a row" comes the closest to the last one, exponentiation. Probably "twice in a row twice in a row" corresponds to multiplication then? I don't know what phrasing would correspond to addition in that case. Something like "twice in a row, then another twice in a row"? This is also related to Quine's Paradox about "'yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation," which is semantically equivalent to "This sentence is false," but seems to be constructed without self-reference. There's some relationship there with Godel's incompleteness theorems, actually, because a big part of those is that you can get a (sufficiently powerful) logical system to answer questions about itself, in a way that doesn't technically use self-reference.

[@harumpher3712]

This is absolutely my favorite math channel

[@RisingSunReviews]

I was able to figure out the 1,1,2,4 just before you went into it! This is fun. And then I was following until you started showing how at the higher levels there might be multiple solutions. That broke me. In a good way. I love math. Something I noticed is that the levels with only 1 solution greater than 6 are all a multiple of 6. And they are all one less than a multiple of 5, where 5 is the first level with multiple solutions. This is fascinating.

[@cerberus0225]

I noticed that it's every odd number equal to or greater than 5 that has a solution with a 3 for the second-to-last digit, while it seems that any number where n mod 3 = 2, n greater than or equal to 5, is where we have a 2,2 for the third- and second-to-last digits. I'm guessing the other solutions repeat on similar patterns, it's just a matter of figuring out when some given sequence first appears?

[@HipsterShiningArmor]

fun fact: not only is sqrt(3)*sqrt(3)*sqrt(3) the same as sqrt(3)+sqrt(3)+sqrt(3), but in both cases the result you get is sqrt(27)

[@Noferrah]

how do the solutions for level 1 at 15:40 work if 1+0 = 1 but 1*0 = 0? same for 2+0 = 2 and 2*0 = 0, 3+0 = 3 and 3*0 = 0, and so on also i feel like the mad scientist persona presented throughout these videos is less of a persona and more an exaggeration of the raw passion you actually have for the topics you cover. and tbh im all for it

[@ComboClass]

There’s no 0 involved in those, Level 1 would be a number on its own, so a single element set could possibly be considered to have its sum or product each just be equal to itself, if you allow that like I decided to. Depends how you define what to do with a single element set

[@Noferrah]

@@ComboClass oh so its less 1 (+*) 0 and more 1 (+*) null? kinda makes sense lol

[@Fire_Axus]

your feelings are irrational

[@pinkunicorns3185]

Would have been kinda funny if the infinite amount of sets would just be (0,0), (0,0,0) and so on as repeating numbers in a set were allowed😅

[@FadkinsDiet]

0 is disallowed bc it's.not positive.

[@Qexilber]

Literally a question I had carried with me since elementary school. Now I feel a bit ashamed, that I never tried to take that on myself, thinking it was too hard of a maths problem to figure that out, because in fact as you have presented here, it is solvable using only rather basic math. 😅

[@maxwchase]

This is probably a stretch, but the varying number of solutions with cardinality, especially the single infinite case with low n, reminded me of the number of regular n-polytopes. And my gut reaction there was to think of them either as ordered sequences of numbers of k-facets, which have an additive constraint in that their alternating sum must be 1 (and the last element is always 1), or an ordered pair of multisets subject to a bunch of cardinality and membership constraints, such that the difference of their sums is 1. The thing is, whatever representation would need a corresponding regularity (regularizability?) constraint that could be generalized to things-that-are-not-individual-polytopes, and I don't know what that could be.

[@AllYourMemeAreBelongToUs]

3:37 “Two oneths.” Two wholes?

[@ojaskumar521]

Amazing video as usual. 16:26 HAHAH loved this one

[@MadclintMusic]

Not sure if that intro was planned but it was genius....

[@castro_rl]

I'm a highschool physics teacher, and I hope to one day have 1% of your chaotic energy. Hope I don't burn the whole building down next class!

[@ImaginaryMdA]

Use any multi-set of numbers, and add '1's until its sum is equal to its product.

[@Hexcede]

This reminds me of a really cool fact I discovered. I asked about it on Reddit and in some other KZbin videos and I saw some people say it was on interview questions for them, but it has no Wikipedia article and I never found a name or any information on it. I've never seen it mentioned anywhere at all, yet I find it so interesting, especially because of one of the realizations it leads to. In the process of finding this, it shows you that -c/(ax + b) = x, a weird way to write a set of simplistic continued fractions, can be solved via the quadratic formula. The square root of 2, the golden ratio (phi), and more can be represented, including an infinite set of numbers where c/x = x - c, like how 1/phi = phi - 1. It is easiest to work backwards and show that multiplying by (ax + b) and then adding c gets you a quadratic equation. And for the really cool part, I encourage you to try yourself, if you notice that the quadratic formula yields x, and the equation is equal to x you may be intrigued by the idea of substituting x. What you get is a second quadratic formula that can solve certain kinds of quadratics (e.g. lines) that the other can't, and they are very similar to eachother in a very cool way! And an unrelated but very cool trick, if you use roman factorials instead of regular factorials, you can extend Taylor series to include negative exponents of x, allowing you to represent functions you otherwise would be unable to. It works by substituting each term with t_d/d!x^d where the factorial is the roman factorial instead, d is the degree, and t_d is the initial value (y intercept) of the dth derivative of the function. This preserves one of the unique properties of Taylor series which is that if you shift all terms left/right by some amount I'll call h (equivalent to adding/subtracting h to/from all d) you will get the derivative or indefinite integral of the Taylor series. This allows you to do calculus on arbitrary polynomials without knowing any rules or anything. And it works for functions which require infinite terms like sine/cosine too!

[@B3Band]

Domotro on Numberphile when?

[@harriehausenman8623]

Oh yeeesss 🤗

[@HomieSeal]

Nice, new video! :D Excited to watch it

[@kittyn5222]

2+2=2×2=2^2=2°2(° being tetration it is same for every "Level")

[@fabriciorodrigues5428]

Yeah, i noticed that too. It's kinda like 2 is an identity for all the algebraic operations, no matter what you do with it, it always equals 4.🤯🤯🤯 Wonder if that holds for pentation and such. @ComboClass should make a episode on that.

[@nartoomeon9378]

Hmm... the general pattern seems like one card solitaire. It has next rules: make the sequence of cards, if you seen 3 consecutive cards where 1st and 3rd has the same (one) property(color) you can remove the 1st card - the sequence become shorter. Next you simply repeat the process. The important is , that first triple on the left has a priority to be done first(removing 1st card). The goal is - obtain 2 cards at the end, it means "win". So, what kind of sequences are winning sequences? If cards has only one property, the answer is easy but interesting(win sequence is binary only and depends on 3 last cards). But if cards has two or more properties, the "game" drasticaly increase their complexity. Im not shure if there is simple formula, as I found for previous case. It called the Medichi solitaire.

[@kevshouse]

Thank yuo for yet another entertaining view into your wonderful world of Mathamatics.

[@azavier-a]

I love combo class so much! You always showcase things in such an intriguing manner

[@Amipotsophspond]

could this/is this being used to speed up multiplication or addition for computers? rather then multiplying 1,2,3 the computer can add 1,2,3, it would likely take too long to check but if the large numbers multiplied basket could be shortly decided if it's along that graph faster then the steps to multiply it might be useful for all. maybe float uncertainty could fudge the numbers in to being along that line with in some acceptable amount of precision.

[@zacharysmith4508]

This channel is literally lawful chaos, which is beautiful.

[@peppermann]

Brilliant! Another super investigative maths lesson I can use at school. 👍

[@lugyd1xdone195]

This is incredible. Simply shows how amazing math can be. I thought it'd be amazing, but this exceeded any of my expectations. Keep it up!

[@bennettpetrik4084]

This is one of my favorite lessons

[@levav8]

Very cool video, I love your vids in general! I feel like I'm getting a side exploration to subjects that I often only encounter on projecteuler, is there any connection between the subjects you choose and the problems on that site?

[@ComboClass]

I actually haven’t explored that site very much, although it seemed cool when I checked it out once and I do have it bookmarked to look into more later since it seemed interesting!

[@levav8]

@@ComboClass Well, I guess it can be a source for if you ever run out of fun things to explore ^^ I'ma plug in two pieces of math that I like while I'm here - 1. Surreal numbers: they are an alternative way to define numbers from scratch (in a similar sense to the classical definition of natural numbers from sets, with 0={}, 1={{}} etc) except instead of the naturals they end up giving us waaaaaaaay more numbers directly. Invented by john conway, surprisingly useful for a specific branch of game theory. 2. This one might be a bit out of scope, but idk, I think its super cool: in general number rings you don't have unique prime factorization (or primes in the usual sense really), but you do have unique prime ideal factorization. Apparently the ideal class group gives a lot of information about how elements factor into irreducibles. for exampls, you have the property Vn that states: for any irreducible a,b in the ring if a*b=c_1*...*c_m where c_i are also irreducible then m< hope it was interesting. Anyhow, Just keep doing what you do! your style is amazing :)

[@hkayakh]

Very exiting ending!

[@JoeBorrello]

Great video! Also, you remind me a bit of ElectroBoom, where he teaches stuff amid chaos and destruction. 🙂

[@LeeSpork]

Although they won't all be positive integers, and the length of the collection will no longer be n, I noticed that the general formula also holds for n = -2, if you assume that a collection of 1s with a length of n sums to n and multiplies to 1 even for a negative n. I wonder if there are other solutions for negative "levels" of this? If that even makes sense? The formula doesn't seem to work for n = -3 or lower though. I'm not sure how it applies to n = 0 or n = 1 either.

[@DavidSartor0]

Fractions?

[@Ing0s]

I don't really have something meaningful to say. I just like your videos a lot and want to push them by writing a comment.

[@harriehausenman8623]

Some more random engagement from me too 😃

[@emilyrln]

"The clock's on fire again." What a madlad 😂

[@Speedy27o7]

2n-3 1’s then a 3 then n gives us for n = 8 (1,1,1,1,1,1,1,1,1,1,1,1,1,3,8) adds to 24 and multiplys to 24

[@flockofwingeddoors]

New combo class! This is not a drill! Obligatory: will this be on the exam? 😋

[@stevenlubick2689]

Yes it will.!! 🙃🙃😁😁👍👍😀😀

[@harriehausenman8623]

Bring your own fire extinguisher! 🔥🧯🚒

[@stevenlubick2689]

@@harriehausenman8623 How much do they cost? 😁😁🙃🙃 And a working clock 🕝🕝🕝🕝🕒🕒🕒🕒. You will have a time limit!! 😬😬😁😁

[@ra1nman_mashups]

Great video Domotro!

[@jordough4495]

When the bubbles show up I pinch myself to make sure I'm not dreaming

[@personal-qs6dz]

I would say the case of b=1 still works: 1 + infinity = infinity = 1 * infinity

[@Don_Albert]

I watched Forrest Gump yesterday and now this guy reminds me of lieutenant dan

[@RendallRen]

Can I come over and take a rake to that back yard? Said with ❤. Great show.

[@nate9672]

Love your videos!

[@BryanLu0]

Can you just start with multiplication and add 1s until it works? I guess this method doesn't guarantee a set length..

[@pedroff_1]

Wow, Big Joel really tried diversifying his video style

[@timknight68]

just became my favourite mathematics professor.

[@JanJeronimus]

Extend a valid seqence with e.g. four new elements -1 , -1 , 1 , 1

[@ronaldc8634]

Thanks again for more interesting facts

[@omerd602]

Honestly doesn't seem too surprising, especially since all solutions take the form of "take any set of non-1 numbers, and then add 1s to the set until the total sum adds up to the product of the non-1 numbers" It's still a pretty cool result, but I think it's valid to say every solution other than (2,2) is trivial

[@dinoeebastian]

I surprised you didn't mention how no matter how far you go the 2 and 2 will still equal 4, like 2^2, and stuff, though I think you might've mentioned it in a different video, I can't remember

[@ComboClass]

Yeah I did mention that in my video about “tetration”/hyperoperations

[@dinoeebastian]

@Combo Class ah, yeah, I remember that, also keep up the great videos.

[@catherinebernard3282]

Does anyone know the name of this sequence, if it has one?

[@ComboClass]

It doesn't really have an official "name" (maybe I'll give it a nickname sometime), but what it's describing is "for a given n, how many multisets of n positive integers have an equal sum and product"

[@morrigancollins2092]

Okay, so hear me out. I'm gonna postulate that this formula works even with a=1. But to do that, we have to attend to the uncomfortable possibility that 1/0 actually IS the identity of infinity. Here we go. A fraction is a numerator over a denominator. A numerator larger than a denominator can be reduced to whole numbers plus a fractional number by subtracting the denominator from the numerator repeatedly until the numerator is smaller than the denominator. The number of times this requires to reach a fraction < 1 is the whole number in front. This is true for all positive numbers, even if the number is an irrational one like pi, although we can't exactly calculate the real number value, only an approximation. (If you dispute this, you dispute the possibility of dividing by pi, e, pick your favorite irrational number). Since 1/0 are discrete values, 1 is greater than 0, and 0 can be called rational or irrational as you please, it still stands in as a valid denominator. The process above is an infinite loop, repeating an infinite number of times, therefore by the formula above 1/0 results in infinity. This means that not only does it hold true for a=1 in your formula, it also provides proof that for whole number values, infinity is equal to 1+ infinity, which has already been proven in OTHER mathematics dealing with comparing scales of infinity (whole number infinities, the infinite hotel, etc.

[@osmium6832]

The problem is that a number divided by 0 is not only infinity. It is also potentially negative infinity, zero, and every real number (integer, fraction, and irrational number) in between those extremes. This is why it's called undefined. You can't determine which of these equally valid series of infinite potential solutions it is. You can slap any value on it that you want to, but it'd be arbitrary. Take the graph of 1/x for example. You probably already know what it looks like, but if not you can Google it or take a look at this video around the 5:30 mark and it resembles that. What is the Y value when x is 0? Well, if you're approaching from the left side of the graph, that value goes down towards negative infinity. If you approach x = 0 from the right side, the value goes up to positive infinity. Negative and positive infinity are about as far apart as you can get in concept and magnitude, but both are equally valid answers to what x could be AT 0. If you look at the equation y = 3x/x, the x's will almost always cancel out and you're left with y = 3. You can use any coefficient, whether it's 3, -5, pi, square root of 7 billion, whatever. In this case, when x is 0, every other part of the graph is just a horizontal line at y = 3 but you have a hole where that line crosses the x axis. In this equation the reasonable guess to what y *should* be when x = 0 is 3 because that's what it is everywhere else. If you solve the problem with calculus and look at the equation when x *approaches* 0 but doesn't ever actually equal 0, then the equation is fully solvable at y=3 and you can get a line with no holes. It's not really that calculus solves the problem where other math can't, it's just that it alters the question slightly to get a sensible and useful answer. Notice that here we divided by 0 but got a normal number answer instead of infinity, negative infinity, or 0. There's better explanations than this out there. Search for "why is dividing by zero undefined?" for some more rigorous and/or easy to understand proofs.

[@StarGarnet03]

Fun fact: the solution of the sum = product question if b is the golden ratio is the golden ratio + 1

[@ComboClass]

Yup and that number (golden ratio plus 1) is also equal to the golden ratio squared! There will be a full episode someday about cool properties of the golden ratio

[@harriehausenman8623]

@@ComboClass Fun stuff!

[@JanJeronimus]

There are even more solutions when allowing imaginary numbers ( where i the root of -1 ). For example a set with 3 numbers 1i , 2i and 3i

[@Alan_Clark]

That is not a solution. The sum is 6i, the product is -6i.

[@JanJeronimus]

@@Alan_Clark Indeed, however with imaginary numbers special sequences are introduced. look at 1i , 1i , 1i , 2i, 5i

[@aonodensetsu]

@@JanJeronimus again, these add to a positive and multiply to a negative imaginary number due to the odd amount of "i"s

[@Keithfert490]

The HBomberGuy feel of this video is fantastic. The math too, obviously

[@slice-the-pi]

i love how janky your setup is!

[@Thrustmaster64]

If this classroom got covered in sediment, and some future archeologist dug it out, they would be very confused.

[@ibrahimali3192]

yes

[@5hape5hift3r]

Level 0 has 0 solutions, an empty set has a product of 1, the sum of an empty set is 0. 0 =/= 1

[@hkayakh]

Man all the different ways you can combine 2 to get 4 is cool, except 2-2

[@kenthartig7065]

But is that combining 2, or 2 and negative 2?

[@hkayakh]

@@kenthartig7065 2+(-2)

[@kenthartig7065]

@@hkayakh precisely

[@DavidSartor0]

2/2 log_2(2) root_2(2)

[@simonruszczak5563]

a = b / ( b - 1 ) b = 1 May prove something divided by zero equals infinity. a + b = a * b a = ∞ ∞ + 1 = ∞ * 1

[@Schnoz42069]

Was just thinking this

[@twelvethousandths1698]

These are so fun!

[@amirmaths3793]

Excellent 💕💕💕💕

[@quintopia]

hmm but are there an infinite number of sets (rather than multisets) of integers with this property?

[@harriehausenman8623]

Suggestion: share your geogebra notebooks! 🤗

[@ComboClass]

These graphs were actually made on Desmos (I haven’t tried Geogebra yet but will sometime in the future). I didn’t save them but they are basically just the equations that I flashed on screen before the graphs or next to them. I should start saving some graphs though, for other cool stuff I find

[@fetch7312]

idk if you still read these comments but this timing is either perfect or terrible because i just found this pattern while trying to work on a math competition practice ws for my school's mu alpha theta club

[@good.citizen]

thank you 20min video > sit through all the combo class.. why are the blues twelve measures >>> 444÷12 = 37 .

[@Jop_pop]

How come every time I watch a video of yours I have an existential crisis?

[@waylonbarrett3456]

I wonder if any of this relates to... (1/n)(1/n+1) = (1/n)-(1/n+1)

[@scali5948]

10/10 video as always

[@АндрейВоинков-е9п]

09:48 it's not a set

[@RSchef]

Great Content and funny character!

[@hrishikeshaggrawal]

y=x/(x-1) is actually just xy=1 or y=1/x translated one unit up and one unit right

[@OrangeDrink]

3,1.5 is basically 2,1 in relation

[@axiomfiremind8431]

All the sets with only one solution are divisible by 6.

[@ibrahimali3192]

*except 2 3 and 4

[@Idekaks]

I love this man

[@CharlesH-de9op]

(1,1,2,2,2) is one that works for 5 number sets and that makes 5 not unique

[@jasimmathsandphysics]

Awesome!

[@Voiding210]

1×2×e≈1+2+e

[@harriehausenman8623]

😅nice one

[@grezamisoit]

Excellent

[@jonjoncs]

This guy have the same voice as the guy form Vsauce2

[@axiomfiremind8431]

There are a lot of primes in those sets.

[@blakeflynn3454]

Did you put the RGB color of your shed on the shed itself?

[@JuanGomez-zj7db]

This is awesome

[@LeetMath]

just add 1 until it works

[@JR13751]

1x2x3 = 2x3 5x6x7 = 14x15 Are there any more that do not involve zero?

[@shiloh_moonfrost]

You can take a number with lots of prime factors and separate these prime factors in different ways

[@ThatOneCrow3]

interesting how all the "levels" with only one solution are one more than a prime number seems like that has something to do with it maybe

[@ThatOneCrow3]

assuming that 1 is a prime number lol

[@andersonseecharan2447]

DESMOS!!

[@skeptica]

All the levels with 1 solution are prime numbers +1, strange…

[@harriehausenman8623]

Wait what?!

[@RibusPQR]

1 isn't a prime number.

[@masterchief5603]

And then you have 2^2 :)

[@evenaxin3628]

a = 1, inf + 1 = inf * 1

[@evenaxin3628]

also, [-1, -1, -1, -2, -5]

[@B3Band]

inf is not a number But even if you allowed it, it's trivial, because a could be 2 also. Or 3, or 4....

[@poorman-trending]

Someone needs to call osha…

[@glowstonelovepad9294]

1 divided by 0

[@harriehausenman8623]

((Pure math) comedy!) Prilliant!

[@joyli9893]

My answer: y=1/(x-1)+1

[@joyli9893]

I used a graph