This one weird trick will get you infinite gold - Dan Finkel

[@brodeyzade1893]

Never thought I'd see TED-ed have a video titled like a Buzzfeed article

[@LuMe.Garage]

True that 😂

[@HealingSunHouse]

I literally said this must be click bait cuz ain't no way the powers that be want us to have access to free gold

[@yuuji3795]

here before this blows up

[@TheOnlyName]

Ikr

[@sanjanaakotyada4753]

😂

[@spacefun101]

For anyone wondering, I found a closed form function for the bag that you could graph on a calculator: f(x) = 2(x-2*3^floor(log_3(x)))+|x-2*3^floor(log_3(x))|+3^(floor(log_3(x))+1). It could probably simplified, but this works for all positive x where f(f(x)) = 3x.

[@iika_a]

how did you do this

[@alasdairsinclair916]

This guy maths

[@danielkellett6981]

Disappointed that there's not a nice way to put it but: When N can be written as 3^x+y where y is less than or equal to 3^x, N transforms to 2*3^x+y Otherwise N transforms to 3y So any N such that N= 3^x+y becomes 2*3^x+y = 3^x+(3^x+y) becomes 3*(3^x+y) = 3N

[@janparkki5704]

I got as far as doing all the steps indicated in the solution except extending the table far enough. I tried to force the general solution out of my brain thinking it had to be something simple and probably do with divisibility by three and modulus. I gave up 15 mins later and am happy now to notice that the proper general form isn't pretty nor totally trivial.

[@MutantChicken7397]

Ulu

[@gre3nishsinx0Rgold4]

Clicked for the curiosity of gold. But I got trapped with a math problem. Touche.

[@aditisk99]

Same

[@daredemptionn]

sameeee i thought it would be about alchemy ahahah

[@ebenezeryeboah7090]

same😀

[@asiaattia2773]

Me too, 😅

[@marckiezeender]

Me signing up for college

[@zoesato3574]

I really appreciate the time effort the animator(s) put into this. As someone who animates casually for fun this is really impressive.

[@t3li5]

It makes sense since it's their job. But of course they are really talented!

[@fruitpl7615]

fr

[@noidea2568]

1:11 that's the first time EVER I heared the TED narrator change his tone in any way.

[@debadityanath4398]

a casual one at 4:14 as well

[@youjustgotgooped]

watch a history on trial video you'll see him in three different accents at once

[@TV-gn9li]

😂😂😂

[@franziska9260]

There's also the traffic video, where he goes "why the F*** is there so much traffic?!"

[@Reletr]

For once I have thought too hard in an attempt to solve a ted-ed puzzle instead of throwing up my hands in confusion. Thank you VSauce for teaching me about Banarch-Tarski

[@mewmew8932]

Same here

[@metal_pipe9764]

Meanwhile i just went to shooting them

[@djdog120]

@@metal_pipe9764 dawg waht

[@metal_pipe9764]

@@djdog120 they can't steal if they're dead

[@lycrowkurato]

It'll probably be easier if he asked what colour were his eyes

[@aisadal2521]

If only all math tests could be like this; I'd be way more invested if they were like this

[@carealoo744]

Still didnt know how to solve it though.

[@K__kelly]

@@carealoo744 me too

[@kompatybilijny9348]

You would get 5% enjoying them and the rest getting frustrated and depressed.

[@amberwerwolfschool8927]

Ikr! The answer should be on the back of the paper! >:( Oh that's waht u meant-

[@erikaz1590]

Yes, if only all math tests had the teacher wearing all the answers on their clothing, so you just need to figure out where to plug everything in

[@noahahmed5821]

I’m a huge fan of the art style of this video! Please have this animation team back!

[@ag-13studios51]

Same. I'm glad I'm not the only one who enjoyed the art style and the animation; very cartoony but also very unique and fun (and other descriptions I can't figure out)

[@Penguinmanereikel]

I hope whoever animated this get jobs at big studios, because I found it so pleasing to watch.

[@nawel991]

yes! it was a mixture of a fairy tale and riddles I used to do when I was a child, it took me back in time, we live for this kind of moments 🤩

[@Airton2]

btu if they get jobs at big studios, we won't have as much ted-ed animations as we have today.....

[@HuskyObscura]

It’s like Netflix level animation, which isn’t that good. I mean the animation is alright, not my favorite at all

[@jehadal-kourdy3129]

1:42 two ez that's twenty to

[@Stickman_Productions]

All I could find was Movult Website

[@LethalPigeon7]

Trying to figure out the general formula, and finding out the answer is "you have a set number, just brute force it until you get to it" is incredibly dissapointing, but, incredibly on theme for riddles, where misdirection and unusual ways of thinking are common tools. Cool.

[@imperator9343]

I actually like this more than just "write the problem as equations and solve with algebra". My issue with most of these riddles is they're just math problems with a thin aesthetic veneer of a puzzle. This one actually requires you to use more basic logic than just "use what you learned in 5th grade" or whatever.

[@subhajitsarkar2272]

Exactly ! I am in depression now..haha

[@Gamesaucer]

There has to be a general formula. Just because it's not easily described in purely mathematical form doesn't mean it doesn't exist. Just like in Fibonacci where "each number is the sum of the previous two numbers" doesn't let you create a formula going from one number to the next, nor can you easily find the Nth number without calculating all the ones before it. If you write down everything in the form A -> B, you just get an index on the left where you can say "the Ath magic bag number is B", and so this is just a series of numbers like any other, just one with a somewhat complicated ruleset. The only remaining question is whether it's infinite or not.

[@imperator9343]

@@Gamesaucer I'm not sure that this completely makes sense. It is entirely possible to create a sequence like this that has no "general" formula, and saying that it might not be infinite (depending on what you mean) contradicts the existence of a general formula. This puzzle was solved using the constraints of a finite band of possibilities. I actually don't think that there is a generalizable f(x) for all x formula to this solution. If the strange man had made this request using, say, 20 coins, I don't think that there is a singular logical answer that can be derived from this premise alone. The constraint that all results must be whole numbers (coins) means that any sort of consistent linear or exponential solution can't be generalized for all numbers. It would require either: a) specific rules for specific sets of inputs, and those rules would necessarily require a number of conditions on the order of the number of inputs (i.e. countably infinite), or b) an initial scheme followed by a "simpler" formula for the rest of the inputs. (a) is not a "general" solution, and both (a) and (b) would not be unique, you would be able to come up with multiple valid ways of constructing it (assuming (b) is even possible, which I'm pretty sure it isn't having played around with the algebra myself).

[@vaxjoaberg]

@@Gamesaucer You might find "Binet's Formula" interesting.

[@user-lf5xq4gu1g]

The way she just DROPPED HER BABY

[@aditisk99]

Yeah 😂

[@asheep7797]

0:59 here

[@GunNNife]

Yeet the baby!

[@raymundolancealfreds1050]

"Were you dropped in the head as a baby" Yes

[@netherite9051]

HAHAHAHAHAHAHHAHAHAHAHAHAHAHHAH 🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣

[@ricofilberto404]

For those who are wondering how to calculate the function without using computation table, here's the function looks like: Lets say k is the largest integer that satisfy 3^k

[@willnewman9783]

For completion, one should also show that there is a unique function satisfying f(f(x))=3x. I do not think this is clear.

[@lennyarms4476]

can you explain the thought process like how did you come up with this

[@ricofilberto404]

@@lennyarms4476 I used a program to generate the computation for the first 1000 value, and then I notice some pattern in f(x)-x, there's a lot of 3^k term showing up.

[@ricofilberto404]

@@willnewman9783 It's easy to compute f(x), for x = 3^k or 2 * 3^k, for example: f(1) = 2 f(2) = 3 f(3) = 6 f(6) = 9 f(9) = 18 f(18) = 27 f(27) = 54 f(54) = 81 f(81) = 162 And now we want to fill in the gap for the value that hasn't be computed, lets see at the gap between f(3^k) to f(2 * 3^k), for example f(9) to f(18), we can uniquely find value for f(10) to f(17) since we know that f(x) must be increasing and produce integer value (for x is positive integer). So f(10) = 19, f(11) = 20, ..., f(17) = 26. Since know we have a function that result in f(x) = 19 to 26, we know can also compute f(x), with x = 19 to 26, since f(f(x)) = 3x. So f(19) = 30, f(20) = 33, ..., f(26) = 51. With this it's can be easily seen that the function f(f(x))=3x can be computed uniquely (with the rule that it's must be increasing and produce integer value).

[@KeithDePew]

@@ricofilberto404 Why is the solution not 26? This would also fit the riddle as far as I can tell if the progression is: Y goes in and the magic adds Y, then 2 Y comes out, when 2Y goes back in, the magic adds Y again, and 3Y comes out...done. As far as I can tell, the riddle only stipulates the rule for two uses of the magic, and not a continuing progression, so this would work for all cases and satisfy that more comes out when more is put it. What am I missing?

[@kriticanamchu4128]

"Also, it's on the back of your shirt." 💀

[@axolotlo2]

bro was flabbergasted 💀

[@jhunelmullanida6852]

😂😂 made me laugh

[@ashwindongre5918]

🤣🤣🤣

[@JoshRendall]

@@ashwindongre5918 I know! That was funny!

[@kyslf1486]

Someone's is snitching on bro. 😂

[@martinarychtarova5339]

4:19 my man got drafted for squid game 💀

[@kingwolf3044]

New lore for the riddleverse? Yay. Clearly she figured it out because he has green eyes which means if he saw two frogs and one said ozo the fuddly must have used the tri source to make the bag for him.

[@pingpong2563]

its always quite unfortunate when your crew of five pirates need to take turns crossing a bridge with only a single lamp to get fuel for your aircraft :(

[@MichaelDarrow-tr1mn]

wow 5 riddles in one sentence

[@redthefoxisWritingUpAStorm]

Wha?

[@MichaelDarrow-tr1mn]

@@uncreative172 6 riddles wow

[@kingvideogames]

Also, it was all on the his nametag

[@millylitre]

Extending the video solution to larger numbers reveals what might be an infinite stack of interleaved sequences, that in total fill the space of all positive integers. Each sequence begins with two values. The pairs of values for starting the first eleven sequences are: (0 0) (1 2) (4 7) (5 8) (10 19) (11 20) (13 22) (14 23) (16 25) (17 26) (28 55) (29 56). To generate the next terms in each sequence you take the last but one number and multiply by three. So for example the sequence starting (4 7) continues as 4, 7, 12, 21, 36, 63, 108, 189, 324 .... The puzzle as posed, for 13 coins, is the start of the seventh sequence 13, 22, 39, 66 etc. This all feels rather Fibonacci. Another presentation of the results is to list in sequence the numbers of coins that come out of the magic bag if you put into the bag 0, 1, 2, 3, 4, etc coins. That sequence starts 0 2 3 6 7 8 9 12 15 ... and (at least) the first 28 terms match "The On-line Encyclopedia Of Integer Sequences" - sequence number A003605.

[@ElfstoneAuriga]

In the Encyclopedia it says "Unique monotonic sequence of nonnegative integers satisfying a(a(n)) = 3n." -- Which is exactly what this is. I also generated the sequence, and found a web page displaying the first 10,000 terms of the sequence, such a curio.

[@karthikeyan020]

And a new sequence is generated by taking a number that doesn't exist in any existing sequence and adding nearest lower power of 3. For e.g. 3^0 for less than 3, 3^1 for less than 9 and so on.

[@chessematics]

This was the most creative way of presenting Banach-Tarsky.

[@TV-gn9li]

😂😂😂

[@kyro7482]

The tone of this one was soo completely different from the usual riddles! So many quick jokes and character breaks, it was very funny

[@JasonMomos]

*Me who clicked hoping to make infinite gold:* I've been tricked, I've been backstabbed and I've been quite possibly, bamboozled. My disappointment is immeasurable, and my day is ruined. 💀

[@carlosgutierrez6970]

You know if u actually looked at the thumbnail you know it was a riddle

[@TV-gn9li]

Yeah

[@computerwundsam]

0:57 did she drop the baby?

[@malininatarajan7948]

Yeah lol

[@getthedunkon9347]

Nah, you crazy.

[@San-lh8us]

you guys should sell pedagogic courses to the schools of the world, because if we were introduced to science and math like this, so many more people would love learning

[@hakimdiwan5101]

Won't be enough for me

[@eBrunoro]

But they do, it's the video's sponsor

[@Bhuvan_MS]

The explanation went above my head lol.

[@cozmin8752]

Did anyone else notice that his name tag disappeared 1:44

[@angelaliao9167]

I love ted-ed riddles... never stop

[@kingwolf3044]

Agreed. What’s your favorite

[@catoctober8005]

Agreed

[@immyownperson1375]

Thought it was a hack, then thought it was a fairytale animation. It turned out to be functions chasing me to the internet 😂

[@johnschmidt1262]

I love this puzzle, it also reminds me of one thing I've always questioned about the what number comes next riddles. At the end of the day you can always make more complicated formulas that will go through all those points. What's more as in this case technically there doesn't have to be a formula at all, a function can simply be looked up from a table. Implicitly they're asking for the simplest continuous function in most cases. But they should say it explicitly to teach the kids what's going on.

[@HyronXVI]

The summary pause screen lacks the fact that you have to put all the coins in the bag in order to make the "always tripled" rule work. With that left out, "using it twice" is still unclear if in the second round you put all the coins or only the initial ones (in which case is trivial since it would be x2)

[@pragyabiswas1562]

Exactly

[@Akronox]

Agreed that I had to rewatch the part before the summary to confirm this.

[@kwaddell]

Ok yeah, I was confused because I got stuck thinking I’m just putting the initial amount back in, which then simply lets you double, then triple that initial number, so I settled on 26

[@SFH2042]

I can't believe that this is the first puzzle I haven't failed!

[@kingwolf3044]

Congratulations

[@Cora.T]

Mind explaining me how it works? Because I can for the life of me not figure it out, like it feels like the explanation makes even less sense

[@andrejors9501]

@@Cora.T the explanation is really clear tho.. you don't have to use any mathematical formula, its just pure logic.. you simply just start from 1-2-3 and keep on doing that until you got 13

[@Cora.T]

@@andrejors9501 how though?? Why is it 3,6,9 and not 3,4,9? Or 3,5,9?

[@Aagames_]

@@Cora.T 1 goes to 2, which goes to 3. 2 goes to X, which goes to 6. Based on the first sentence, X equals 3. Because of this, 3 goes to 6, and since the outcome is triple, 3 goes to 6 goes to 9. The same logic can apply to the rest.

[@bowboi]

Bruh, can't y'all just put the baby in the bag and get like, 2 or 3 babies? Then the Tarski guy can take the other two away, and everything would be fine!z

[@nawel991]

im not sure I got this right are you speaking bunny dialect?

[@mick4563]

That probably wouldn't be ethical to the babies though.

[@luxtempestas]

The bad part is that you would have ended up with one more baby too! 😂

[@bowboi]

@@nawel991 ah yes sry it's ohio dialect so i was saying just duping the baby would work if the guy wants it

[@bowboi]

@@luxtempestas you'd have to use the magic twice to do that, the 1 coin was turned into 3 after the magic was used twice.

[@Sir_shorts_a_lot]

1:44 everyone else: what a good riddle. Me: bros name tag went to the backrooms

[@ManicDymanic]

fr i just saw that 💀

[@aisadal2521]

I'm getting Infinite 1-Ups from jumping on a Koopa Shell vibes from this 😂

[@NOTZeroBlank]

3D world was amazing

[@ChaosZTC]

@@NOTZeroBlank fr

[@unrellated]

Banach-Tarski: Has a bag that gives him infinite gold. Also Banach-Tarski: Can't afford to forgive your debt.

[@waiyisit]

He just wants to make people suffer.

[@Sumirevins]

I love TED-ed riddle series, I am not able to solve these but I do always like watching them. It fascinates me😂

[@ehtacoguy4079]

1: Read the back of the little man's shirt. 2: Confirm you have green eyes 3: Ask the guard if you can leave 4: Steal the secret sauce recipe 5: Lick the male frog 6: Pick the Churrozard disk 7: Cheat death 8: Get your guitar from the drumset box 9: drop the worthless egg from story 34 10: Separate the Fire dragons from the Ice dragons 11: Write down the jousting tournament scores 12: Ask "If I had a burrito for lunch, would you say Ozo"? 13: Light all of the giant's birthday candles 14: Keep the Keystone 15: Put in the charged batteries in the giant iron 16: Cut the werewolf antidote into five squares 17: Make the Professor and the Janitor cross the bridge together 18: Program the multiverse teleportation robot 19: Stop going on youtube because you watch WAY too much Ted-Ed riddles 20: Have a nice day!

[@josephwade8593]

How dare they trick me to doing math at 6:00 in the morning

[@Pyrogecko08]

I thought I had a different solution, but it turned out to be wrong. My idea was to multiply odd numbers by 2 and multiply even numbers by 1.5, and in two uses that does triple most numbers you put into it, but it turns out that it fails when you start with a multiple of four.

[@christiannielsen725]

Got the same “solution”. How does it fail again?

[@Pyrogecko08]

@@christiannielsen725 because two uses only triples most numbers, but it is supposed to triple any number you start with. Starting with four, or any multiple of four doesn't work the way it's supposed to.

[@christiannielsen725]

@@Pyrogecko08 yeah ok, ty

[@dwaraganathanrengasamy6169]

​@Pyrogecko08 mate, I thought the same and worked around it. Let's say we are currently having x gold coins. 1) If x is odd, we double the coins. 2) If x is divisible by 2 but not 4, multiply x by 3/2. Following these 2 steps, we will have almost all numbers in its distinct loop. Note that if number x already lies in a loop, next number is found by tripling the number at previous step of the loop. Initially, we might have left a few numbers that are divisible by 4. Form loops of 2 with such nearby numbers and continue the loop by tripling the previous element. Doing this, we are able to uniquely map each element in its distinct loop such that the number is always tripled 2 steps ahead. Loops for first few numbers are, 1 - 2 - 3 - 6 - 9 - 18 - 27 -... 4 - 8 - 12 - 24 -... 5 - 10 - 15 - 30 -... 7 - 14 - 21 -... 11 - 22 - 33 -... 13 - 26 - 39 -... 17 - 34 - 51 -... 19 - 38 - 57 -... This logic too works. So, it makes me wonder, are there infinitely many such possible functions on whole numbers which when applied twice, triples a number. So the next number looped with 13 is 26 AM I WRONG ANYWHERE..?!

[@pianoplayer1262]

The issue with this strategy is 7->14 but 8->12 which is less. This is due to using two different multipliers, as two adjacent numbers will get mapped non-monotonically (fail to satisfy the condition that the more you put in, the more you get out).

[@jeconiahjoelmichaelsiregar7917]

I can't be the only one who realized the riddle's backstory references Rumpelstiltkin, even all the way up to the lady promising the little man his firstborn and the little man riddling her to guess his name.

[@pasta_eeee]

proud to say i noticed this too:D

[@janTesika]

it's only one of my the most famous fairy tales in the world, of course you aren't the only one who noticed. I hope... or maybe I'm just one of very few people who read Grimm's Fairy Tales...

[@Inkyminkyzizwoz]

Must admit I didn't remember the bit about the baby

[@elfpiesomeanotherword]

I only watch the video show I will try to get Grimm Fairy Tales though to read @@janTesika

[@spooks188]

Serious question, why do the numbers in the second column have to increase? What's stopping 4 from becoming 5 and 5 becoming 12?

[@carlosgutierrez6970]

It's honestly really satisfying having our character have a line in 1 of these let alone 3-5

[@SonicLoverDS]

So if I understand correctly, this is a function defined by induction, meaning f(x) is defined in terms of f(y) where y

[@jaromeltuzerad9267]

"I'm thinking of a number between 1 and 3, not including 1 or 3." "M!"

[@paul7944]

Function ( y(n) ): 1. Calculate ( x = floor(log3(n)) ). 2. If ( n ) is within [ ( 3^x ), ( 2 * 3^x ) ] : [ y(n) = n + x ] 3. Otherwise: [ y(n) = 3 * (n - 3^x) ] I put it in the simplest way possible :)

[@AA-100]

x = 3^floor(log3 n) y(n) = n + x + 2(n mod x)(floor(n/x - 1)) Edit: even simpler is y(n) = 2n - x + |n - 2x|

[@jonathanlevy9635]

what's even more exciting about this function is that it is describes it's value for every integer, as big as we want, with those few simple rules. also, it's easy to notice it is altering from jumping 1 step at a time to 3 steps at a time in increasing lengths. One can explain this last phenomenon geometrically by looking at it between the graphs of f(x)=x and g(x)=3x. When drawing a line from the x axis at one point (n,0) to it's value obtained by the said function to (n,h(n)) and then taking a perpendicular to (h(n),h(n)) you can always close continue it by taking it to (h(n),h(h(n)) and the great surprise is that the alternating behaviour of the function leads it to return back to (n,3n)=(h(h(n))/3,h(h(n)) by going back horizontally to g(x)=3x! this is much better explained visually so I highly encourage anyone who wants to understand how this function behaves

[@laincoubert7236]

yeah, i was surprised there's only one function f: N -> N that satisfies those rules

[@Muhahahahaz]

@sussykanyeballs176f and g refer to 2 simpler related functions, and h refers to the function from the video h(n) = #coins after using the magic bag on n coins (once)

[@DangAlfa]

If we simplify the function as an algebraic function, and the multiplier as √3= 1.726. we still get 22 removing the fractional part.

[@melodeyh970]

*solves a complicated riddle* "..also it's on the back of your shirt."

[@andrejvelickovski639]

x is the starting value of the coins, y is the multiplier used in every action we start with the base X and multiply it with Y (First action) afterwards, we multiply it again with the same multiplier as in First action (Second action) -> with that said we can create the formula y^2 * x = 3x with easy access to y we can solve that the multiplier is 1.732, which answers the question and saves the baby.

[@rioc2802]

Doesn't work. 6 * 1.732 = 10.392. 10 * 1.732 = 17.32 Your formula says that 6 * 1.732^2 should equal 18, therefore putting 10 coins in the big should get you 18 coins back. However, your formula also says that 10 coins put in the bag also returns 17 coins, which contradicts the first rule saying the bag works consistently.

[@yokaiwatcher8500]

“Sometimes an enigmatic man is going to pose you a riddle, that’s life” Who do you think I am Ted-Ed, Professor Layton?

[@Mynonono]

“Fortunately for you, a strange little man appeared.” 😂

[@cardinalhamneggs5253]

I love that his name is Banach-Tarsky, after the mathematical paradox that theoretically allows you to split a sphere into 2 spheres which are perfectly identical to the original.

[@isaacsayolpiedra5563]

Would this reasoning also be valid? n = the amount that goes in each time. - if n is even, multiply it by 1,5 - if n is odd, multiply it by 2 13 --> 26 --> 39 12 --> 18 --> 36

[@Thaplayer1209]

Rule 2 says that putting in more coins will give out more coin than in you put in fewer. e.g. putting in 4 coins would give you more coins than putting in 3 coins. But you solution has both 3 and 4 giving 6 coins.

[@relpi7538]

At first I thought it was some myth story. But it was my favorite riddle video again. Thanks!

[@k3ose45]

The Banach-Tarski paradox-The idea that any whole object can be cut in such a way that it can be put together into two objects congruent to the first

[@voidwarden1413]

i think they already patched it, doesn't work

[@adventureboy444]

I thought this is gonna be a story time and end up letting my guard down

[@gabrielamaral978]

I would have answered correctly even tho i did the wrong way. I've multiplied 13 for square root of 3, it was 22,49... And i like "Oh, so the bits of gold just don't get generated into a coin, the bag will not have enough gold to make 23, so the answer must be 22"

[@chixenlegjo]

Haven’t watched the solution, but if you put in n coins, one possible (but not necessarily the only) rule could be that you get back (7n-cos(πn))/4 coins. n=13 evaluates to 26. Edit: I did not see that the function must be strictly increasing.

[@PaulBaird-Smith-h5x]

Why isn’t the answer f(x) = { 3/2x for x even, 2x for x odd } where x is the number of coins put in the bag? In the even case, x (even) -> 3/2x (odd) -> 3x. In the odd case, x (odd) -> 2x (even) -> 3x. This is a strictly increasing function that will always triple after 2 uses. The answer according to this would be 26.

[@syphon5899]

Exactly my thoughts, and because of such, I have become confused as to why the answer was 22 instead of 26 As a arithmetic series

[@Hendrik_F]

With those rules you also get 4 -> 6 -> 9, which doesn't triple the initial number@@syphon5899

[@suspiciousshadeofgreen]

That "Also it's on the back of your shirt" at the end killed me

[@vex3091]

actually root of 3 does work, if you assume its around 1.7, it comes out to 22.1 coins, which rounds up to 22.

[@CastoriAlter]

.... But... 13 times 2 is 26....

[@capnfail5807]

"Also, it was written on the back of your shirt," killed me. This is the worst-prepared magical baby snatcher I've ever heard of

[@mathguy37]

Never thought I’d see “worst prepared magical baby snatcher” in my life but here we are

[@ahalfeatenpotato463]

I'm like half sure this riddle has slight nods to Rumpelstiltskin. Could just be me being a literature nerd though.

[@kingwolf3044]

Pretty sure it’s the inspiration. There’s a lot of obvious inspiration in the riddleverse

[@timeme5460]

it definitely is

[@chashubokchoy8999]

not even inspiration, it’s literally a rendition of it

[@thenovicenovelist]

I thought of Rumplestiltskin too. Maybe they're related 😉

[@mrosskne]

only a literature nerd would know the very obscure tale of ... Rumpelstiltskin

[@erikaz1590]

Okay wow, I got that wrong. I thought it was 26, since if it triples at 2, maybe it only double at 1. But then they brought out the 1-2-3 into 2-3-6, so I started writing that out and went 'oh, okay maybe it's a 'if,then' computer function, where if your starting number is odd, you add itself, but if it's even, you add half of itself. But that doesn't work for anything not in the 3's family. tldr, I would definitely have needed the 3 guesses....or knowing that this guy keeps using his answer keys as his clothing fabric.

[@sahasrakondapalli50]

Ya, I thought you would multiply by root 3 until the no bits rule came up, then I tried to associate it to sets since in those relations, you don't need to know the function itself and this is how they get the answer, but my mind drew a blank and went: 26, even though you initially disproved it.

[@ΣοφίαΣτουπά-ρ2τ]

I had a similar idea, I thought that if you put in some gold then you get that gold doubled but if you put that same gold in again you only get that initial amount doubled so in this example it would go 13, 26, 39, 52 and so on. But doesn't make much sense so yeah

[@cavox5205]

I got the same idea but then realized any number divisible by 4 wouldnt work lol

[@dwaraganathanrengasamy6169]

Guys, I thought of the same idea and worked around it. Let's say we are currently having x gold coins. 1) If x is odd, we double the coins. 2) If x is divisible by 2 but not 4, multiply x by 3/2. Following these 2 steps, we will have almost all numbers in its distinct loop. Note that if number x already lies in a loop, next number is found by tripling the number at previous step of the loop. Initially, we might have left a few numbers that are divisible by 4. Form loops of 2 with such nearby numbers and continue the loop by tripling the previous element. Doing this, we are able to uniquely map each element in its distinct loop such that the number is always tripled 2 steps ahead. Loops for first few numbers are, 1 - 2 - 3 - 6 - 9 - 18 - 27 -... 4 - 8 - 12 - 24 -... 5 - 10 - 15 - 30 -... 7 - 14 - 21 -... 11 - 22 - 33 -... 13 - 26 - 39 -... 17 - 34 - 51 -... 19 - 38 - 57 -... This logic too works. So, it makes me wonder, are there infinitely many such possible functions on whole numbers which when applied twice, triples a number. So the next number looped with 13 is 26 AM I WRONG ANYWHERE..?!

[@ilyakam]

@dwaraganathanrengasamy6169 I came up with the exact same logic as you. Weird to find it this deep in the comments. It works for all cases as far as I can tell.

[@itsamemoo1372]

I love the animation! It looks so refined and animated compared to the old style 😀

[@Agamemnonoverhead]

The secret to the bag is that there's a mini universe inside of it and he's giving the little people in there loans at a 200% interest rate

[@KingMatthewXV]

Can you solve the infinite gold riddle? would be a better title.

[@AA-100]

There is actually a closed form solution for x number of coins (however it's quite complicated and much simpler forms may exist) First let y be the value of x rounded DOWN to the nearest power of 3, or y = 3^floor(log3 x) E.g. If x=3, y=3 If x=13, y=9 If x=80, y=27 And now the result f(x) from x coins is: f(x) = x + y + (2 × (x mod y) × floor(x/y - 1)) Note the result is usually just x+y since the right part usually becomes 0 Edit: Much simpler form is: 2x - y + |x - 2y|

[@Klick404]

I love the art style in this one! The animation is stellar

[@nickburns4341]

The sequence f(x) where f(f(x)) = 3x, can also be described as the list of numbers who's base 3 representations begin with a 2 or end in a 0 (starting from 2). Proving the equivalency is left as an exercise to the reader.

[@jaliyahkane5127]

How would 1-2-3 work if they said more gold in will mean more comes out? If you put 1 in, you get 2 out. Meaning you get a gain of 1 coin. So if you put in 2 coins, you must need greater than 1 coin to come back out which must be greater than 3? That’s how I interpreted the problem

[@kohwenxu]

They provided an example. (In 1:54, when they showed the rules.) Basically the number of coins you get back putting 3 coins in the bag has to be greater than the number of coins you get back putting 2 coins in the bag.

[@johnwhinston4626]

@@kohwenxu but it should work the same for 2 coins vs 1 coin and 3 coins vs 6 coins but in both cases you get the same amount of additional coins as the previous term in the series which breaks the rule

[@johnwhinston4626]

@Personal Anonymous 1:58 rule #2 at best it's bad wording at worst it's straight up false

[@nostillno1912]

Wait... why cant you just put the coin in once to double it and putting it in twice would triple it? Is he expecting three alternate solutions to his riddle so he gives you three guesses to guess the actual amount in his hand, or does this somehow break one of the rules?

[@nHans]

I don't want to spoil this for anyone. But the "one weird trick" he's talking about-is to gain access to a magic bag that increases the number of gold coins that you put into it. And since it increases the gold coins only by a finite amount, you also need an infinite amount of time (and/or the ability to work with the magic bag at infinite speed). So-two weird tricks.

[@tristanridley1601]

With the price of gold today, a bag that doubles your gold (if you put in the right number of coins at a time) with only the effort of picking it up and putting it down sounds close enough to infinite.

[@nHans]

​@@tristanridley1601 Agreed, any of us would be ecstatic to possess a gold-doubling bag 😂. But it's still a *_magical_* bag, right? Can any *_existing_* technology or practice do the same - *_consistently?_* Bank accounts, stocks, bonds, mutual funds, ETFs, commodity, forex, derivatives (futures, options, CDS, CDO), hedge funds, private equity, angel investing, entrepreneurship, real-estate, tulips, art, collectibles, crypto, NFT ... anything?

[@tristanridley1601]

@@nHans Doubling vs inflation? Not reliably. :P I actually wrote a dialogue once about how *any* magic that breaks the rules of physics could be used by a science-based society like ours to get basically infinite gain. Magic is magic, and magic plus cleverness is OP.

[@Winnable_ALU]

"This one weird trick will get you infinite gold" : When Ted-Ed uses clickbait

[@JustinGrant-jf5gv]

I enjoy these riddle videos for not just the riddles, but also the scenarios that come with them. My guess for this riddle was 26 coins, well I was close.

[@Muhahahahaz]

Unfortunately, that doesn’t work because the bag has no way to “know” you started with 13 coins before the doubled magic (rather than 26 being your starting point) You want 13 -> 26 to yield 39, but if you apply the same reasoning when starting with 26, then you would be wanting to triple it via 26 -> 52 -> 78 This is inconsistent, since 26 cannot yield both 39 and 52 at the same time

[@James2210]

1:44 dude's nametag just nopes out of there

[@ANormalChannel-TheFirstOne]

"also its on the back of your shirt" got me rolling on the floor

[@ΚΩΝΣΤΑΝΤΙΝΟΣΚΟΤΣΑΡΙΝΗΣ]

How do we know that if we put in 2, it will give us 3? Can't it just be 4 or 5? And how do we know that 6 gives back 9?

[@uknownada]

This one is fun because it tricks you into thinking it's a math puzzle. But it isn't math, it's just logic!

[@zmaj12321]

People often say that, but IMO math and logic are too similar to make a clear distinction. For example, even though the function given in the video was never completely defined, you can use math to figure out the exact nature of the function (indeed, many people have done this in the comments).

[@jonathanlevy9635]

well, this is math

[@Muhahahahaz]

Can it be solved with basic algebra? No But it’s still math. Not everything can be solved with a simple formula (though there is a rather simple algorithm to solve the puzzle for any number of coins, it’s just not a closed algebraic formula like some might assume)

[@vaxjoaberg]

I thought I had it by defining a piece-wise function: f(x) = 2x, if x is odd; 3x/2 if x is even That function meets all the requirements except for the second one ("the more gold comes in, the more comes out"). In my system, f(5) = 10, but f(6) = 9. Since 9 < 10, it fails. I'm a little disappointed that the answer wasn't a nice, elegant closed-form expression but rather merely a set of arbitrary mappings. I guess that's why I'm more interested in mathematics than magics.

[@MichaelDarrow-tr1mn]

it is closed-form, it's just piecewise

[@spacefun101]

This also wouldn’t work for multiples of 4 because you would end up multiplying by 1.5 twice

[@maxwellhavoc6996]

Well, you’re in a real pickle 0:07

[@DumAndSmart]

No she’s in a fake one

[@maxwellhavoc6996]

Hmmmm I seee

[@Woodledude]

Yeah, this one was nuts. I saw a comment saying "I thought too hard," and I think the conclusion I've come to in my own attempt is that I, instead, didn't WORK hard enough. I found a pattern starting at 1, as the video suggests to do first: 1 -> ?? -> 3 must have 2 as an intermediate step. And once you have three steps of a pattern, you can just extend it - So, 1, 2, 3, 6, 9, 18, 27, 54, and so on. Just triple the number of the step previous to the current one to get the next step. There is a problem, of course - The "trivial" pattern skips right over 13 and never looks back. So, I tried 4 and 5. With 4, I ran into trouble immediately. 4 MUST go to 12 - That, and I'd noticed a convenient pattern in the "trivial case". That being, it always alternated between doubling, and multiplying by 1.5, and it also alternated between odd and even numbers. 4 immediately breaks this. It is an even number, half of which is also an even number. So, it must multiply by 1.5 instead, right? Except that gets us 6. And then we have to get to 12 on the next step to successfully triple 4, but that would imply that 6 goes to 12, and we know FOR A FACT from the trivial pattern that 6 goes to 9. You can't get a different result out of the bag putting the same number of coins in, so 6 is a bust. So I applied the pattern the other way - 4 goes to 8, I guess, which then goes to 12 via 1.5x, then we continue on to 24, 36, 72, 108; it all seems to fit the pattern. I move on to 5, and find a pattern that matches pretty easily - 5, 10, 15, 30, 45, 90. And it's fitting my nice little "alternating between 1.5 and 2 multiplication" trick. So, I've run into a few snags, ironed them out, and now I'm seeing more work ahead, and I think to myself "Do I really need to do anything else? I'm not sure I totally understand the case for even numbers, but 13 goes to something goes to 39 has two odd numbers, it should probably fit the trivial pattern. Multiplying by 1.5 gives us a partial coin; multiplying by 2 gives us 26. I've been avoiding 26 like the plague, because he wouldn't be so smug if it were that simple or that easy to guess, but I did the math. It's the only thing that makes sense." No. No, I did not do the math. I tricked myself into thinking I'd done enough math, without actually committing to the rigor, and I failed to check my answer. I would have needed more than one guess. Brute forcing it would have been more effective, and I let myself take a risk I didn't technically need to by demanding the universe conform to simple rules, instead of understanding it for what it was. A VERY good riddle, I'd say.

[@adityadutta5324]

One of the few ones I managed to solve. At first I tried to figure out the logic behind the function but couldn't so I just ended up doing it exactly the same way showed in the video. Genuinely can't believe they did it the same way. Never felt smarter 🙈🙈🙈😍😍😍

[@physicbrain6318]

but that was no functional connection but a logical one...

[@shadowmancy9183]

My answer was 26, and also seems to satisfy the wording of the riddle. 13-26-39, and as the third time is always triple, and more coins yield more out, then 1-2-3, 2-4-6, etc still hold true to the wording of the riddle.

[@tacohouse01]

same

[@diedoktor]

does 2 become 3 or does 2 become 4?

[@MCTrapsandTutorials]

Your solution doesn't follow the consistent rule stipulation. If 1-2-3 works then 2-4-6 does not, because for the first one you put in 2 and got 3 and for the second you got 4.

[@shadowmancy9183]

@@MCTrapsandTutorials 3 is triple of 1, 4 is double of 2. It's consistent.

[@tacohouse01]

@@diedoktor well we didnt understand that part had to be considered

[@mikekazz5353]

She just drops the baby on the ground.

[@sketchyskies8531]

The title sounds like a video game hack Edit: I actually came really close this time

[@nawel991]

▶

[@halocraft314]

I love this animation style! It's so dynamic, and it's very consistent. I hope you use this style more!

[@tomwilkinson7139]

I would really like if they brought back the demon of reason

[@EthanConstantinescu-nl1nz]

the title makes this video sound like some sort of clickbait "this one weird trick will solve your problem" "do these 5 things to get this" "click this video if you have x y z"

[@mmayne0dadaydreamer873]

Glad the baby is ok after she drop it 😂

[@DoctorRobertHand]

I started out the same, but then noticed a pattern of: 2x-0, 2x-1, 2x-0, 2x-1, 2x-2, 2x-3, 2x-2, 2x-1, 2x So I extrapolated the next loop down from 2x-0 to 2x-5 and back again, making 13 land at 2x-4, or 22.

[@jacobgoldman5780]

I understand why 1 coin becomes 2 then 3 but why does 2 become 3 then 6 not 4 then 6 or 5 then 6?

[@mwolfe99]

Because of the first rule. "PUTTING IN A GIVEN AMOUNT AT ANY POINT WILL ALWAYS PRODUCE THE SAME RESULT." So if 1 -> 2 -> 3 (which is the only possibility given "more goes in, more comes out" and putting in 1 coin over two iterations must be 3*1 = 3) that implies by the first rule that putting in a 2 will always produce 3. And we also know that putting in 3 must produce 6 because putting in 2 (which always produces 3 because of above) and then putting all coins back in again must be 3x the number of coins you originally put in, which is 2 * 3 = 6. So 2 -> 3 -> 6. And so on.

[@rampion]

I definitely didn’t get this right on my first try (I had to finish the video to find my error), but I eventually found a piecewise definition for the bag function, f(x) f(0) = 0 f(3ᵃ + b) = 2•3ᵃ + b, 0 ≤ b ≤ 3ᵃ f(2•3ᵃ + b) = 3ᵃ⁺¹ + 3•b, 0 ≤ b ≤ 3ᵃ Note that the piecewise definition is consistent on the overlapping cases where b = 3ᵃ as f(3ᵃ⁺¹) = 2•3ᵃ⁺¹ = 3ᵃ⁺¹ + 3•3ᵃ = f(2•3ᵃ + 3ᵃ), and f(2•3ᵃ) = 3ᵃ⁺¹ = 2•3ᵃ + 3ᵃ = f(3ᵃ + 3ᵃ). It’s easy to see from the definition that f(x) > x for all x > 0. It’s less obvious that f(x) is monotonically increasing (that is, f(x + 1) > f(x) for all x, or “the more coins you put in, the more you get out”). The overlapping cases of f(x) make proving this easier, though. If x = 3ᵃ + b, 0 ≤ b < 3ᵃ, then f(x) = f(3ᵃ + b) = 2•3ᵃ + b f(x + 1) = f(3ᵃ + (b + 1)) = 2•3ᵃ + (b + 1) since 0 < (b + 1) ≤ 3ᵃ so f(x+1) - f(x) = 1 > 0. If x = 2•3ᵃ + b , 0 ≤ b < 3ᵃ, then f(x) = f(2•3ᵃ + b) = 3ᵃ⁺¹ + 3•b f(x + 1) = f(2•3ᵃ + (b + 1)) = 3ᵃ⁺¹ + 3•(b + 1) since 0 < (b + 1) ≤ 3ᵃ so f(x+1) - f(x) = 3 > 0. Finally, to prove f(f(x)) = 3•x, it suffices to examine it by cases. If x = 0, then f(f(x)) = f(f(0)) = f(0) = 0 = 3•0 = 3•x. If x = 3ᵃ + b with 0 ≤ b ≤ 3ᵃ, then f(f(x)) = f(f(3ᵃ + b)) = f(2•3ᵃ + b) = 3ᵃ⁺¹ + 3•b = 3•(3ᵃ + b) = 3•x. If x = 2•3ᵃ + b with 0 ≤ b ≤ 3ᵃ, then we know 0 ≤ 3•b ≤ 3ᵃ⁺¹, so f(f(x)) = f(f(2•3ᵃ + b)) = f(3ᵃ⁺¹ + 3•b) = 2•3ᵃ⁺¹ + 3•b = 3•(2•3ᵃ + b) = 3•x.

[@mathguy37]

and then you realize to do 1 2 3 and then go from there

[@rampion]

@mathguy37 The nice thing about having a formulaic representation of the function rather than a linear algorithm is that we can calculate f(x) without first calculating f(y) for y < x. For example, consider f(1,000,000,000). One billion = 2·3¹⁸ + 225,159,022 So f(1,000,000,000) = 3¹⁹ + 3·225,159,022 = 1,837,738,533.

[@AzertyWasTaken]

I watched all TED-Ed riddles and loved them :)

[@jackmcnally8706]

Smart how you chose a name for the guy based on a paradox based on duplicating things. That led me down quite the rabbit hole… Now if only I can figure out where the heck his name tag magically disappeared to at 1:45….probably on the side of his shirt.

[@sorsocksfake]

Improper answer, since it doesn't show there's a consistent set of rules as per #1. The most obvious rule that would fit most would be x2 if odd, x1.5 if even. This will result in an alternating set of odds and evens that do match the x3 rule (obviously). However, it will not satisfy condition 2, since for instance 9->18, 10->15. The system that the riddle uses, appears as such: (1 becomes 2) then the next 1 goes +1 then the next 1 goes +3 then the next 3 go +1 then the next 3 go +3 then the next 9 go +1 then the next 9 go +3 then the next 27 go +1 etc Stated differently: by default, it's +1. But for the following numbers already passed, there's an additional +2 each: 2-3, 6-9, 18-27; 54-81, etc. 3,9,27,81 forming 3^x, while 2,6,18,54 form 2*3^(x-1).I'm sure there's some way to write it as a formula, but I'll let the real mathheads do that :). It's mainly interesting that 1->2. That's a +1. I think it's because it only gives the +2 for half a step, making it a +1.

[@sairamsk3206]

These such theories of wonders that matchs with physics is really auspicious within the interconnection of the fictional magical world to sense making practical world. Awesome right!

[@guillermorosalesgonzalez1308]

I found this riddle to be misleading. When he said "we already know that 2 becomes 3" I was confused. Nowhere is it ever stated that the way the magic works on the coins when you use it for the first time is the same as when you use it for the second time.

[@mackerelmafia2898]

See 1:55, hint 1.

[@Ujjwalseth2412]

If bag works as a function what would be the equation of that function

[@sieevansetiawan4792]

A function does not necesarrily have an equation.

[@XCM666]

That is also how I approached the problem and I got stumped. There has to be some internal logic to that bag, a function that for input x returns output y. The solution suggests that it's just a lookup table and solves the problem through deduction. It works for solving the riddle, but I suspect the approach will fail for higher numbers.

[@vladislav_sidorenko]

@@XCM666 A function is essentially defined as a lookup table tbh, assigning exactly one value of the output set to each value within the input set.

[@ForteGX]

@@XCM666 I found that you can define a sort of piecewise recursive formula that covers all natural numbers in this function. f(x+1)=f(x)+1 or f(x+1)=f(x)+3. The first case is followed if 3^k

[@kohwenxu]

Had one I did (piecewise function) f(x) = x + 3^[floor(log_3(x))] if x < 2 * 3 ^[floor(log_3(x))] = 3x - 3^[floor(log_3(x))] otherwise.

[@androidpietowski]

Actually with the method of "square root of 3" you would get almost precisely 22 (22,52), so if you use logic that non whole numbers will get rounded downwards (you can't have half of coin), you would also get the right number!

[@Swagpion]

I tried it in a calculator, and some numbers didn't work