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@bosmoth

Жыл бұрын

Isn't this the same as the mcnugget problem?

@V3Aura

Жыл бұрын

@@bosmoth yeah, I think vsauce made a video about it

@MJ-lo1me

Жыл бұрын

Word problem not riddle. this clickbait disrespects mathematicians and writers.

@Amitmalaghan

Жыл бұрын

A

@V3Aura

Жыл бұрын

@@Amitmalaghan thank you sir, very inspiring

Easy solve: Go up to the approaching army and say that at least one of them has green eyes. This will buy you enough time to build your defenses up as they wait days to determine who has green eyes

@cedriclee8724

Жыл бұрын

H a

@EvTheBadConlanger

Жыл бұрын

Nice

@alfi2531

Жыл бұрын

Smort

@erikaz1590

Жыл бұрын

Alternate answer: go to one army and ask "If I asked 'is the army to your right led by Arr', would you answer 'ozo'?" The armies will be so confused by the question, you'd have time to build two walls.

@maapauu4282

Жыл бұрын

I would tell the dino that they have green eyes

The moment I heard "6 9 20" I immediately thought "43 CHICKEN MCNUGGETS" and realized why the enemies are all based on fast food chains.

@jonramos2676

Жыл бұрын

True

@maxwang2562

Жыл бұрын

smart

@roideschiffres6760

Жыл бұрын

I thought the same ting.

@manavshah8335

Жыл бұрын

@@roideschiffres6760 ting ting dong in hole, triplets came out, all abort

@jartistsimpression2291

Жыл бұрын

Please explain?

60% of ted ed riddles can be summed up as “hope you know your prime numbers nerd”

@itsphoenixingtime

Жыл бұрын

I mean, they ARE pretty much what mathematicians obsess about.

@d1v94

Жыл бұрын

@@itsphoenixingtime true.... but then letters came along and made it worse( or better)!!

@imperator9343

Жыл бұрын

"riddles"

@Noname-67

Жыл бұрын

@@itsphoenixingtime only number theorists

@ganbarimasu9446

Жыл бұрын

And the rest is about repeating odd and even numbers...

When I saw the siege weapons, I realized this was the Chicken McNuggets problem. Thank you for animating it. That solution was actually quite interesting. Used guess and check to get the answer, but this method is much more elegant.

"I fart in your general direction." That's the single most hilarious quote I've ever seen in a TED-Ed video

@antonioscendrategattico2302

Ай бұрын

It is a Monty Python and the Holy Grail reference.

Here's how I solved it: The greatest common divisor of 6 and 9 is 3, and the largest safe wall of size 3n is 3. This means that every wall of size 3n that is larger than 3 can be destroyed. Therefore, a bigger wall would have to be 3n+1 or 3n+2. The siege weapons can only reach 3n+2 numbers by using the 20 once or 3n+1 times, and since they can reach every multiple of 3 larger than 3, every 3n+2 wall larger than 20+3 can be destroyed. Similarly they can only reach 3n+1 numbers by using the 20 twice or 3n+2 times, so every 3n+1 wall larger than 20+20+3 can be destroyed. 43 is safe and no larger safe walls can exist, so 43 must be the answer.

@vishakh6929

Жыл бұрын

same here

@rozafisheikh7968

Жыл бұрын

Damn you lot smart. I just didn't want the cute little dinosaurs to die XD

@fejfo6559

Жыл бұрын

This is essentially the same approach as in the video but you only used 3 columns instead of 6

@Serjgap

Жыл бұрын

But you can not make a 1000 length wall out of blocks of length 43. 1000 / 43 = 23,2558... The closest you can get is 43x23 = 989. Stating that "you can plug gaps at the ends with loose boulders" without stating max allowed length of the gaps just makes the riddle bad

@reevus2558

Жыл бұрын

Ez , get more wall fabricator .

Alternatively, point the wall fabricator at your enemies and tell it to produce a 100 000 m long wall. Due to the machine producing any size of wall at equal speed, this huge wall will be shot out of the machine at outstanding speed, crushing your opponents. Unless it, uh, requires raw materials or something.

@anbubu

2 ай бұрын

If the wall fabricator's input unit is in inches, then the required amount of inches to be inputted for the fabricator to be fabricating so many inches of wall at once that it would be travelling at 99% the speed of light, then you would need to input about 1.43 times 10^10 inches

For the first time watching one of these riddle videos I actually decided to try and solve it and got the answer by 1. Listing every number up to 60 2. Removing every multiple of 3 except 3 because they can be made with combinations of 6's and 9's 3. Repeating step 2 but starting from 20 then removing 26, 29, 32, 35 etc. 4. Notice a pattern and realise that I don't need to go beyond 60 5. Highest number left is 43! It's not the most elegant solution but that doesn't mean I can't be proud of myself.

@kfirnachum4928

Жыл бұрын

They can get 6.01×10^52

@redthefoxisWritingUpAStorm

Жыл бұрын

Impressive

@ie8118

Жыл бұрын

What a coincidence I also listed every number up to 60 probably cause none in 50-60 range work

@neofitoff

Жыл бұрын

43 factorial is a pretty large number indeed

@Misteribel

17 күн бұрын

You basically used the sieve of Eratosthenes, tailored to your situation. I find it very elegant. 😊

0:04 This video had the best philosophical quote yet

@-nightlyfireflies-9561

Жыл бұрын

Ah yes, "i fart in your general direction" the most philosophical quote ever

@Jake28

2 ай бұрын

@@-nightlyfireflies-9561 philosophy not psychology

@-nightlyfireflies-9561

2 ай бұрын

@@Jake28 thanks didn't even notice till you pointed it out lol

Love it when TED-Ed makes these puzzle videos! Most of the time, there's a mathematical concept that's hidden in these puzzles. So excited to watch!

@EnriqueLaberintico

Жыл бұрын

Funnily enough, this time it's the Chicken McNugget Theorem.

@almophant1673

Жыл бұрын

@@EnriqueLaberintico Was so funny seeing all the Easter eggs about the chicken McNugget Theorem. With all the other armies being inspired by fast food chains and the unhappy meal at the end.

@imperator9343

Жыл бұрын

You have it backwards. Most of the time, it is a simple mathematical mechanism that they've dressed up as a riddle. Impressive to laypeople, kinda boring and silly to anyone that has passed college freshman level math.

@JaybeePenaflor

Жыл бұрын

@@imperator9343 Yes, I agree.

@commander8625

Жыл бұрын

@@imperator9343 it annoys me because it really isn't isn't riddle. Don't get me wrong, I like the math and all, but I came to try to solve a puzzle, not to be told about some mathematical equation or concept that hasn't been used in 1000 years.

I love how it just opens with “BAD NEWS”

Usually I have no idea how to solve these and just watch the solutions, but for the first time, I actually managed to figure it out on my own!

@altersami9660

Жыл бұрын

What approach did you use?

@LEMOO-ds2mz

Жыл бұрын

I..... Uhh... Congratulations

I got it via minor brute force and guessing because 6 and 9 are both divisible by 3. You're screwed if one siege designer misheard and made one 5 or 7 wide, though.

@frozenfeet4534

Жыл бұрын

nah, they use fabricators too, nothing to mishear

@therealelement75

5 ай бұрын

boss: 5 wide is useless you're fired them after seeing wall of 43 being 20 + 5 + 3(6): nvm you're unfired 5 is useful

@AffectionateGraphingCalc-bq6cb

5 күн бұрын

😂

Finally I have solved my ever first Ted-Ed Riddle.

@Purple_Mesh

Жыл бұрын

I have solved my first one too! Inspired

@kohikappu

Жыл бұрын

Nice, I failed for the first time, I went for 34 and didn't notice it should be the highest.

Found it even faster, using a different method! Any multiple of 3 can be divided by a combination of 6s and 9s, except 3. If a number isn't divisible by 3, then it WILL become divisible by 3 if you remove 20 or 40 from it. (because it changes the rest, and changing it either once or twice always work). This means that any number higher than 40 is either: 1) Divisible by 3 2) Will be divisible by 3 if you remove 20 3) Will be divisible by 3 if you remove 20 twice (so, 40) And if the number is divisible by 3 and it's above 3, it's no good. (the added 20 don't matter because one of the wallbreakers can take the 20s down). So if you remove 20 or 40 from the number, it can't ever give you a number divisible by 3, that is above 3. But as you're looking for the highest possible solution, then it has to be the solution where the number you get from removing 20 or 40 gives you EXACTLY 3; Between 23 and 43, you obviously pick 43, and that's the highest possible solution! Anything above that will either divide by 3, or divide by 3 if you remove 20 or 40 from it (and if it divides by 3, it means you can get it with the 6s and 9s, no matter what it is - as long as it's not 3, but we already picked the solution that gives you 3, i.e. 43).

@existentialchaos8

Жыл бұрын

That's what I did!

@lubu682

Жыл бұрын

**has a mental breakdown**

@hash5996

Жыл бұрын

that is a great solution!

@dhien2103

Жыл бұрын

Geines

@shreyanray7624

Жыл бұрын

Same method, imo their method is not the best or even efficient as it involves some hit and trial

43? Okay, time to check my answer... Yay! I got it :) I realized that 3 is the greatest common factor between 6 and 9, so all multiples of 3 will be eliminated from 6 onward. If you add 20 to this, then you end up at 26, which eliminates all multiples of 3 minus 1 from 26 onward. Once you get to 46, then you eliminate all multiples of 3 plus 1. At this point, you can create any number. The highest number that wasn't eliminated is multiples of 3 plus 1, but less than 46. Which is 43. Fun math problem!

@adrianwoodruff1885

Жыл бұрын

This is how I solved it.

@aarongrooves

Жыл бұрын

@@adrianwoodruff1885 Great minds think alike!

@devangsingla5411

Жыл бұрын

But 43 is 4×6+9+10

@pietroparini2008

Жыл бұрын

I also used this approach.

@VNYoshi

Жыл бұрын

Basically Chinese Remainder Theorem.

Hope can update this series more often. One of the very few on the entire internet that can provide quality riddles

Step one: make sure that the invaders have green eyes. Step two: tell them to leave

@Dexaan

Жыл бұрын

Ulu

@kishorikumari9014

Жыл бұрын

an explaination would be helpful

@lambadajewo.4143

Жыл бұрын

@@kishorikumari9014 that is a running joke on the channel, ever since the green eye prisoners dillema video was released!

@lambadajewo.4143

Жыл бұрын

@@Dexaan ozo!

@pageturner2958

Жыл бұрын

Glad this joke is being kept alive

In rule 2, it states that the opening is one kilometer wide, & that loose boulders can plug up any gaps. So, wouldn't it be easier to use a prime number, say 997, & plug up the 3 meters of holes? Yes, you could use several boulders, but that seems impractical.

@raphaellaalexander8853

Жыл бұрын

While 997 is prime, you can still make that number from two 20s, one 9, and 158 6s, since it falls in the first column in the chart that they made to solve the problem

@joshuatinyforest1204

Жыл бұрын

@@raphaellaalexander8853 Ah okay, I see where I went wrong. Thank you.

@aadityamohta6966

Жыл бұрын

I had a similar idea and then realsied it was wrong lol

@TheWatcher328

Жыл бұрын

@@raphaellaalexander8853 How about 998?

@IH8Reflorished

Жыл бұрын

@@TheWatcher328 that’s a mutible of 2

The chicken Mcnugget problem! I remember seeing it on Numberphile what feels like a decade ago.

I had a similar sort of elimination method: With 6 and 9 you can make any multiple of 3 besides 3 itself, and only multiples of 3 with those 2 numbers 20 is 1 less than a multiple of three so, above 20, you can add 6s and 9s to get 1 less than any multiple of 3 besides 23 40 is 2 less than a multiple of 3, so similar to the last step, beyond 40, you can add 6s and 9s to get 2 less than any multiple of 3 besides 43, effectively eliminating all numbers past 43 since you've already eliminated multiples of 3, one less than multiples of 3, and 2 less than multiples of 3

@randompastahandle

Жыл бұрын

That is what I did also.

@invisiblejong

Жыл бұрын

same, i also used this method, so I was fairly surprised seeing prime number is used

@jezer8325

Жыл бұрын

Yess, I did it exactly the way you did

I really like the aesthetic theme the graphics of this video take.

given there's no max limit, and it doesn't matter how big the wall is, just make the number several quadrillion or so. I kinda doubt hundreds of siege weapons are sufficient for that.

@dennisevanko6152

10 ай бұрын

Oh yeah that’s right… Wait then how will you get the resources for that big of a number?

@aldar8240

10 ай бұрын

@@dennisevanko6152 the fabricator isn't stated to have any resource requirements, and it's already an arbitrary magic device with unclear rules, so it can be assumed it just handles that somehow

Got it with a different method after a few takeaways. - Realized you can get all multiples of 3 (except for 3 itself) with 6 and 9 alone. - If you can prove that you can find all lowest multiples of three such that 3A=20B+x for every integer x from [1,20], you could basically make any number greater than the greatest of those multiples (ex. for x=4 you have 3*8=20*1+4). This works since if we have a representation like this for each number from 1-20, we can just add multiples of 20 to one of these to get any other natural number greater than these representations we find. Note that A and B must be natural numbers. - It turns out that you can find those lowest multiples of three in the first 20 multiples. The greatest of these low multiples is 63 (for x=3 we have 3*21=20*3+3). - Since 63 is the lowest number we can find for a representation of x=3, there must be no representation for 43 (since it is 20*2+3) or 23 (since it is 20*1+3). Since 63 is the greatest among these "low" multiple representations, 43 must be the greatest number without a representation made up from 6,9,20. How did you all do it?

I may not be able to solve the riddle of the wall, but I can solve the riddle of the kingdoms: Royalburg : Burger King Circuslandia: McDonald's Redheadonia: Wendy's And I feel that is a real achievement

@frozen_spider

Жыл бұрын

As someone from a country not having any of these - thanks! I didn't know about Wendy's at all.

@sharonannco8713

Жыл бұрын

You forgot about jurassic bite

@samudralaanu3429

10 ай бұрын

Dunno, don' care

Just do 1,000,000,000 because they only have hundreds of weapons

@survivordave

14 күн бұрын

That's why the opening is 1 kilometer. You can be short by a bit and fill in the ends with boulders but you can't go over 1,000 m

@123idk

14 күн бұрын

Oh

I actually solved this one pretty quickly thanks to the fact that I've seen it before in a different video and remembered the answer. I'm gonna try and use this tactic more in the future.

Now this was creative. Hats off to the people who can come up with such unique logic problems.

@darthgumby2406

Жыл бұрын

They don't make the problems. they find an obscure mathematical solution someone made up at some point in history and create a silly scenario that would use that solution.

@HunterOfWumpus

Жыл бұрын

This is a rip off of the “what is the largest number of chicken nuggets you can’t order from McDonald’s”

My solve was a little different. The first thing I realized was that with 6 and 9, you could combine them to make any multiple of 3 above 6 (6+6=12, 6+9=15, etc.). Then using the 20, you would be able to offset it down one number, so multiples of 3, minus 1 (20+9= 30-1). So offsetting it a second time, with 40, would remove every potential number. So the highest available number would be a number less than 46, that is offset from a multiple of 3 by -2, 45 - 2 = 43. Not the solution they hit on, but I'm proud of it.

my favourite videos here please keep churning these riddles up :p

-can you solve this riddle ? -yeah but that's gonna take forever and also i would've missed this awesome video

If the dinos are sought after as a food source, is there something special about the food source or are the invading forces starving? The latter seems more likely and you could simply create rows of walls and wait until they go hungry. The walls would also have to either be fabricated "into place" from the side of the machine, or physically moved into place. If it's the latter, the invaders could just move the walls. But all of that ignores that the camped forces could attack the engineers in the video's scenario or mine through scout parties in the night. (And that the machine is apparently able to fabricate entire walls in equivalent degrees of time regardless of the size)

I used the divisible-by-three insight some other commentors mentioned, knowing that only 20's could change the divisibility by three. Knowing that it would take up to two 20's to make the number easily divisible by three, I started with 39, the biggest number you couldn't fit two 20's into, and counted backwards from there until I hit a number that couldn't be made any other way. Thus, I got the answer 37. ...apparently I overlooked the fact that the number 3 itself couldn't be made, only greater multiples thereof.

@PragmaticAntithesis

Жыл бұрын

I made the same mistake.

@adarshmohapatra5058

Жыл бұрын

Same dude. And then the animation at 2:37 showing a lot of 37's made me think, "aha" I've hit the right answer.

이번 문제는 아마 처음으로 제 힘으로 풀어 본 TED-ed 수수깨끼인 것 같습니다. 혼자서 풀어보니까 성취감도 느끼고 더욱 재미있네요. 앞으로도 이런 영상 많이 올려주세요! 좋은 영상 감사합니다.

The kingdom: Does not want dinos to become food. Also the same kingdom: Names itself jurassic bite.

Now I remember. Numberphile made a video about this with McDonald’s nuggets

Ted: Can you solve the fortress riddle? Me: *No.*

Thank goodness I watched the numberphile video about this first! This is the only time I’ve ever managed to solve a Ted-Ed riddle without pausing!

The sun shines when Ted-ed makes riddles

I love how this riddle was inspired by the "how many chicken mcnuggets can't you get?" problem, and everything got burger-themed.

0:25 Mc Donalds , Burger king, Wendy's

Feels nice when you create a program that can solve the question

This is my favourite riddle, Ted Ed! You should make more!

Ted always provide everyone with a wide range of knowledge and also inspire science to the world

Teded: Can you solve the riddle? Me: Can you ask me a riddle? Teded: Yes. Me: An actual riddle, not a convoluted math problem? Teded: Yes . . . . . . .. . . .. Me: Then I'll give it a shot. Teded: It's math. Me: For the love.

@desrenaud100

Жыл бұрын

Thank you! Really glad I'm not the only one to feel this way. Everytime a new one drops, I keep hoping maybe this time it'll be an actual logic puzzle like the pirate riddle, or the medieval tournament or green eyed islanders, and every damn time it's just some long, convoluted math problem.

i love teded narrator and these riddles

There are so many people referencing old riddles and I love it

1:58 “[…] the siege weapons, which your enemies have hundreds of” So not thousands. Anything that’s 35,000 or longer is safe.

Love ted ed riddles, because they feel real and honest, the text holds all the information you need. But I wish they werent mainly mathematically driven

I used a different method. 6 and 9 are both multiples of 3, so the 20s have to be used to make up for any remainder when divided by 3, so if you let them use three or more 20s, they can replace three of the 20s with 6s and 9s to get more options, but if you keep it to no more than two 20s, they don't have this option. Now if you divide 6 and 9 by 3, you get 2 and 3, which together can cover any number larger than 1, since for even numbers you just use 2s, and for odd numbers you use 2s to get to the even number below your target and then replace the the last 2 with a 3. Thus, the only multiple of 3 that 6 and 9 can't cover is 3 (1x3). Add the two 20s (that are the most 20s you can use without letting them start from a lower number of 20s) and you get 43.

I did this in a fairly roundabout way- first you figure out the smallest combos of 6s and 9s to get to various numbers in the 1s spot. The largest number combo of 6s and 9s is 33 to get a 3 in the 1’s spot, so it would be advantageous then, to have a 3 as the last number, since it’d take up more of the entire number to get it to a multiple of 10, would then be divided by 20s, or an additional 30 if it’s an odd number of 10s. With all that in mind, only 43 would work, as 53 gives you a 20, 63 a 30, and literally everything else can be solved by all 20s or a bunch of 20s + 30. So it was indeed 43.

Love that you guys made a Monty python and the holy grail reference

0:07 I see...meme potential

I tried with other numbers along 6, and it works like magic.

5:00 and a herd that wont become *unhappy meals*

1:06 urgh, already bored. Weaponise your dinos, murder your enemies in their beds, drink from their skulls.

I can see this as an actual math examination question, not too hard but still challenging

Appreciate that the wallbuster sizes (6/9/20) are perfectly matched to chicken nugget serving sizes

whoever makes these riddles you're legendary

This is by far the best quote we get at the beginning of any TED-Ed videos.

If anyone has ever heard of the McDonald's 43 nuggets problem, then they should have instantly known what the answer was. Exact same problem, exact same numbers, just simply given a different coat of paiint

This is the reason im subbed just for the riddles

Your solve was very pretty and impressive In my solve I found the lowest number combination possible for each unit digit and then added ten because there's 20 and 2x(6+9) = 30 the highest one was 33 and I added 10 for 43

That feeling of superiority when you recognize the TedEd quote.

"It is possible to solve this problem with trial-and-error" Not when we only have *one* kingdom it isnt.

@alex2005z

Жыл бұрын

Just yolo it

Damn, the quote at the beginning was especially inspiring this time: "I fart in your general direction."

Dude, awesome riddle!

Reached the 70th riddle! We have come a long way.

0:20 is that mcdonalds burger king and wendys!

Great content love it keep up the good work

A riddle is a statement, question or phrase having a double or veiled meaning, put forth as a puzzle to be solved. Riddles are of two types: enigmas, which are problems generally expressed in metaphorical or allegorical language that require ingenuity and careful thinking for their solution, and conundra, which are questions relying for their effects on punning in either the question or the answer

I love that you decided to just throw in parodies of fast food chains into this riddle for no reason.

@JoeThomas-lu6fy

Жыл бұрын

The numbers are based on the former McNugget serving sizes.

2:10 infinite in size obviously

@TheWorldsLargestOven

7 ай бұрын

There's a time limit

@hatimbharmal8537

5 ай бұрын

​@@TheWorldsLargestOven1:43

@Wezryx

4 ай бұрын

I like your style better than my answer of a billion kilometers.

@Wezryx

4 ай бұрын

But they will mention that safe weapons and there will just be more wall materialized Infinitely instantaneously

oh my god this is the mcnugget problem disguised as a ted-ed riddle

lesss goo-finally got one of these riddles

Easy solve: if the builder always takes the same time, and the enemies have an big number, although finite number of walls breakers, just make an 9999999999 Meter wall

@unrellated

Жыл бұрын

But the opening is only a single kilometer wide.

@fandroid6491

Жыл бұрын

Imagine making a 9-billion meter long wall...

@ie8118

Жыл бұрын

Yeah speaking of the opening being only 1km wide, my thought on why that wouldn't work is that if it was that long the wall and wall machine would start going into enemy territory, and they'd destroy the machine, stop it from generating more and then just take time to destroy it

I thought a wall segment of 1 was 1 meter, and we had to fill 1 km or 1000m. So the answer I got was 997. On the bright side, this is the first time I solved a ted-ed riddle.

@merolivia

Жыл бұрын

997-49=948. 948÷6=158

@ie8118

Жыл бұрын

Yeah it does say each wall segment is in meters, cause all the weapons are also in meters. I got confused the same way I thought it was asking how can we make the longest wall in 1 print and I missed the thing about equal segments

The way I figured this out goes like this: The 6 and 9 share a factor, so they’re easier to work with; turns out they can make any multiple of 3 higher than 3 itself. (Once you have 6, you can substitute a 6 with a 9, or a 9 with two 6s, to count up 3s to whatever you need.) Then to add in the 20s, there are 3 cases we need to consider, based on the remainder of the number mod 3. For multiples of 3, you can make anything above 3, as mentioned. For multiples of 3 minus 1, 20 is one less than a multiple of 3, so you need to include one 20 and then can make the rest with 6s and 9s; this lets you make 20 itself, and then 26, 29, etc, with the highest safe number here being 23. Then for multiples of 3 plus 1, you need to include two 20s to get to 40 (=3*13+1), and then you can add 6 and then more 3s; the highest safe number in this category is 43, for similar reasons to above. Since the highest safe number in any of the 3 categories is 43, that’s the answer. I honestly think this is a bit better than what you have here, since yours still has some trial and error, and it may be easy to miss a combo, but mine is more difficult to extend when you have more than 2 distinct factors.

@KnakuanaRka

5 ай бұрын

And in fact, there is a more systematic way: Start with writing out the numbers in columns based on the smallest number as you did; write 0 above the rightmost column and circle it. For each factor other than the smallest: For each circled number in the grid, add the factor to the number and circle that number; repeat until you would circle a number that is already circled or underneath a circled number. If you would circle a number above another circled number, erase the number underneath afterwards. After doing this for all factors, the unsafe numbers are precisely the numbers that are circled or below a circled number in the column. Check all the numbers directly above circled numbers; the largest of these is the best safe number. If a column has no circled numbers, there is no solution because the factors are not relatively prime.

6 and 9 can be simplified into 3 and 2. Since 3*1 - 2*1 = 2*2 - 3*1 = 1, 3 and 2 can give anything but 1, so we need to solve this: 20n + 3k while k != 1, so we can add 3 to the solution if there's no limit for k. That would be 38. Anything 39 or above can be obtained with 20n + 3k. If we add 6 to them, it's all guaranteed. 44 = 20+24, so it's 43.

When you are not able to solve any Ted ed riddle but then also watch every Ted ed riddle video.

My Feedback: So I did this by listing numbers up and basically through manual testing them all to see if they could be sieged. The problem is I didn't clearly get that you were going to make 1 wall out of segments of the same length and the point was to use the longest segment length possible in order to minimize the wall making time/number of segment used. I thought that you could only do 1 print total and that was going to be the whole wall, because it talked about the machine taking the same amount of time to print any length. I was just under the impression you only had time to do 1 print before the enemies got here, totally forgot about what it said at the start about how it can make segments of a specific whole number length you set. I think the main thing that would help is that instead of the question at 2:00 being "What wall length will save your kingdom?" say "what wall segment length" instead. I have a suspicion this is the main thing that confused me. Also, at 1:54 when it shows the segment, it could show it as a part of many other segments of similar size to emphasize you want to figure out the segment length not the whole wall length. (I know the voice says segment but I must have missed that). The clock being shown to take a full 12 hour cycle to print 1 segment also might have been misleading, not by itself but in combination with the other things like the question wording. The sentence "Adjacent wall segments reinforce each other if struck simultaneously" is just confusing. It basically just means "A siege weapon cannot take down multiple segments at once- only 1 segment at a time." and I think this would be much clearer.

@ronaldiplodicus

Жыл бұрын

It does say that the final wall will be 1km wide

@ie8118

Жыл бұрын

@@ronaldiplodicus Yeah if it said it would be exactly 1km wide then I'd know the answer was the segment size. But since it said at the countdown that it doesn't have to be exactly 1km because you can fill the ends with rocks, I was still able to keep the wrong impression that the answer is the total wall length

This is the earliest I’ve ever been to one of these riddles

I did a less elegant version of this. I realised that once 6 consecutive numbers were discovered, all numbers above could be made. So gave a way to identify an upper limit.

Imagine being able to bust down all walls of length 44 and greater but not being able to bust down a wall of length 1...

@kohwenxu

Жыл бұрын

Or not being able to bust down the wall one shorter (43)

@kohwenxu

Жыл бұрын

42 and 44 are bustable, but not 43

@ie8118

Жыл бұрын

I think cause it must contact into both side pillars Actually I don't think wall length 1 would be possible if each must have 2 pillars

Fact: It's your habit to check the comments after watching videos. No matter WHAT

@AnthonyDentinger

6 ай бұрын

That’s not true! I haven’t- oh.

this ones quote is fire🔥

This was awesome!

I want to see more fortresses from our history and some riddles like this

But since your walls are all in a line, the enemy can still break through. Since some of the siege weapons would overlap onto another wall. Thus allowing the enemy to break through.

@N0noy1989

Жыл бұрын

That's not allowed in the problem as stated in the rules. Adjacent walls reinforce each other is what it said. Has to be exact as per rules of the Riddle.

Ted Ed: Can you solve this riddle? Me: no also me: *clicks video*

FINALLY! A NEW RIDDLE VIDEO!

None of these riddles are ever really riddles. They’re just maths problems with a story

Step 1: Confirm you have green eyes Step 2: Ask the enemies to leave

i am addicted to these!! :D

The method I came up with was realizing that with 6 and 9, you can get any multiple of 3 above 3. 20 gets rid of everything that's congruent to 2 moduleo 3 above 23, and 40 gets rid of everything congruent to 1 modulo 3 above 43. Which I guess is pretty similar to your method, but 3 wide instead of 6.

Lol I thought I got it right on the first try cuz I did 997 lol

@GandWizard

Жыл бұрын

Me too! 1 km = 1000 metres, so I sought the largest prime under 1000, 997 metres. As no maximum size is given, I don't understand why this answer is not correct.

@EnriqueLaberintico

Жыл бұрын

I recognised the Chicken McNugget Theorem and ended up finding a formula that gave me 151, ab - a - b. Turns out it can be obliterated by 5 redheads, 5 royals and a clown.

@EnriqueLaberintico

Жыл бұрын

@@GandWizard 47 redheads, 5 royals and 2 clowns. 47•20 + 5•9 + 2•6 = 940 + 45 + 12 = 997

@GandWizard

Жыл бұрын

@@EnriqueLaberintico Thanks for pointing out the flaw in my assumption!

How are walls of 43s gonna be immune to an attack? Can't the enemies just crack down multiples of 43 walls at once?

@itsphoenixingtime

Жыл бұрын

Technically only a single wall of 43 works, assuming the wall is just 43 segments.

@danielyuan9862

Жыл бұрын

The video never said the enemies could do that (they can't).

@samspiano7070

Жыл бұрын

nope, they specified it in the ‘rules’ before the solution

Greatest starting proverb ever in a ted ed video😂

You guys are like the Riddler in terms of what constitutes a riddle.