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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
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.
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.
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.
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.
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
@@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.
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.
@@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.
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.
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
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).
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!
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.
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?
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.
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
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
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.
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 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.
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
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.
If they have hundreds and not thousands, the maximum number of siege weapons would be 2,997. Assuming the worst, all of those end up being Redheads. 2997x20= 59,940. The 58,941 is unreachable. Numbers equal to or greater than 58941 are unbreakable.
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 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
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.
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.
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.
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.
Just make a wall that’s less than 6 There no rule saying that they can’t removes walls below the break freashold So if you wanted you could make a bunch of 1 length walls and be chillin
Better Idea : Build a wall so ridiculous big Like 10000000000000 Lengh, it would literally take then years to break it (the fastest they could go is 20) And the wall maker says it can create ANY wall in the SAME speed REGARDLESS of lengh
i did something similar. 6 and 9 can sum to any multiple of 3 besides 3 because 6 = 2 * 3 and 9 = 3 * 3 We can always increment our multiple of 3 by subtracting 6 and adding 9 which is just adding 3. Then, we can find that 20 and 30, 21, 12, 33, 24, 15, 6, 27, 18, and 9 can find any 1s digit we like, so if we just start at 50, every number after that can be reached by adding some amount of 20s and 30s and then one of the digits numbers. if the digit number is 20 something like 24 or 27, we can add 30 to get to 50, and add two twenties instead to get to 60, and so on. you can always increment by subtracting 20 and adding 30. if you run out of 20s, simply turn two 30s into one 60. this logic will lead you to conclude that all numbers past 50 are lost, as well as all multiples of 3. 49 is also obviously lost, 48 is 3n, 47 is 27 + 20, 46 is 6 + 40, 45 is 3n, 44 is 24 + 20, and you end at 43
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
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
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
What I'm not clear on is the underlying principle for solving for prime numbers using the chart. While eliminating the factors of 2,3,5, and 7 are good through 120, you'd need to also run that elimination for the higher primes to eliminate numbers like 187, which is 11 * 17. Unless it's all diagonals all the way down, which would be harder as you have to start skipping lines.
The nice thing about 6 columns is that all primes except for 2 and 3 are actually of the form 6n+1 or 6n-1. This does in fact make it diagonals all the way down. And I mean ALL the way down. Every single one of them. Yes the slopes would be much steeper but they're still straight lines.
Another solution: mentally make the grid have length 20, but we won't have to write it down. Since we have the numbers 6 and 9, every multiple of 3, except 3 itself, can be made, so we can remove those columns. The next row is the first numbers +20, which means the numbers we just crossed out are no longer divisible by 3 (digits add to 2, 5, or 8), so we can cross out all of the multiples of 3 in this range. In the next row, a similar story, except now we've just out all numbers whose digits sum to 2, 5, or 8; are about to cross out all numbers that sum to 3, 6, or 9, and in the first round crossed out almost everything that now sums to 1, 4, or 9. The only survivor is the 3 we couldn't make, which is now 43, and is the largest as 63 is crossed out in the next row.
I seem to be missing something... The width to be defended is 1000, which isn't divisable by 43, so that leaves a gap of 11 through which the enemies can get in? Wouldn't it be simpler to produce 997 + 3 instead of 23 * 43? Or if you want to stick to 43, produce 1 extra piece of 11 to close the gap?
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.
Fun fact the best way to apologize if this is an accident is: I’m sorry I didn’t know how much those dinos meant to you. Fun fact: The best way to apologize if this is on purpose (which it is) How did you feel? I felt sad. Sorry I broke my promise.😮 true it’s on Ted Ed channel best ways to apologize.
I solved this riddle by writing a program in C which goes through every number from 1 to 1000 and goes through every combination of weapons to check whether this number is suitable. I also got the answer to be 43 but in just 2.802 seconds.
I kinda found an alternate more elegant soln I)All nos of the type 3n can be broken ( except 3) II)Now every no 20+3n can also be tore down( except 23) III)And every no 40+3n behave the same way(except 43) Now we can see as this series continues after 40, 40 belongs to (III) category 41 to (II) and 42 to (I) And since these categories go with a difference of 3 every no after 40 is counted in either one of the categories EXCEPT 43 and thus that is the answer
There is a similar question pratically: In Hearthstone in China, the players are only allowed to invest the whole number combinations of RMB 30, 60, 128, 328, 388 or 488, of which only 30, 128 and 328 practically makes sense. For instance, some virtual products are sold at 98 RMB, which is an "undestructible" amount. I have run a brute force code that gives out the result that the largest "undestructible" even number in this case is 1010. I do not know if there are any better algorithms better than brute force in similar questions.
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.
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
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 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
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.
@@dieselboy.7637 A story with a problem that requires logic to solve... so while not all riddles are math problems with a story, the inverse is almost always true
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
We really got Ronald McDonald, Burger King, and Wendy teaming up to eat some dinosaurs. Nice to know that despite the fact that they're constantly at war with each other they're willing to team up to eat some baby dinosaurs.
This would be a good bonus challenge: Make an algorithm for determining the max. wall length for 3 types of siege engines, of arbitrary length. For a _bonus_ bonus riddle: do the above problem with an arbitrary number of the types of siege engines. Oh, and you get extra points for your algorithm taking reasonable time(proportional to the size of the numbers, so a century is reasonable if, say, the numbers were in the billions)
الحُزن! يقول ابن كثير رحمه الله: "لا تُستنار البصيرة إلا بالحزن، فعندها يرى المرء حقيقة كل شيء؛ حقيقة نفسه، وحال قلبه وصحبته وأهله، حقيقة الدنيا على حالها، فيجعل الله من كل ذرّة حزن في نفس العبد نورًا يضيء به بصيرته حتى يدرك هوان الدنيا برغم جمالها وحقيقة الأشياء".
With the clown you can take down any wall size that is a multiple of 6, but you need the other two to get non-multiples of 6. If you add just one royal to the mix you can take down any wall of size 6c+3 (c is the number of clowns) except for wall size 3. Adding more than one royal is not necessary, because any two royals can be replaced with three clowns. If you add one redhead then you can destroy walls of size 6c+2 or 6c+5 if you also added a royal to the mix. With two redheads you can destroy walls of size 6c+4 or 6c+1 if you add a royal to the mix. Having more than two redheads is unnecessary, because every three redheads can be replaced by ten clowns. This covers all the possible positive integers, but only if they are higher than a certain number x. This number x is the one we are looking for and x = -6 + 9 + 20x2 = 43. Why? Clearly x cannot be destroyed by any combination of clowns, royals, and redheads, because 43=6x7+1=6c+1 and any number of this type can only be found with a royal and two redheads, but those put together are already too big. Since wall size x cannot be destroyed we do not have to look for a lower number. Numbers 44, 45, 46, 47, and 48 require either less redheads or less royals or both. This means that these numbers will definitely be possible to destroy, because there is room left to add clowns in the mix. Wall size 49 can be destroyed with a royal and two redheads and any number higher can obviously be destroyed as well, because any wall size that is 6 less can be destroyed as well, so it is just a matter of adding more clowns. So 43 is the biggest wall size that cannot be destroyed!
This is easier than I thought. Remember 6 and 9 are factors of 3 so it's simplified to no 3x numbers. They can only use 1 or 2 20s as 3 times 20 is 60 which is duplicate with 3x. Since 20 is in 3x+2 form with both 20 and 40 (which is 3x+4 or in another word 3x+1) they can tear down any number. Count down from 40 and you found 37 not suitable for the enemies.
Damn 43 is also suitable. It took advantage of 40. Yes this remaining is divisble by 3 but since it's made with 6 and 9s the 3 has became a dead angle where the siege weapons won't reach.
Keep honing your riddle-solving skills by visiting brilliant.org/TedEd and check out Brilliant’s 60+ courses in math, logic, science, and computer science. They feature storytelling, code-writing, interactive challenges, and plenty of puzzles for you to solve. And as an added bonus, the first 833 of you to use that link will receive 20% off the annual premium subscription fee.
Isn't this the same as the mcnugget problem?
@@bosmoth yeah, I think vsauce made a video about it
Word problem not riddle. this clickbait disrespects mathematicians and writers.
A
@@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
H a
Nice
Smort
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.
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.
True
smart
I thought the same ting.
@@roideschiffres6760 ting ting dong in hole, triplets came out, all abort
Please explain?
60% of ted ed riddles can be summed up as “hope you know your prime numbers nerd”
I mean, they ARE pretty much what mathematicians obsess about.
@@itsphoenixingtime true.... but then letters came along and made it worse( or better)!!
"riddles"
@@itsphoenixingtime only number theorists
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.
The Chicken McNuggets Problem? What's that?
@@buildertherobloxian4731 This video sums it up: ua-cam.com/video/vNTSugyS038/v-deo.html
VSauce2 also has a good video explaining what it is.
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.
They can get 6.01×10^52
Impressive
What a coincidence I also listed every number up to 60 probably cause none in 50-60 range work
43 factorial is a pretty large number indeed
You basically used the sieve of Eratosthenes, tailored to your situation. I find it very elegant. 😊
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.
same here
Damn you lot smart. I just didn't want the cute little dinosaurs to die XD
This is essentially the same approach as in the video but you only used 3 columns instead of 6
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
Ez , get more wall fabricator .
"I fart in your general direction."
That's the single most hilarious quote I've ever seen in a TED-Ed video
It is a Monty Python and the Holy Grail reference.
Your mother was a hamster and your father smells of elderberries
I love how it just opens with
“BAD NEWS”
When you forgot to press save before you close your game
BAD NEWS
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!
Funnily enough, this time it's the Chicken McNugget Theorem.
@@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.
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.
@@imperator9343 Yes, I agree.
@@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.
0:04 This video had the best philosophical quote yet
Ah yes, "i fart in your general direction" the most philosophical quote ever
@@-nightlyfireflies-9561 philosophy not psychology
@@Jake28 thanks didn't even notice till you pointed it out lol
"I fart in your general direction
-sersues of Athens
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.
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
@@anbubuthe input is in meters evidenced by the labels on the siege weapons
Did you… not see the Big Funnel of Rocks on top of the Fabricator?
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).
That's what I did!
**has a mental breakdown**
that is a great solution!
Geines
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!
This is how I solved it.
@@adrianwoodruff1885 Great minds think alike!
But 43 is 4×6+9+10
I also used this approach.
Basically Chinese Remainder Theorem.
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!
What approach did you use?
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.
nah, they use fabricators too, nothing to mishear
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
😂
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?
Finally I have solved my ever first Ted-Ed Riddle.
I have solved my first one too! Inspired
Nice, I failed for the first time, I went for 34 and didn't notice it should be the highest.
Hope can update this series more often. One of the very few on the entire internet that can provide quality riddles
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.
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
@@raphaellaalexander8853 Ah okay, I see where I went wrong. Thank you.
I had a similar idea and then realsied it was wrong lol
@@raphaellaalexander8853 How about 998?
@@TheWatcher328 that’s a mutible of 2
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
That is what I did also.
same, i also used this method, so I was fairly surprised seeing prime number is used
Yess, I did it exactly the way you did
The chicken Mcnugget problem! I remember seeing it on Numberphile what feels like a decade ago.
-can you solve this riddle ?
-yeah but that's gonna take forever and also i would've missed this awesome video
Now this was creative. Hats off to the people who can come up with such unique logic problems.
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.
This is a rip off of the “what is the largest number of chicken nuggets you can’t order from McDonald’s”
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 really like the aesthetic theme the graphics of this video take.
Step one: make sure that the invaders have green eyes.
Step two: tell them to leave
Ulu
an explaination would be helpful
@@kishorikumari9014 that is a running joke on the channel, ever since the green eye prisoners dillema video was released!
@@Dexaan ozo!
Glad this joke is being kept alive
The kingdom: Does not want dinos to become food.
Also the same kingdom: Names itself jurassic bite.
Just do 1,000,000,000 because they only have hundreds of weapons
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
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.
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
As someone from a country not having any of these - thanks! I didn't know about Wendy's at all.
You forgot about jurassic bite
Dunno, don' care
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.
Instructions unclear I built a 922336079872633807 meter wall
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.
I love how this riddle was inspired by the "how many chicken mcnuggets can't you get?" problem, and everything got burger-themed.
If they have hundreds and not thousands, the maximum number of siege weapons would be 2,997.
Assuming the worst, all of those end up being Redheads.
2997x20= 59,940.
The 58,941 is unreachable. Numbers equal to or greater than 58941 are unbreakable.
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.
Oh yeah that’s right…
Wait then how will you get the resources for that big of a number?
@@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
The wall is 1km large. It's assumed that the armies have infinite of each weapon.
my favourite videos here please keep churning these riddles up :p
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.
I made the same mistake.
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.
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.
Damn, the quote at the beginning was especially inspiring this time: "I fart in your general direction."
The sun shines when Ted-ed makes riddles
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.
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.
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!
oh my god this is the mcnugget problem disguised as a ted-ed riddle
There are so many people referencing old riddles and I love it
Just make a wall that’s less than 6
There no rule saying that they can’t removes walls below the break freashold
So if you wanted you could make a bunch of 1 length walls and be chillin
Never mind, I completely ignored Rule 1 :(
Better Idea : Build a wall so ridiculous big
Like 10000000000000 Lengh, it would literally take then years to break it (the fastest they could go is 20)
And the wall maker says it can create ANY wall in the SAME speed REGARDLESS of lengh
@@onetwothreefour386 The gap is only 1km, so that wouldn't fit.
@@JoeThomas-lu6fybut still, a 1km wall will hold fast, @onetwothreefour386 might be onto something
i did something similar. 6 and 9 can sum to any multiple of 3 besides 3 because 6 = 2 * 3 and 9 = 3 * 3
We can always increment our multiple of 3 by subtracting 6 and adding 9 which is just adding 3.
Then, we can find that 20 and 30, 21, 12, 33, 24, 15, 6, 27, 18, and 9 can find any 1s digit we like, so if we just start at 50, every number after that can be reached by adding some amount of 20s and 30s and then one of the digits numbers. if the digit number is 20 something like 24 or 27, we can add 30 to get to 50, and add two twenties instead to get to 60, and so on. you can always increment by subtracting 20 and adding 30. if you run out of 20s, simply turn two 30s into one 60. this logic will lead you to conclude that all numbers past 50 are lost, as well as all multiples of 3. 49 is also obviously lost, 48 is 3n, 47 is 27 + 20, 46 is 6 + 40, 45 is 3n, 44 is 24 + 20, and you end at 43
Ted: Can you solve the fortress riddle?
Me: *No.*
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
1:58 “[…] the siege weapons, which your enemies have hundreds of”
So not thousands. Anything that’s 35,000 or longer is safe.
Appreciate that the wallbuster sizes (6/9/20) are perfectly matched to chicken nugget serving sizes
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
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.
Twenty three 43's, and an 11. That combination will protect you.
0:20 is that mcdonalds burger king and wendys!
What I'm not clear on is the underlying principle for solving for prime numbers using the chart. While eliminating the factors of 2,3,5, and 7 are good through 120, you'd need to also run that elimination for the higher primes to eliminate numbers like 187, which is 11 * 17. Unless it's all diagonals all the way down, which would be harder as you have to start skipping lines.
The nice thing about 6 columns is that all primes except for 2 and 3 are actually of the form 6n+1 or 6n-1. This does in fact make it diagonals all the way down. And I mean ALL the way down. Every single one of them. Yes the slopes would be much steeper but they're still straight lines.
5:00 and a herd that wont become *unhappy meals*
I finally found their video with the funniest quote that I thought never existed
When you are not able to solve any Ted ed riddle but then also watch every Ted ed riddle video.
Another solution: mentally make the grid have length 20, but we won't have to write it down. Since we have the numbers 6 and 9, every multiple of 3, except 3 itself, can be made, so we can remove those columns. The next row is the first numbers +20, which means the numbers we just crossed out are no longer divisible by 3 (digits add to 2, 5, or 8), so we can cross out all of the multiples of 3 in this range. In the next row, a similar story, except now we've just out all numbers whose digits sum to 2, 5, or 8; are about to cross out all numbers that sum to 3, 6, or 9, and in the first round crossed out almost everything that now sums to 1, 4, or 9. The only survivor is the 3 we couldn't make, which is now 43, and is the largest as 63 is crossed out in the next row.
Love that you guys made a Monty python and the holy grail reference
I seem to be missing something...
The width to be defended is 1000, which isn't divisable by 43, so that leaves a gap of 11 through which the enemies can get in?
Wouldn't it be simpler to produce 997 + 3 instead of 23 * 43? Or if you want to stick to 43, produce 1 extra piece of 11 to close the gap?
The instructions say you can fill the gaps in with boulders
997 could be broken with 3×9+5×6+47×20. And making 23 walls of 43 and 1 of 11 is the fastest way possible
I love that you decided to just throw in parodies of fast food chains into this riddle for no reason.
The numbers are based on the former McNugget serving sizes.
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.
✨TInY dInO cReAtUrEs✨
I tried with other numbers along 6, and it works like magic.
lol there's an animation error at 4:54 where it makes the same wall twice
Fun fact the best way to apologize if this is an accident is: I’m sorry I didn’t know how much those dinos meant to you. Fun fact: The best way to apologize if this is on purpose (which it is) How did you feel? I felt sad. Sorry I broke my promise.😮 true it’s on Ted Ed channel best ways to apologize.
Reached the 70th riddle! We have come a long way.
I solved this riddle by writing a program in C which goes through every number from 1 to 1000 and goes through every combination of weapons to check whether this number is suitable. I also got the answer to be 43 but in just 2.802 seconds.
This is by far the best quote we get at the beginning of any TED-Ed videos.
I kinda found an alternate more elegant soln
I)All nos of the type 3n can be broken ( except 3)
II)Now every no 20+3n can also be tore down( except 23)
III)And every no 40+3n behave the same way(except 43)
Now we can see as this series continues after 40, 40 belongs to (III) category
41 to (II) and 42 to (I)
And since these categories go with a difference of 3 every no after 40 is counted in either one of the categories
EXCEPT 43 and thus that is the answer
Now I remember. Numberphile made a video about this with McDonald’s nuggets
There is a similar question pratically: In Hearthstone in China, the players are only allowed to invest the whole number combinations of RMB 30, 60, 128, 328, 388 or 488, of which only 30, 128 and 328 practically makes sense. For instance, some virtual products are sold at 98 RMB, which is an "undestructible" amount. I have run a brute force code that gives out the result that the largest "undestructible" even number in this case is 1010. I do not know if there are any better algorithms better than brute force in similar questions.
I can see this as an actual math examination question, not too hard but still challenging
This is my favourite riddle, Ted Ed! You should make more!
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.
997-49=948. 948÷6=158
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
Feels nice when you create a program that can solve the question
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.
It does say that the final wall will be 1km wide
@@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
You guys are like the Riddler in terms of what constitutes a riddle.
Plot twist- they climb the mountains-
I figured I could just start plugging numbers, but hearing how the solution gets set up is much more interesting.
0:25 Mc Donalds , Burger king, Wendy's
unhappy meals
That’s such a creative way to think about this
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.
None of these riddles are ever really riddles. They’re just maths problems with a story
fr
So... a riddle?
@@sinnerthesinful552 math problems are not necessarily riddles, do you even know what a riddle is?
@@dieselboy.7637 A story with a problem that requires logic to solve... so while not all riddles are math problems with a story, the inverse is almost always true
whoever makes these riddles you're legendary
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
We really got Ronald McDonald, Burger King, and Wendy teaming up to eat some dinosaurs. Nice to know that despite the fact that they're constantly at war with each other they're willing to team up to eat some baby dinosaurs.
Ted always provide everyone with a wide range of knowledge and also inspire science to the world
This would be a good bonus challenge: Make an algorithm for determining the max. wall length for 3 types of siege engines, of arbitrary length.
For a _bonus_ bonus riddle: do the above problem with an arbitrary number of the types of siege engines.
Oh, and you get extra points for your algorithm taking reasonable time(proportional to the size of the numbers, so a century is reasonable if, say, the numbers were in the billions)
0:08 this part literally said BAD NEWS
yes
الحُزن!
يقول ابن كثير رحمه الله: "لا تُستنار البصيرة إلا بالحزن، فعندها يرى المرء حقيقة كل شيء؛ حقيقة نفسه، وحال قلبه وصحبته وأهله، حقيقة الدنيا على حالها، فيجعل الله من كل ذرّة حزن في نفس العبد نورًا يضيء به بصيرته حتى يدرك هوان الدنيا برغم جمالها وحقيقة الأشياء".
1:06 urgh, already bored. Weaponise your dinos, murder your enemies in their beds, drink from their skulls.
With the clown you can take down any wall size that is a multiple of 6, but you need the other two to get non-multiples of 6. If you add just one royal to the mix you can take down any wall of size 6c+3 (c is the number of clowns) except for wall size 3. Adding more than one royal is not necessary, because any two royals can be replaced with three clowns. If you add one redhead then you can destroy walls of size 6c+2 or 6c+5 if you also added a royal to the mix. With two redheads you can destroy walls of size 6c+4 or 6c+1 if you add a royal to the mix. Having more than two redheads is unnecessary, because every three redheads can be replaced by ten clowns. This covers all the possible positive integers, but only if they are higher than a certain number x. This number x is the one we are looking for and x = -6 + 9 + 20x2 = 43. Why? Clearly x cannot be destroyed by any combination of clowns, royals, and redheads, because 43=6x7+1=6c+1 and any number of this type can only be found with a royal and two redheads, but those put together are already too big. Since wall size x cannot be destroyed we do not have to look for a lower number. Numbers 44, 45, 46, 47, and 48 require either less redheads or less royals or both. This means that these numbers will definitely be possible to destroy, because there is room left to add clowns in the mix. Wall size 49 can be destroyed with a royal and two redheads and any number higher can obviously be destroyed as well, because any wall size that is 6 less can be destroyed as well, so it is just a matter of adding more clowns.
So 43 is the biggest wall size that cannot be destroyed!
How are walls of 43s gonna be immune to an attack? Can't the enemies just crack down multiples of 43 walls at once?
Technically only a single wall of 43 works, assuming the wall is just 43 segments.
The video never said the enemies could do that (they can't).
nope, they specified it in the ‘rules’ before the solution
This is easier than I thought. Remember 6 and 9 are factors of 3 so it's simplified to no 3x numbers. They can only use 1 or 2 20s as 3 times 20 is 60 which is duplicate with 3x. Since 20 is in 3x+2 form with both 20 and 40 (which is 3x+4 or in another word 3x+1) they can tear down any number. Count down from 40 and you found 37 not suitable for the enemies.
Damn 43 is also suitable. It took advantage of 40. Yes this remaining is divisble by 3 but since it's made with 6 and 9s the 3 has became a dead angle where the siege weapons won't reach.
Why can't they just Eat their Siege Weapons? I mean, they're Giant Food.
>Lisa: Poor predictable Bright, always uses brute force
>Me: Good old brute force, nothing beats that!