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The digit from first to last represents the quantity of 0,1,2,3........ In that number Suppose 1st digit is 5 then the totap quantity of 0 is 5 2nd digit is x then the quantity of 1 is x 3rd digit is y then the quantity of (n -1)= 2 is y
@@mohitkumar-wl8dz because when you get to the 10th digit, the one that tells you how many nines there are, you'd have to have 1-and then you don't have 9 zeros anymore.
@Vedanth 2020 is autobiographical, 2010 is not. 2020: two zeros, zero ones, two twos, zero threes. While 2010 says that there is: two zeros (correct), zero ones (wrong), one two (correct), and zero threes (correct)
@@ninjahunterx7497 because we forgot to say how many 1s there are (1) but writing it makes 2 1s, but hold on, writing 2 ones makes the number of 1s...1 so it's a catch 22 situation
@@neonrider6599 No I figured it out, so if I write 8000000010 it's not that cause like you said we forgot to tell that there is a 1 in the number, so if you put a 1 there, the number becomes 8100000010 but as you can see now there are seven, 0's not eight, so what if we change it to 7, (the placement of 1 should be changed to the 7th position as well) , then the number becomes 7100000100 this seems correct but there are a total of 2 ones in the number so we have to put 2 there so if we put that the number will be 7200000100 but as you said this doesn't make any sense cause now there is only one 1's in the number but it says there are two, and to add to that the number doesn't say there is a single 2 in it. So we have to put 1 in the 2nd position as well, so the number becomes 7210000100 but the problem now is there are only six 0's, not eight(the placement of 1 should be changed to the 6th position as well) so finally the number becomes this and it's the only correct answer, 6210001000
No no no there are no such thing such as a wrong formula. Just because you solved it through an unorthodox manner means that you solved it wrong. It just implies that you think uniquely. I'm not saying being unique already equates to being the most effective, but come on, this is just simple maths, it's not about philosophy and moral values. Therefore, any solution that comes up with the right answer is correct and one could say in this context, "The ends justifies the means."
@@sophiap.8859 this is why I said above that being unique does not already equate to being the most effective. Yes you might have coincidentally got the correct answer through unconventional means but that doesn't mean that a method like that works for other numbers and variables. Hence, it is not very effective. These are only my 2 cents of opinions, maam, and anyways, thank you.
how i found out the 3rd code is by logic.. kinda 9000000000 "hmm. there's 9 zeroes, and that's right, but it shows that there's "no 9s". i'll fix that" 9000000001 "now the number of 9s is correct, but the count of 0s isn't." 8000000001 "wait.. now i have to change the position of the 1, since the 9 isn't a 9 now, and the number of 1s." 8100000010 "wait.. now the number of 1s is also a 1, making it not accurate" 8200000010 "wait.. now the number of 0s is not correct, but this'll change the 1's position, and i have to notate how many 2s there are. oh well" 6210001000 "wait.. WAIT that's the answer!" and uh yeah, that's how i found out
6210001000? Edit: OMG, I got it right! I’ve never solved Autobiographical Numbers in my life but I just used simple process of elimination: Step 1: 0000000000 Step 2: 9000000000 Step 3: 9000000001 Step 4: 8000000010 Step 5: 8100000010 Step 6: 7100000100 Step 7: 7210000100 Step 8: 6210001000
@@nobody-11550 trial and error. Step one, with 9 zeros, but then you'd have to have a one in the last position, and you no longer have 9 zeros. Step 2, 8 zeros, a 1 in position eight, but then you'd need a 1 in position one, which would have to be a 2, and you'd need a one somewhere else, not leaving enough zeros. So, just keep adjusting the numbers, making sure they follow the rules, and you'll quickly get to the only possible solution.
I finally answered a riddle correctly! Although, I actually used a different method. Here's how I did it: I decided that the simplest way of finding the number was to first consider the amount of zeros. First possibility: there are 10 zeros. This is impossible since that would require a 10 as the first digit. Second possibility: there are 9 zeros. This solves the problem of there being no room for the zero counter, allowing for 9 zeros and a 9 as the first digit to count the zero. However, this would require a 1 as the 10th digit, which is impossible since this would give 11 digits Third possibility: there are 8 zeros. This solves the problem of there being no room to count the zero counting digit, giving 8000000010. However, there is a new problem: there is no way to count the ones, so this too is impossible Fourth possibility: there are 7 zeros. This solves the problem of there being no way to count the ones, allowing for 7100000100. However, this then causes a paradox: if there is a 1 as the second digit, then there are two ones and you need a 2 as the second digit, and if there is a 2 as the second digit, then there is one one and you need a 1 as the second digit. There is a solution though: 7210000000. This gives you two ones and one two. One two and one one are already used, but you still have the one one you need to cover the one seven. However, one seven, two ones, one two, and seven zeros, giving 11 digits. Thus this is impossible Fifth possibility: there are 6 zeros. This final possibility frees up the room needed to place the second one, giving us 6210001000. There are no contradictions inherent in this number, meaning that it is the right one
Shubham Ahirwar yeah, it’s easier if you start trying numbers from the end. you basically now most big numbers are 0, and so you work your way to the front like that. so not totally random, but kind of.
"It would take forever." I literally just started trying combinations starting with 9000000000 and found the answer in less time than it took to explain the calculated answer.
Did the exact same thing. Our solution is faster for a limited amount of digits to discover, his solution is faster for a big number of digits to discover (if you assume to use computational power). That's the difference between "human algorithm" and "computational algorithm".
It's a very clear explanation, but the concept is difficult. You have to pause after almost every sentence and think for a minute or two to really get it.
Seems like a whole lot of reasoning when you can just start with a 9 at position 0 and continuously fix your number with logical thinking till you arrive at a valid solution. Since there are only 6 possible choices for the first digit and the rest follow immediately it is guaranteed to be easy to find.
@@hazardousteam2899 Well of course you can start with whatever you want. But it's better to start with a 9 because then you are counting the number of zeros already and it becomes a more logical sequence of steps in my opinion.
Yeah. Or similarly, you start with N 0s. Therefore, you now have one N. Therefore you now have one 1, but that's a paradox. This paradox can be resolved by saying you have one 2 and two 1s. This just leaves X 0s, which is what N should be and also decided the position of the second 1. There, done, no need for a four minute explanation :P
That's exactly what I did! This explanation was pretty long and overwhelming, but I just started at 9, then placed a 1 at the end, then a 2 for the second digit, then a 1 for the third digit. Then I changed the 9 to 6 and moved the 1 at the end to the "Six slot." It took about 10 seconds.
The zero rule: First: 3000 >i have a 3> 3001 >i have a 1> 3101 >i have two 1s> 3201 >i have just one 0> 1201 >i have one 2> 1211 >i have zero 3> 1210 Second: 6000000 >i have one 6> 6000001 >i have one 1> 6100001 >i have two 1s> 6200001 >i have one 2> 6210001 >i have only three 0s> 3210001 >i have zero 6> 3210000 >i have one 3> 3211000 Third: 9000000000 >i have one 9> 9000000001 >i have one 1> 9100000001 >i have two 1s> 9200000001 >i have one 2> 9210000001 >i have only six 0s> 6210000001 >i have zero 9> 6210000000 >i have one 6> 6210001000
Just start from either end using 0s and u fill find a pattern or series how these numbers work so u could find any of this type of number whenever now. This method is muchhh better than the solutuion provided according to me. It took me only a min or two.
Did I do something wrong? He said there was only one answer but after looking over the rules and method again and again I don't see why 9 billion doesn't also work. Can someone please explain?
God Bear 9000000000 wouldn’t work because according to how autobiographical numbers work we would need to add a 1 to the last digit because we have 1 9 but if we were to do that we don’t have 9 0’s now we have 8 😂 it’s complicated
As TED-Ed riddles go, this one is fairly easy. I solved it using an instinctive trial-and-error method in less time than it took to watch the video. (The key insight is that there should be as many 0s as possible, to simplify the number.) Judging from the comments, many people used the same approach. Apparently no one used the "official," pseudo-deductive method given. A TED-Ed answer should track how humans actually think, not merely set forth an algorithm that can be grasped in hindsight but that is more of a proof than a practical solution.
A variation, somewhat less likely to destroy the combination (assuming you seriously _cannot_ be bothered to look up the Wikipedia entry "Autobiographical Numbers"): 1. Find yourself a paper conservation laboratory. You're the protagonist of a TED riddle video; there's probably one _somewhere_ in your bizarrely-specific contacts list. If there isn't, try asking a library with a rare books collection. 2. Pay them to analyze the stain and spend all of ten minutes figuring out a way to remove the stain without damaging the paper or the writing. It might be a bit pricey, but money should be no object. Whatever's in _"The Lost Vault of Leonardo DaVinci" must_ be worth _something._ 3. Profit. A lengthy book of an old man's last ramblings written _entirely_ in backward Renaissance-era Italian might not interest _you,_ but dozens of universities, museums, historians, and other scholarly types (not to mention independent tech billionaires) will pay through the nose to so much as _see_ it, let alone _own and analyze_ it, and while you wait for their checks to clear, you can write the story of how you found the Vault and live for the rest of your life on royalties checks and appearance fees.
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It's not all good and something else matters in these problems - speed and efficiency. The purpose is not only finding the combination, but doing it in the least time possible, which is why you look for a pattern, constraints, formula, relationship. One could spend hours or days going over possible numbers one by one, but that's what we want to avoid. 😊
"It would take too long to try all the possible combinations" Whoops. Guess I got lucky then! Lesson learned - use EVERYTHING that you are given. Nice riddle!
1210 3211000 6210001000 9210000010000 Just keep adding 3 to the first one, the next ones are always 2 and one, then fill with 0's except for the one representing the 1st number. Boom solved all of them.
@@eugenemoreno482 because of your answer to that person, he only asked her if she completed it or not and then you tell him it took her a day to solve it only based on her comment, THAT'S WHY.
I solved this riddle in a completely different way from the one described in the video. It's always great to see there are so many ways to solve the same problem, and it's encouraging for a math goat like me!
This riddle's explanation is hopelessly convoluted. The unnecessary set-up and pattern inferences were confusing and unintuitive. I solved it in about a minute by heuristically realizing that most of the number would have to be 0s. Trial and error from 9 to 8 to 7 to 6 found me the answer. The longest autobiographical number is 10 since you can't count double digit+ numbers, so using a formulaic approach doesn't achieve anything, unlike discovering some pattern between prime numbers, or perfect square numbers etc. Still a fun puzzle though.
For the bonus riddle is debatably more efficient, especially when considering you need to be certain you have them all. Also, this riddle is likely aimed at people with far weaker problem solving skill for whom the idea that looking for patterns and narrowing in on and answer instead of randomly guessing can be an important take away, even if you are correct that eventually people get beyond that and can learn to evaluate the complexity of a problem and tell whether, practiced guessing or proper pattern seeking is more efficient.
Sorry, but your answer is wrong. Here is a autobiographical number with 11 digits: 72100001000 The upper limit thus is 13 digits. Also you miss some autobiographic numbers if you generalize the proof, because not every autobiographic number requires to have exactly two ones. For example the autobiographic number 2020 does not. The proof in case of ten only works because we know from our calculations before that there are at least three zeros. A better proof regarding autobiographic numbers would be the following. Step 1: There has to be at least one zero. Proof: If there were no zero then the digit for the numbers of zero would be zero. That is a contradiction. Step 2: If n_i is the number of digit i, define k_i as follows: k_i=1 if n_i>0 and k_i=0 if n_i=0 This means: either n_i -k_i >0 or n_i- k_i =0 Also n_i -k_i >0 means that n_i is at least 2. Step 3: Let d be the number of digits and m=d-1, and let s be the number of non-zero digits. We know that n_0+n_1+n_2+...+n_m=m+1 and n_1+n_2+...+n_m=s since n_0 is NOT zero (see step 1) we know there are exactly s-1 nonzero digits among n_1, ...,n_m this means that l_1,...l_m from step 2 is exactly 1 for s-1 of these l_i. and this again means that (n_1-l_1)+(n_2-l_2)+...+(n_m-l_m)=s-(s-1)=1 (!) since all of the n_i-l_i are nonnegative integers (see step 2) this means there is one and exactly one of the n_i that is two for i>0, while the other n_i for i>0 are either ones or zeroes. If one of the n_i for i>0 is three or larger this would mean that the equation marked with (!) would not be true anymore, no matter what the other n_i were. Also n_i can only be two for i=1 or for i=2. If n_i=2 for i>2 then there would be two numbers of at least three. One could be the number of zeroes, but the other one would then be one of the numbers of non-zeroes. So one of the n_i for i>0 would be three or larger, which would contradict the statements before and the equation (!). So there are two cases now. Either n_1=2 or n_2=2. Step 4: Checking case n_2=2 If n_2=2 then there are two 2s, one of them describing the number of 2s according to our assumption. The other HAS to describe the number of zeroes, or (!) would be false. So there are 2 zeroes and two twos. Because every non-zero number only appears zero times, once, or twice, and 1 is not appearing twice according to our assumption, this means one is either appearing once or one one is not appearing. Because there are exactly two zeroes, any number bigger than two can not appear either or equation (!) would be wrong again. This means for this case there are two possible solutions: 2020 and 21200 Step 5: Checking case n_1=2 Since n_1 is the only n_i with n_i > 1 and i>0 (see step 3) and the digit 2 appears this means n_2=1. So we know there are exactly one 2, which is the number of ones, and two ones, one of them representing the number of twos. Either the number of zeroes is one, which then leads to one zero, two ones and one two, with the rest of the numbers being zero. But because there is only one zero, this has to be a 4-digit number. And this is the number 1210 . The other possibility is that there are more than one zero. This means from the two ones in the number, one is giving the number of twos, the other one has to be one of the remaining n_i with i>2, with the other n_i being zero (we know the number of ones and twos and according to step 3 for i>0 n_i can only take values one, two or zero). But this means then that s as defined in step 3 is always 4 in this case. Thus d-s=d-4 (with d being defined in step 3 as number of all digits) and that is both the amount of zeroes and the remaining i for which n_i equals 1. This means the number consists of: d-4 zeroes 2 ones 1 two 1 number d-4 the restriction is that d-4 is not higher than 9 otherwise you cannot express the number of zeroes in a single digit. Thus d is 13. Also d-4 has to be at least one according to step 1, thus d has to be at least five. However for d=5, d-4=1 and then there had to be both exactly one one and two ones, which is impossible. This means d has to be at least six, but then d-4 is two. The number would then only have one one, despite the digit, giving the numbers of ones being 2. This means d has to be at least 7. This gives the numbers 3211000 42101000 521001000 6210001000 72100001000 821000001000 9210000001000 So there are -nine- ten (wow I can solve this but I cannot count to ten) autobiographic numbers in total: 1210 2020 21200 3211000 42101000 521001000 6210001000 72100001000 821000001000 9210000001000
@@brunobar4808 you're obviously very into mathematical proofs and I am not as enthusiastic about math to get into your ridiculously long UA-cam comment but I will say that you are wrong about autobiographical numbers. The longest autobiographical number for base 10 numbers is 10, and this is a definition. If you don't believe me, just look up self-descriptive numbers on Wikipedia. So the ones you found after that for how many times 11, 12 and 13 show up are not autobiographical numbers. Also, the longest one of those is 36 digits long, so maybe you can expand your proof.
AxielFan "The longest autobiographical number for base 10 numbers is 10, and this is a definition." That is changing the definition from autobiographical numbers from the video and even then still incorrect/unprecise. The video author provides the definition of _Autobiographical Numbers_ as numbers with a flexible length in base 10 (no other base mentioned) where the digits give the number of digits corresponding to that position. Or in other words: In the definiton of the video there is no restriction regarding the length of a number. For this case AND ONLY for this case my proof holds although it is esay to gerneralize that proof (see further below). Since the task at the end of the video is to find ALL autobiographical numbers, it seems plausible that we only should have a finite amount of them. Now to your statement. You say the longest autobiograpical numer for base 10 is 10, but 10 is actually not a autobiographical number. If you mean that the _self-descriptive_ number of base 10 always has the length 10 according to the definition of _self-descriptive numbers_ from wikipedia (and by the generalization defined by wikipedia this would be the highest autobiographical number), then the statement is true, but trivial. Also according to the wikipedia definition there of course then exist self-descriptive numbers of length 11, 12, etc. just not in base 10. But then your argument still is wrong, because in your initial comment you claim, that you yourself assume to numbers to be written in base 10 (which is not reasonable when refering to the wikipedia definition, but you assume a longest self-descriptive/autobiographic number to exist). And then there are not only 36 self-descriptive numbers but infinite. For example in base 1000 000 the self-descriptive number would be 999 996 zeroes (in base 1000 000 the number 999 996 is a single digit), two ones, one two and one 999 996. This disproves, that the longest autobiographic number is 36 digits long. My proof would only have to change at one place to be valid for the wikipedia definition of self-descriptive numbers and by generalization to fewer digits autobiographic numbers: I simply have to drop the upper limit for the amount af zeroes. This would always be valid, because the amount of zeroes cannot exceed the base=length of numbers and thus be a single digit in that base. In no other case the base is relevant for my proof, as long as the base is at least 3 (because otherwise the digit 2 would not exist). However for base 3 or lower there is no autobiographic number. And the autobiographic number remains the same when we look at the same digit positions and number length in a higher base, which is the bridge between the video definition and the wikipedia definition. Compare the numbers i found with my method to the wikipeida numbers until base 13. Why are they the same? Well the higher bases have MORE digits than the lower bases. So if a digit combination in the lower base exists, the same digit combination will also exist in a higher base (but the number represented by the digit combination is not the same). And even in base 13 the highest digit, that appears is the digit 9, a digit which also appears in base 10. Also the trial and error method is flawed, because such a method can only prove at best that a solution exists (by finding one) not that the solution is unique (unless you check ALL combinations, which would take longer than proving my stuff). This is what justifies a formulatic approach. the mathematician often is not interested in the problem itself, but in the solution of a class of problems.
This took me 2.5 whole hours of dedication and ended up with a different process (not hit and trial), but the same solution. My brain is drained, my day is destroyed.
Coop Cooper Indeed it took me something less than 10 minutes, because half of what he says comes easily once you start to test combinaisons. You realize rapidly that except for the first positions "0" or "1" or "2", if you put big numbers like a 5 or so, you will have to write that number too many times ! Ex : 5 _ _ _ _ _ _ _ _ _ _ 0 1 2 3 4 5 6 7 8 9 I now need to pull 5 numbers "4" in my number which is hard ! Cause it means, they are "at least" 4 zéro, 4 one, 4 two, and 4 three ! And that's even more numbers to put into my number (the 4 two, the 4 three etc.) I don't have enough digits ! ==> So you end up pretty fast trying to put the more Zéros you can in your number, and with that, the solution comes after maybe 50 trials at most.
I solved this in a different way that could be considered as going through every possible answer but technically isn't: 1. I started with 0000000000 2. I insert 10 for the first digit before taking away 1 since 10 would replace the 0: 9000000000 3. I inserted a 1 as the last digit (since there is one 9) thus decreasing the amount of 0 to 8: 8000000001 4. I changed the position of one to 8: 8000000010 5. I put a 1 at the second digit (representing the amount of 1s) but now there are 2 ones so I placed a 2 instead, I also decreased the amount of 0s to 7 and moved the 1 to spot seven : 7200000100 6. Finally I put down a one at spot 2 thus making the 2 at spot 1 correct, this meant I had six 0s so I moved the 1 for spot 7 to spot 6 and got the same result as the video: 6210001000
Here is an unoptimized brute-forced one-liner in python: print(next(n for n in range(10**9, 10**10) if all(str(n).count(str(i)) == int(c) for i, c in enumerate(str(n)))))
Theres actually a much easier way of doing this. I found this riddle intriguing so i sat down with a notebook to solve it and it took me only like a min to do that. And i didn't do it anything like they did in the video.
@@sanjogagarwal1532 in these kind of problems u usually start from the end or the start. For example starting with nine 0s and one 9 u realize u have to put one 1 too and like this u continue further. Its a cute little pattern. It would work for some other cases too.
This was probably the easiest Ted-ed riddle i've seen so far, yet it has the most overcomplicated solution of them all... It's fun how he said "trying combinations would take forever" because that was much simpler than the solution shown in the video
@_ DeathStr0ke _ they meant the sign on top of the door saying “Leonardo Da Vinci’s Secret Vault,” saying that it wasn’t much of a ‘secret’ since there’s a sign stating it right there.
I worked it out hard way. Went from left and right side, assumed there would always be a zero somewhere, so I put "9" in slot 0. Then "1" in slot 9. Then "1" in slot 1, "2" in slot 1 and "1" in slot 2. - This was a little sneaky, because it's a short paradox state, when there were two "1", for a moment after shifting it to "2", there was suddenly only one of them, but setting "1" for slot 2 fixes that paradox. Anyway, after all that, I recounted number of "0" left, set slot 0 to "6", shifted "1" from slot 9 to slot 6, which did not change the number of "0", solved. I assume you might have done the same, if you didn't work it out through the intended solution either.
My process was to guess randomly with 5 and went as follows 5 ? ? ? ? 0 0 0 0 0 "ok, so 1 5 and 1 1" 5 1 ? ? 0 1 0 0 0 0 "Ok, so 2 1s now, and then 1 2" 5 2 1 0 0 1 0 0 0 0 "Ok, so there's 6 0s. If I move the 1 in the 5 slot to the 6 slot, (and make the 5 a 6), the sequence retains the same number of 1s, 2s, and 0s." 6 2 1 0 0 0 1 0 0 0 "double check? Yep. Checks out"
This is an ideal puzzle for a general math audience. The setup is easy for a layman to understand. The solution takes some thought, but requires only elementary methods. Perhaps best of all, the puzzle immediately suggests generalizations and variations (What if the 0's are counted in the last digit instead of the first? What happens using other bases? When is one number the biography of another? What happens if we iterate: take the bio of the bio of the bio...?) Hours of math entertainment here!
As much as i love these, i much prefer phylosophical riddles which require you to think outside the box. Mathematical riddles have a tendancy to spoil themselves as soon as you know the formula to solve it. Much like a rubik's cube, i used to be interested by them when i was younger, until i found out that you only need to memorize a few algorithms, then i completely lost interest. Sure, with practice and skill you can learn shortcuts, but i find puzzles and riddles much less interesting when you can just follow a formula to solve them.
But philosophical questions tend to be subjective, and it is difficult to find an answer that everyone agrees on, which is, I'm sure, not the way Ted-Ed likes to present riddles.
Also people do not just "follow a formula." The formula Ted ed made was custom made to help solve this riddles. Formulas are simply a way of writing down logic and reasoning so we can understand it better and do more complex problems. The 'formulas' for these problems are just explanations of the logic.
If someone found Da Vinci's autobiography, they might have found one of the greatest leaps in pre computer science in history. Significantly more important than a Corona virus vaccine
I solved it starting from the end and realizing there could be no 9's since a 9 would result in 9 digits of that number (namely 0's), but then the number of 1's would result as 0, which contradicts the 1 in position 9. I did a similar reasoning for 8's and 7's. Then when I reached 6 I realized there could be a 6, and 6 obviously had to appear in position 0. Cool riddle.
6210001000 I started to think about how many times the 0 must occur. If you make the number small, you a few kinda high numbers which have to occur at least once. Wherever you place these high numbers, you need to place a high amount of digits, so you wont have enough space to place all these digits. After you try a 6 at the first spot the solution becomes very obvious. I think that most of these autobiographic numbers will have lots of 0 in general.
Max Mustermann I took the second number 3211000, added 3 0s to the end to make it 10 digits, changed the 3 to 6, moved the 1 from the 3rd spot onto the 6th spot and I ended up with 6210001000.
Jau Jo Doesnt really matter how much digits you try. The principle is always the same. I dont know if there is alsways only 1 possible solution (like for the 10 digit number), but the principle X210000000...00000001000 for an X+4 digit number always works, up to infinity. At first I wasvery excited about the complexity of an autobiographic number, but as I realized that, it got kinda lame.
My solution: 1. Figure out that the digits add up to 10 2. Consider any combinations of non-zero digits adding to 10 (there are 39 total combinations, but many you can eliminate right off the bat) 3. Count the number of zeros that would be in each combo (for example, for combo 5, 3, 2 there would have to be be 7 zeros) 4. Eliminate combos where number of zeros you found from #3 is not already in your number combo (left with 7 possibilities) 5. Of the 7 possible combos it is pretty easy to see that it's 6 zeros, 2 ones, and 1 two.
@@youstolemyhandleyoutwat I know what joke looks like. It wasn't joke and if you still think it is so think it I don't care and for God's sake don't explain me joke 😑😑💆
@@hasnain9654 ok, so jokes go like "knock, knock." Then you ask "Who's there?" And i answer "sugma." And you ask "Sugma who?" And i answer "your neighbor. Half of your house is on fire. Maybe we should call the fire department." And that, is a joke.
I got 6210001000 too Just type 0000000000 and find errors 0000000000 9000000000 9000000001 8000000001 8000000010 8100000010 7100000010 7200000010 7210000010 6210000010 6210001000
Just started with 9000000000 and worked backward. After that I put a one in the nine spot and changed the 9 to 8. Then I moved the one to the 8 spot. Then put a one in the one spot and changed the 0 spot to 7. Then had to change the # of ones to 2 and put a one in the 2 spot. Then had to changed the # of zeros to 6 and moved the one in the 7 spot to the 6 spot. Done. Gets you 6210001000
This explanation was pretty long and overwhelming, but I just started at 9000000000 and then used common sense for the rest. I started at 9, then placed a 1 at the end, then a 2 for the second digit, then a 1 for the third digit. Then I changed the 9 to 6 and moved the 1 at the end to the "Six slot." It took about 10 seconds.
A 1 digit autobiographical number = a paradox The first number represents the number of 0s but if it is zero then it means there are no 0s but there has to be a zero to represent that etc.
I first wrote 5210010000, and then i noticed it wasnt correct so i put a 6 and put the 1 one forward and it worked. One of the easiest ted talk riddles ever! Nice
I solved it but by using a completely different style of logic. - I did the same thing knowing that all the numbers would have to add up to 10 as my starting point. - I knew that the 9 HAD to be a 0. If it was a 1 (or higher), then that would mean another number would have to be a 9, which would be impossible without having to add other numbers. One digit solved! - Following the logic, I next checked the 8. If the 8 was 1, then that means that the 0s could be an 8. That then leaves a paradox, though, where the 1 can't say 1 without making it a 2, which would cause the 2 to be a one, which pushes it up to 12. Therefore, 8 also HAD to be a 0. - The above then made me realise that the sweet spot was X 0s, Two 1s, One 2, and One X's. - 2 + 1 + 1 + X = 10, so X must be 6. Double checked my work to make sure I was right, and that it all made sense, and got 6 2 1 0 0 0 1 0 0 0
I actually figured this out in a way that was weird even to me. I started by looking at 9000000000 (it was obvious that zeroes would be most of the answer). From there, I just changed values to reflect what was showing. The steps I followed looked something like this: 9000000001 8100000001 7200000100 7210000100 6210001000 [SPOILERS] I also solved the bonus puzzle, but not using the ridiculously complicated method from the video. I'd say the typical answers (with 10 digits being considered the longest) are these seven numbers: 1210, 2020, 21200, 3211000, 42101000, 521001000, 6210001000. However, I'd also say that this could extend even farther (where digit places past the tenth would obviously be 0 automatically because multiple digits can't fit in one place). The extended ones follow the exact same pattern (where n is the number of digits and is greater than 6, the only autobiographical number of that length will consist of the digits [n-4], 2, 1, an [n-7] number of 0s, 1, and three 0s) and still, technically, describe themselves: 72100001000, 821000001000, 9210000001000. So that gives us ten autobiographical numbers in base 10, but going into higher bases can make even more: A2100000001000, B21000000001000, etc. So I'd say that, even though I can't possibly type them all, I truly found all autobiographical numbers.
Okay, so, a piece of paper later I think the number is 6210001000. Cuz there are 6 zeroes, two ones, one two and one six. I hope I'm right Edit: first riddle I manage to get right xD I'm so happy :D Edit 2: also the way you get to the solution is way too complicated. Someone called Luca Perju explained an easier way. I did something like that.
I'm so surprised i got it. All i did was put ten blanks, started testing a few combinations until i came up with one that had me marking 6 in the zero spot... also, your explanation was waaay more confusing than it needed to be. Like i said you know you need a big digit in the zeros place so you just test out with 5, since that is the minimum possible, see if you can add numbers to equal the sum of 10, and if not move on
1:16 I did it more or less blindly (just a bit of logic), and it took me a lot less time than the length of this video (Didn't time it, but under a minute). For obvious reasons, this number has a lot of 0. Thus the first number has to be large (don't know how large, that's not important yet). There can't be 2 numbers that large, for the same reason the video stated, it would take too many digits. So that first digit (whatever it is) is listed only once, therefore it has a "1". Now, can that be the only 1? Of course not. If it was, there would be a "1" under 1, and that'd be the second "1". So there's at least 2 "1". If there's more than 2 (say, 3) then there's a "1" under three, but only 2 ones are used to there needs to be another 1 somewhere, say "4", which means there needs to be 4 of something, and so on, we quickly see that it doesn't work. So there can only be 2 "1". So we know, at this point, that the number is 'large digit _ 2 _ 1 _ some zeros _ a 1 under the large digit _ more zeroes". So there has to be 6 zeroes, which tells us what the large digit it, as well as the full number, 6, 2, 1, (three zeroes), 1, (three zeroes).
eh it wasn't too hard, i just started with 9 as the first digit and the rest zeroes and then checked it. if it was incorrect, i'd adjust the list to make it correct. Thought process: 9000000000 - there isn't a 1 to count the number of nines 8000000100 - made it so that there is a 1 to count the number of nines but i realized that would bring the 9 to an 8 since there is one less 0 7100001000 - made it so that there is a 1 to count that 1 that counts the leading digit 6100010000 - made it so that there is a 1 to count the number of ones 6110010000 - made it so that there is a 1 to count the number of ones again 6210010000 - changed it to a 2 because changing that to a 2 would result in 2 1's and 1 2 balancing it out
Dammit, I thought I was so clever with 9 000 000 000 then realized there's one 9 in there. So it can't be 9 000 000 001. 8 000 000 010 is wrong too. Bedtime.
I started with 9000000001 Went back realized I need 2 1's 1 for 2 and 1 for the number of 0's added them to 4 realized I only needed 6 at that point and just moved the 1 from 9 to 6 and Came to the answer
Yeah, so via the conditions you established, we know that for any case such that t_0>2 any sequences must look like k,2,1,0,...,1,...,0. For lower values this doesn't work cuz k can be included in the 1/2 count. Some case checking shows that n6 the above sequence works, with K=n-4. 3211000,42101000,521001000. The first digit increments, you slide a 0 into the middle and it works. Basically what you've given is a necessary condition, (and it confines us to only one answer per n), and some case checking can give us enough sufficient conditions to have the full answer
Who cares about the method they use? We each think differently (even though most, if not all, in the comment section used the same method). What matters is that the puzzle is solved.
Figuring out the ten digit one is easy. Finding a six digit one, using the same process, I found is extremely difficult. Edit: there are no six digit autobiographical numbers.
So for the confusing part, let’s say you have a sum that we’ll call S. It has exactly n terms, and S = n + 1. All terms are positive, meaning they’re all at least 1. If they were all equal to 1, then S = n, which is a contradiction. This means at least one term is >= 2. If two or more terms are at least 2, then S >= n + 2, which is also a contradiction. This means at most one term is >= 2. So there is exactly one term that is at least 2. All the rest have to be 1. If that one term is greater than 2, then S > n + 1, which is a contradiction, so it has to be 2. Finally, we have that S has exactly one 2, and all the rest are 1’s.
Get the solution to Bonus Riddle here: brilliant.org/tededautobiographical/! Also, the first 833 of you who sign up for a PREMIUM subscription will get 20% off the annual fee. Riddle on, riddlers!
Could androids become a possibility? And if so could they go rouge?
TED-Ed how do snakes not bite themselves?
@@ankushpatanwal6535 nope. The 9 digit appears once, so the sequence should be 9100000001
The answer is 6210001000
@@ankushpatanwal6535 no as there is 1 nine
I don’t understand anything but his voice is smart and soothing
His voice is irritating
Yeah, I was trying it out for luck and got it after one minute. Luck was with me
😂😂thank God am not alone
I Think he uses Voice modulation to make it sound more like an AI
The digit from first to last represents the quantity of 0,1,2,3........ In that number
Suppose 1st digit is 5 then the totap quantity of 0 is 5
2nd digit is x then the quantity of 1 is x
3rd digit is y then the quantity of (n -1)= 2 is y
I solved the riddle for about 2-3 minutes on my own, after I watched the explanation I was confused for about an hour
I am glad i wasn't the only one 😂
same here
Its 90000000
Please tell me why it cant be 9000000000?
@@mohitkumar-wl8dz because when you get to the 10th digit, the one that tells you how many nines there are, you'd have to have 1-and then you don't have 9 zeros anymore.
Too complicated, I'm bombing the door.
Yeah explosive chemistry is way easier than this *sarcasm intensifies*
TSAR BOMBA OR NAPALM NUKE?!?!?!
As a chemist, I agree. Chemistry is so much simpler than this.
@@avplays214 TSAR BOMBA.GLORY TO РУССКИЙ
What if it still doesn't open?
Remember guys, 2020 is an autobiographical number.
How?
It is not but 2010 is
@Vedanth 2020 is autobiographical, 2010 is not. 2020: two zeros, zero ones, two twos, zero threes. While 2010 says that there is: two zeros (correct), zero ones (wrong), one two (correct), and zero threes (correct)
no
@@magipanda7810 2120 and 2110 are autobiographical I think
Riddles: Who was the murderer?
TED-Ed Riddles: *laughs in planck's constant*
Can someone please tell me why it can't be 8000000010 ?
@@ninjahunterx7497 because we forgot to say how many 1s there are (1) but writing it makes 2 1s, but hold on, writing 2 ones makes the number of 1s...1 so it's a catch 22 situation
@@neonrider6599 No I figured it out, so if I write 8000000010 it's not that cause like you said we forgot to tell that there is a 1 in the number, so if you put a 1 there, the number becomes 8100000010 but as you can see now there are seven, 0's not eight, so what if we change it to 7, (the placement of 1 should be changed to the 7th position as well) , then the number becomes 7100000100 this seems correct but there are a total of 2 ones in the number so we have to put 2 there so if we put that the number will be 7200000100 but as you said this doesn't make any sense cause now there is only one 1's in the number but it says there are two, and to add to that the number doesn't say there is a single 2 in it. So we have to put 1 in the 2nd position as well, so the number becomes 7210000100 but the problem now is there are only six 0's, not eight(the placement of 1 should be changed to the 6th position as well) so finally the number becomes this and it's the only correct answer, 6210001000
@@ninjahunterx7497 start it with 4 and you'll get easier to the solution
@@ninjahunterx7497 the total number tells how many digits they are
Title: Can you solve....
Me: Probably not. But I'll watch the 5 minute video anyway.
chikka woahwa LOL
So relatable
RE-LA-TA-BLE
#Relatable
Despite the fact that this is one of the easiest riddles they have?
When you got the right answer using the wrong formula
No no no there are no such thing such as a wrong formula. Just because you solved it through an unorthodox manner means that you solved it wrong. It just implies that you think uniquely. I'm not saying being unique already equates to being the most effective, but come on, this is just simple maths, it's not about philosophy and moral values. Therefore, any solution that comes up with the right answer is correct and one could say in this context, "The ends justifies the means."
Jan Jacob Fernandez what if I solved 2^2 because I said 2+2=4
@@sophiap.8859 this is why I said above that being unique does not already equate to being the most effective. Yes you might have coincidentally got the correct answer through unconventional means but that doesn't mean that a method like that works for other numbers and variables. Hence, it is not very effective. These are only my 2 cents of opinions, maam, and anyways, thank you.
@@JaimeAMSilva genius
@@janjacobfernandez5527 Tell that to schools...
A rare riddle from TED-Ed where the solution is easier to get by trial-and-error than to understand the "explanation".
how i found out the 3rd code is by logic.. kinda
9000000000 "hmm. there's 9 zeroes, and that's right, but it shows that there's "no 9s". i'll fix that"
9000000001 "now the number of 9s is correct, but the count of 0s isn't."
8000000001 "wait.. now i have to change the position of the 1, since the 9 isn't a 9 now, and the number of 1s."
8100000010 "wait.. now the number of 1s is also a 1, making it not accurate"
8200000010 "wait.. now the number of 0s is not correct, but this'll change the 1's position, and i have to notate how many 2s there are. oh well"
6210001000 "wait.. WAIT that's the answer!"
and uh yeah, that's how i found out
But the explanation let's us easily find all infinitely many autobiographical numbers (assuming we are dealing with an arbitrarily large base).
excellent @@marcinjanoska3345
@@marcinjanoska3345 Understood through your method, now this is the real way of finding this out in such a short period of time.
".... Inside you find DaVinci's long lost autobiography"
*disappointed face*
Maybe it could be sold for more than a million dollars
@@murtazahamid6141 more than a million... yeah. Like multi billions dude.
It would tell the story of his friendship with Ezio Auditore
Yeah, uh, if you don't want it I'll happily take it off your hands.
@@b2spirit35 And how he gave Ezio pretty much all of his cool equipment...
6210001000?
Edit: OMG, I got it right! I’ve never solved Autobiographical Numbers in my life but I just used simple process of elimination:
Step 1: 0000000000
Step 2: 9000000000
Step 3: 9000000001
Step 4: 8000000010
Step 5: 8100000010
Step 6: 7100000100
Step 7: 7210000100
Step 8: 6210001000
Can you please explain this? I am a little girl but mathlover.
@@nobody-11550 trial and error. Step one, with 9 zeros, but then you'd have to have a one in the last position, and you no longer have 9 zeros. Step 2, 8 zeros, a 1 in position eight, but then you'd need a 1 in position one, which would have to be a 2, and you'd need a one somewhere else, not leaving enough zeros. So, just keep adjusting the numbers, making sure they follow the rules, and you'll quickly get to the only possible solution.
@@SaraMargaridaP thanks but i figured out on my own.
T
i figured it out quickly like this too :)
4:27 *when the teacher says the homework only has 3 questions*
4:30 *when you find out each question has 10 parts*
I finally answered a riddle correctly! Although, I actually used a different method. Here's how I did it:
I decided that the simplest way of finding the number was to first consider the amount of zeros.
First possibility: there are 10 zeros. This is impossible since that would require a 10 as the first digit.
Second possibility: there are 9 zeros. This solves the problem of there being no room for the zero counter, allowing for 9 zeros and a 9 as the first digit to count the zero. However, this would require a 1 as the 10th digit, which is impossible since this would give 11 digits
Third possibility: there are 8 zeros. This solves the problem of there being no room to count the zero counting digit, giving 8000000010. However, there is a new problem: there is no way to count the ones, so this too is impossible
Fourth possibility: there are 7 zeros. This solves the problem of there being no way to count the ones, allowing for 7100000100. However, this then causes a paradox: if there is a 1 as the second digit, then there are two ones and you need a 2 as the second digit, and if there is a 2 as the second digit, then there is one one and you need a 1 as the second digit. There is a solution though: 7210000000. This gives you two ones and one two. One two and one one are already used, but you still have the one one you need to cover the one seven. However, one seven, two ones, one two, and seven zeros, giving 11 digits. Thus this is impossible
Fifth possibility: there are 6 zeros. This final possibility frees up the room needed to place the second one, giving us 6210001000. There are no contradictions inherent in this number, meaning that it is the right one
What
I did it exactly like this. Although you forgot to put a 1 in the 72100001000
i did exactly that, but i didnt really have strategy, i just messed with things until something worked
Well explained, far better than the video
@@RandomGoldieStuff lol same
Sweet another awesome Riddle I can't solve
Ikr!
I'm mesmerized by the flickering white background :D
i feel u man
You just don't want to. It can be done by trial and error, if you start for example with 000000...00.
It's not actually difficult, just really time consuming. We're simply too lazy
I literally got it faster by "try from the highest zeros possible" way. It was faster than the 2 minutes explanation!!
Me too
They need to make you understand, that's why it's long
Right!!??
Howw it took me an hour and I still didn't get ittt
same like after seeing how savagely 7 failed i tried 6 and the answer came on its own
video: “trying different combinations would take forever”
me: found the answer in about 1-2 minutes by randomly plugging in numbers
Is that true??
Shubham Ahirwar yeah, it’s easier if you start trying numbers from the end. you basically now most big numbers are 0, and so you work your way to the front like that. so not totally random, but kind of.
@@laralopez1645 Wow,
Quarantine has done amazing things to your mind
Shubham Ahirwar guess it’s not all bad 😂
@@laralopez1645 😂😂
By the way,for how many days is the lockdown in your country??
In mine,it's upto 14 april or may increase
"It would take forever."
I literally just started trying combinations starting with 9000000000 and found the answer in less time than it took to explain the calculated answer.
stronk computer lol
I did the same thing. Is this wrong?
@@NicoPanasiuk yes it is, there should be a 1 at the place where the count of 9 needs to be placed. Which in turn increases the count of 1.
Did the exact same thing. Our solution is faster for a limited amount of digits to discover, his solution is faster for a big number of digits to discover (if you assume to use computational power). That's the difference between "human algorithm" and "computational algorithm".
I did this too. Just keep fixing everything that is wrong, and you end with 6210001000
Wait, did he realize that the third code was missing, JUST when he got there?
Not that it would matter, since the wooden door has rotted away.
Haha, true!
Oh yea
Ye, but ye
@@liamgoldbeck????
（∵）？（´～｀)
That was a confusing explanation
Where's Richard Feynman when you need him.
It's a very clear explanation, but the concept is difficult. You have to pause after almost every sentence and think for a minute or two to really get it.
No kidding. There is an easier solution to this
Luca Perju's comment offers a better solution
It's annoying when you're sure you found the answer after a minute or so and have to sit through an explanation that doesn't match your approach.
Seems like a whole lot of reasoning when you can just start with a 9 at position 0 and continuously fix your number with logical thinking till you arrive at a valid solution. Since there are only 6 possible choices for the first digit and the rest follow immediately it is guaranteed to be easy to find.
It also works with a 1 on the first two spots. You adjust until you get the answer.
@@hazardousteam2899 Well of course you can start with whatever you want. But it's better to start with a 9 because then you are counting the number of zeros already and it becomes a more logical sequence of steps in my opinion.
Yeah. Or similarly, you start with N 0s. Therefore, you now have one N. Therefore you now have one 1, but that's a paradox. This paradox can be resolved by saying you have one 2 and two 1s. This just leaves X 0s, which is what N should be and also decided the position of the second 1. There, done, no need for a four minute explanation :P
That's exactly what I did! This explanation was pretty long and overwhelming, but I just started at 9, then placed a 1 at the end, then a 2 for the second digit, then a 1 for the third digit. Then I changed the 9 to 6 and moved the 1 at the end to the "Six slot." It took about 10 seconds.
The zero rule:
First:
3000 >i have a 3>
3001 >i have a 1>
3101 >i have two 1s>
3201 >i have just one 0>
1201 >i have one 2>
1211 >i have zero 3>
1210
Second:
6000000 >i have one 6>
6000001 >i have one 1>
6100001 >i have two 1s>
6200001 >i have one 2>
6210001 >i have only three 0s>
3210001 >i have zero 6>
3210000 >i have one 3>
3211000
Third:
9000000000 >i have one 9>
9000000001 >i have one 1>
9100000001 >i have two 1s>
9200000001 >i have one 2>
9210000001 >i have only six 0s>
6210000001 >i have zero 9>
6210000000 >i have one 6>
6210001000
I’ve watched so many of these math type riddles and I decided to actually sit down and try for once and I got it! Feels nice
I got the correct answer and I swear I have never been so happy in my entire life
What's the ans then
i feel the same way with you my friend
Same omg 😂😂😂 i feel like a genius
Me too. this is the first time I solve a riddle on this channel and I've never felt smarter. I'm so happy djsjhaikghais
Same
I got lost at the explanations but I got the answer by trial and error
Yeah just add random numbers until it starts to make sense
Just start from either end using 0s and u fill find a pattern or series how these numbers work so u could find any of this type of number whenever now. This method is muchhh better than the solutuion provided according to me. It took me only a min or two.
WHAT....
Did I do something wrong? He said there was only one answer but after looking over the rules and method again and again I don't see why 9 billion doesn't also work. Can someone please explain?
God Bear 9000000000 wouldn’t work because according to how autobiographical numbers work we would need to add a 1 to the last digit because we have 1 9 but if we were to do that we don’t have 9 0’s now we have 8 😂 it’s complicated
step 1: confirm you have green eyes
step 2: ask the combination to let you in
Wrong riddle
r/wooosh @KENYGAMER302
@@HotaruLancelot302 ...
@@timtam. I'm lonely ☹️
@@HotaruLancelot302 being so lonely you make fun of yourself
As TED-Ed riddles go, this one is fairly easy. I solved it using an instinctive trial-and-error method in less time than it took to watch the video. (The key insight is that there should be as many 0s as possible, to simplify the number.) Judging from the comments, many people used the same approach. Apparently no one used the "official," pseudo-deductive method given. A TED-Ed answer should track how humans actually think, not merely set forth an algorithm that can be grasped in hindsight but that is more of a proof than a practical solution.
0:10
Leonardo Da Vinci's Secret Vault with a sign saying "Leonardo Da Vinci's Secret Vault"
SEEMS LEGIT
Well, not many people in Florence spoke English back then.
47Jaspers He means it is not very secret if it has a giant sign that says Leonardo Da Vinci’s Secret Vault
Secret
I HAVE A BETTER EXPLANATION.
Grab a big eraser and try as hard as you can to erase the scribble.
If its markerTHEN TRY HARDER!!!
A variation, somewhat less likely to destroy the combination (assuming you seriously _cannot_ be bothered to look up the Wikipedia entry "Autobiographical Numbers"):
1. Find yourself a paper conservation laboratory. You're the protagonist of a TED riddle video; there's probably one _somewhere_ in your bizarrely-specific contacts list. If there isn't, try asking a library with a rare books collection.
2. Pay them to analyze the stain and spend all of ten minutes figuring out a way to remove the stain without damaging the paper or the writing. It might be a bit pricey, but money should be no object. Whatever's in _"The Lost Vault of Leonardo DaVinci" must_ be worth _something._
3. Profit. A lengthy book of an old man's last ramblings written _entirely_ in backward Renaissance-era Italian might not interest _you,_ but dozens of universities, museums, historians, and other scholarly types (not to mention independent tech billionaires) will pay through the nose to so much as _see_ it, let alone _own and analyze_ it, and while you wait for their checks to clear, you can write the story of how you found the Vault and live for the rest of your life on royalties checks and appearance fees.
Then you ripp the paper and they'll start to call you Jack.
what if part of the brochure was torn???
r/woosh
Or try a rubber from an erasable pen
@@staceykhoo5971 try paper it works for animation!
I have no clue what the explanation is but I got the answer on my own and that's what matters to me
same
The explanation is just one way of solving it. As long as you get to the answer, it's all good.
100% powerful => Use the Threelly SmartView Chrome Extension to add your favorite slices to this video. Get it
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It's not all good and something else matters in these problems - speed and efficiency. The purpose is not only finding the combination, but doing it in the least time possible, which is why you look for a pattern, constraints, formula, relationship. One could spend hours or days going over possible numbers one by one, but that's what we want to avoid. 😊
Well I spent 1 min getting the answer and the solution lasted for around 3 so I think i'm fine
"It would take too long to try all the possible combinations"
Whoops. Guess I got lucky then! Lesson learned - use EVERYTHING that you are given. Nice riddle!
1210
3211000
6210001000
9210000010000
Just keep adding 3 to the first one, the next ones are always 2 and one, then fill with 0's except for the one representing the 1st number. Boom solved all of them.
Following your first rule. "1"210
1 + 3 = 4 so 3211000 is wrong?
It's supposed to be 4211000?
42101000
@@kieferplayzftw1324 which rule?
Solved this in less than 2 minutes😎
I also solved it that way 😂😂😂😂 in 1 min
Step 1 of robbing and highly security guarded bank
Hire the Ted-Ed riddle guy
and LockPickingLawyer
This is why TEd Ed is so special!! These highly analytical riddles are enough to make one's day!
Were you able to solve this
@@eugenemoreno482 he said to make one's day which is also used to say that it makes you have a good day
@@eugenemoreno482 because of your answer to that person, he only asked her if she completed it or not and then you tell him it took her a day to solve it only based on her comment, THAT'S WHY.
@@eugenemoreno482 no, maybe you are unaware of how and when to use sarcasm😒
Your comment does not even relate to sarcasm lol
I solved this riddle in a completely different way from the one described in the video. It's always great to see there are so many ways to solve the same problem, and it's encouraging for a math goat like me!
Can you solve this riddle? Nah..But I'll watch it anyway.
yep
Ban All Pineapples same lelz
Same
Same my homie, same
This riddle's explanation is hopelessly convoluted. The unnecessary set-up and pattern inferences were confusing and unintuitive. I solved it in about a minute by heuristically realizing that most of the number would have to be 0s. Trial and error from 9 to 8 to 7 to 6 found me the answer. The longest autobiographical number is 10 since you can't count double digit+ numbers, so using a formulaic approach doesn't achieve anything, unlike discovering some pattern between prime numbers, or perfect square numbers etc. Still a fun puzzle though.
For the bonus riddle is debatably more efficient, especially when considering you need to be certain you have them all.
Also, this riddle is likely aimed at people with far weaker problem solving skill for whom the idea that looking for patterns and narrowing in on and answer instead of randomly guessing can be an important take away, even if you are correct that eventually people get beyond that and can learn to evaluate the complexity of a problem and tell whether, practiced guessing or proper pattern seeking is more efficient.
I used the exact same method to get my number,but I eliminated 9 as there would be 9 zeros,and
9th zero would be 0 which is logically incorrect.
Sorry, but your answer is wrong. Here is a autobiographical number with 11 digits:
72100001000
The upper limit thus is 13 digits.
Also you miss some autobiographic numbers if you generalize the proof, because not every autobiographic number requires to have exactly two ones. For example the autobiographic number
2020
does not. The proof in case of ten only works because we know from our calculations before that there are at least three zeros.
A better proof regarding autobiographic numbers would be the following.
Step 1: There has to be at least one zero.
Proof: If there were no zero then the digit for the numbers of zero would be zero. That is a contradiction.
Step 2: If n_i is the number of digit i, define k_i as follows:
k_i=1 if n_i>0 and
k_i=0 if n_i=0
This means:
either n_i -k_i >0
or n_i- k_i =0
Also n_i -k_i >0 means that n_i is at least 2.
Step 3: Let d be the number of digits and m=d-1, and let s be the number of non-zero digits.
We know that
n_0+n_1+n_2+...+n_m=m+1
and
n_1+n_2+...+n_m=s
since n_0 is NOT zero (see step 1) we know there are exactly s-1 nonzero digits among n_1, ...,n_m
this means that l_1,...l_m from step 2 is exactly 1 for s-1 of these l_i.
and this again means that
(n_1-l_1)+(n_2-l_2)+...+(n_m-l_m)=s-(s-1)=1 (!)
since all of the n_i-l_i are nonnegative integers (see step 2) this means there is one and exactly one of the n_i that is two for i>0, while the other n_i for i>0 are either ones or zeroes. If one of the n_i for i>0 is three or larger this would mean that the equation marked with (!) would not be true anymore, no matter what the other n_i were.
Also n_i can only be two for i=1 or for i=2. If n_i=2 for i>2 then there would be two numbers of at least three. One could be the number of zeroes, but the other one would then be one of the numbers of non-zeroes. So one of the n_i for i>0 would be three or larger, which would contradict the statements before and the equation (!).
So there are two cases now. Either n_1=2 or n_2=2.
Step 4: Checking case n_2=2
If n_2=2 then there are two 2s, one of them describing the number of 2s according to our assumption. The other HAS to describe the number of zeroes, or (!) would be false. So there are 2 zeroes and two twos. Because every non-zero number only appears zero times, once, or twice, and 1 is not appearing twice according to our assumption, this means one is either appearing once or one one is not appearing. Because there are exactly two zeroes, any number bigger than two can not appear either or equation (!) would be wrong again.
This means for this case there are two possible solutions:
2020
and
21200
Step 5: Checking case n_1=2
Since n_1 is the only n_i with n_i > 1 and i>0 (see step 3) and the digit 2 appears this means n_2=1. So we know there are exactly one 2, which is the number of ones, and two ones, one of them representing the number of twos. Either the number of zeroes is one, which then leads to one zero, two ones and one two, with the rest of the numbers being zero. But because there is only one zero, this has to be a 4-digit number. And this is the number
1210
.
The other possibility is that there are more than one zero. This means from the two ones in the number, one is giving the number of twos, the other one has to be one of the remaining n_i with i>2, with the other n_i being zero (we know the number of ones and twos and according to step 3 for i>0 n_i can only take values one, two or zero). But this means then that s as defined in step 3 is always 4 in this case. Thus d-s=d-4 (with d being defined in step 3 as number of all digits) and that is both the amount of zeroes and the remaining i for which n_i equals 1.
This means the number consists of:
d-4 zeroes
2 ones
1 two
1 number d-4
the restriction is that d-4 is not higher than 9 otherwise you cannot express the number of zeroes in a single digit. Thus d is 13. Also d-4 has to be at least one according to step 1, thus d has to be at least five. However for d=5, d-4=1 and then there had to be both exactly one one and two ones, which is impossible. This means d has to be at least six, but then d-4 is two. The number would then only have one one, despite the digit, giving the numbers of ones being 2. This means d has to be at least 7.
This gives the numbers
3211000
42101000
521001000
6210001000
72100001000
821000001000
9210000001000
So there are -nine- ten (wow I can solve this but I cannot count to ten) autobiographic numbers in total:
1210
2020
21200
3211000
42101000
521001000
6210001000
72100001000
821000001000
9210000001000
@@brunobar4808 you're obviously very into mathematical proofs and I am not as enthusiastic about math to get into your ridiculously long UA-cam comment but I will say that you are wrong about autobiographical numbers. The longest autobiographical number for base 10 numbers is 10, and this is a definition. If you don't believe me, just look up self-descriptive numbers on Wikipedia. So the ones you found after that for how many times 11, 12 and 13 show up are not autobiographical numbers. Also, the longest one of those is 36 digits long, so maybe you can expand your proof.
AxielFan
"The longest autobiographical number for base 10 numbers is 10, and this is a definition."
That is changing the definition from autobiographical numbers from the video and even then still incorrect/unprecise. The video author provides the definition of _Autobiographical Numbers_ as numbers with a flexible length in base 10 (no other base mentioned) where the digits give the number of digits corresponding to that position. Or in other words: In the definiton of the video there is no restriction regarding the length of a number. For this case AND ONLY for this case my proof holds although it is esay to gerneralize that proof (see further below). Since the task at the end of the video is to find ALL autobiographical numbers, it seems plausible that we only should have a finite amount of them.
Now to your statement. You say the longest autobiograpical numer for base 10 is 10, but 10 is actually not a autobiographical number. If you mean that the _self-descriptive_ number of base 10 always has the length 10 according to the definition of _self-descriptive numbers_ from wikipedia (and by the generalization defined by wikipedia this would be the highest autobiographical number), then the statement is true, but trivial.
Also according to the wikipedia definition there of course then exist self-descriptive numbers of length 11, 12, etc. just not in base 10. But then your argument still is wrong, because in your initial comment you claim, that you yourself assume to numbers to be written in base 10 (which is not reasonable when refering to the wikipedia definition, but you assume a longest self-descriptive/autobiographic number to exist). And then there are not only 36 self-descriptive numbers but infinite. For example in base 1000 000 the self-descriptive number would be 999 996 zeroes (in base 1000 000 the number 999 996 is a single digit), two ones, one two and one 999 996. This disproves, that the longest autobiographic number is 36 digits long.
My proof would only have to change at one place to be valid for the wikipedia definition of self-descriptive numbers and by generalization to fewer digits autobiographic numbers: I simply have to drop the upper limit for the amount af zeroes. This would always be valid, because the amount of zeroes cannot exceed the base=length of numbers and thus be a single digit in that base. In no other case the base is relevant for my proof, as long as the base is at least 3 (because otherwise the digit 2 would not exist). However for base 3 or lower there is no autobiographic number.
And the autobiographic number remains the same when we look at the same digit positions and number length in a higher base, which is the bridge between the video definition and the wikipedia definition. Compare the numbers i found with my method to the wikipeida numbers until base 13. Why are they the same? Well the higher bases have MORE digits than the lower bases. So if a digit combination in the lower base exists, the same digit combination will also exist in a higher base (but the number represented by the digit combination is not the same). And even in base 13 the highest digit, that appears is the digit 9, a digit which also appears in base 10.
Also the trial and error method is flawed, because such a method can only prove at best that a solution exists (by finding one) not that the solution is unique (unless you check ALL combinations, which would take longer than proving my stuff). This is what justifies a formulatic approach. the mathematician often is not interested in the problem itself, but in the solution of a class of problems.
This took me 2.5 whole hours of dedication and ended up with a different process (not hit and trial), but the same solution. My brain is drained, my day is destroyed.
but at least u solved it!
How does it go??
Coop Cooper Indeed it took me something less than 10 minutes, because half of what he says comes easily once you start to test combinaisons.
You realize rapidly that except for the first positions "0" or "1" or "2", if you put big numbers like a 5 or so, you will have to write that number too many times !
Ex :
5
_ _ _ _ _ _ _ _ _ _
0 1 2 3 4 5 6 7 8 9
I now need to pull 5 numbers "4" in my number which is hard ! Cause it means, they are "at least" 4 zéro, 4 one, 4 two, and 4 three ! And that's even more numbers to put into my number (the 4 two, the 4 three etc.)
I don't have enough digits !
==> So you end up pretty fast trying to put the more Zéros you can in your number, and with that, the solution comes after maybe 50 trials at most.
I solved this in a different way that could be considered as going through every possible answer but technically isn't:
1. I started with 0000000000
2. I insert 10 for the first digit before taking away 1 since 10 would replace the 0: 9000000000
3. I inserted a 1 as the last digit (since there is one 9) thus decreasing the amount of 0 to 8: 8000000001
4. I changed the position of one to 8: 8000000010
5. I put a 1 at the second digit (representing the amount of 1s) but now there are 2 ones so I placed a 2 instead, I also decreased the amount of 0s to 7 and moved the 1 to spot seven : 7200000100
6. Finally I put down a one at spot 2 thus making the 2 at spot 1 correct, this meant I had six 0s so I moved the 1 for spot 7 to spot 6 and got the same result as the video: 6210001000
I didn't even bother to solve the riddle bc I don't even know the meaning of the question. Lmao
Hi army
larisa army
same
Same and hi army
Hey Army, imma gonna guess ur bias is the one and only MIN YOONGI I purple u
"Can you solve Leonardo Da Vinci's riddle?"
Me: Nope but hecc yes I'll watch
I can
S
“Blindly putting in numbers would take forever.”
Wrote a java program to find the number for me. Can confirm.
lul can you put down the codes xD i am just curious
Code please
confirm pls
Here is an unoptimized brute-forced one-liner in python:
print(next(n for n in range(10**9, 10**10) if all(str(n).count(str(i)) == int(c) for i, c in enumerate(str(n)))))
This was one of the few riddles I’ve ever figured out in this whole series :,)
That's a crazy riddle! I think riddles actually help prevent dementia in people due to the increase in brain stimulation and activity.
k
Theres actually a much easier way of doing this. I found this riddle intriguing so i sat down with a notebook to solve it and it took me only like a min to do that. And i didn't do it anything like they did in the video.
@@ishwantsingh955 same
@@ishwantsingh955 can u tell me. I did it by trial and error and it took me some time.
@@sanjogagarwal1532 in these kind of problems u usually start from the end or the start. For example starting with nine 0s and one 9 u realize u have to put one 1 too and like this u continue further. Its a cute little pattern. It would work for some other cases too.
I think it would just be easier to just bulldoze the door.
The Mix.... agreed
I think it would be better to just kick down the old and rotting wooden door
Abahsfjzgs
I actually got it :D
started with 9000000000
continued with
8000000010
and
7100000100
and got to the solution:
6210001000
did exactly the same
I did the same! High five!
same
I did it the same way!
I wrote a whole damn comment mentioning this without scrolling down lmao
This was probably the easiest Ted-ed riddle i've seen so far, yet it has the most overcomplicated solution of them all...
It's fun how he said "trying combinations would take forever" because that was much simpler than the solution shown in the video
Try the elemental Crystal riddle if you think this one’s easy
The video is titled everything changed when the fire Crystal was stolen
What I understood:
ZEEEEEEROOOOOOOO
Me : *See the title*
Also me : Hahaha, of course I can.....'t
0:09
It has a sign, so it isn’t that much of a secret
@@SrGris-oq9ce lol, tri lerning gramemr befure u every thinnc of inculting some1
I forgot about this comment that I made, I don’t even know what I mean too
@@akashtiwari8044 I think he's referring to the Vitruvian Man painting on the door
@_ DeathStr0ke _ they meant the sign on top of the door saying “Leonardo Da Vinci’s Secret Vault,” saying that it wasn’t much of a ‘secret’ since there’s a sign stating it right there.
Everything is okay till you hear..."click on three, nothing on four" 😂😂
Well I found the answer myself.
But when I saw their solution I forgot how my answer came.
I worked it out hard way. Went from left and right side, assumed there would always be a zero somewhere, so I put "9" in slot 0. Then "1" in slot 9.
Then "1" in slot 1, "2" in slot 1 and "1" in slot 2. - This was a little sneaky, because it's a short paradox state, when there were two "1", for a moment after shifting it to "2", there was suddenly only one of them, but setting "1" for slot 2 fixes that paradox.
Anyway, after all that, I recounted number of "0" left, set slot 0 to "6", shifted "1" from slot 9 to slot 6, which did not change the number of "0", solved.
I assume you might have done the same, if you didn't work it out through the intended solution either.
seems legit
My process was to guess randomly with 5 and went as follows
5 ? ? ? ? 0 0 0 0 0 "ok, so 1 5 and 1 1" 5 1 ? ? 0 1 0 0 0 0 "Ok, so 2 1s now, and then 1 2" 5 2 1 0 0 1 0 0 0 0 "Ok, so there's 6 0s. If I move the 1 in the 5 slot to the 6 slot, (and make the 5 a 6), the sequence retains the same number of 1s, 2s, and 0s." 6 2 1 0 0 0 1 0 0 0 "double check? Yep. Checks out"
i stopped watching when i saw 'NUMBERS' word
Larry you did good mate!
Ahsfsuf
Why? can't understand some math?
As soon as I read the title I was like, “No” but I clicked it anyways
This is an ideal puzzle for a general math audience. The setup is easy for a layman to understand. The solution takes some thought, but requires only elementary methods. Perhaps best of all, the puzzle immediately suggests generalizations and variations (What if the 0's are counted in the last digit instead of the first? What happens using other bases? When is one number the biography of another? What happens if we iterate: take the bio of the bio of the bio...?) Hours of math entertainment here!
As much as i love these, i much prefer phylosophical riddles which require you to think outside the box. Mathematical riddles have a tendancy to spoil themselves as soon as you know the formula to solve it. Much like a rubik's cube, i used to be interested by them when i was younger, until i found out that you only need to memorize a few algorithms, then i completely lost interest. Sure, with practice and skill you can learn shortcuts, but i find puzzles and riddles much less interesting when you can just follow a formula to solve them.
But philosophical questions tend to be subjective, and it is difficult to find an answer that everyone agrees on, which is, I'm sure, not the way Ted-Ed likes to present riddles.
Also people do not just "follow a formula." The formula Ted ed made was custom made to help solve this riddles. Formulas are simply a way of writing down logic and reasoning so we can understand it better and do more complex problems. The 'formulas' for these problems are just explanations of the logic.
MisterXeolan
I agree not fully but quite substantially, philosphy is more interesting!
Alright Albert
Let's be honest - 95% of us watch these riddle videos without trying to solve them. *And yet we still watch them because they're so addicting ffs*
98%*
99.9%
Honestly...I'm in rest 5%
I was just interested in what was behind the door.
My Snapple me too
4:27 I would have an equally disappointed face if I found a long lost DaVinci's autobiography instead of a corona virus vaccine.
If someone found Da Vinci's autobiography, they might have found one of the greatest leaps in pre computer science in history. Significantly more important than a Corona virus vaccine
@@sykeassai but that's not going to save millions of lives
I solved it starting from the end and realizing there could be no 9's since a 9 would result in 9 digits of that number (namely 0's), but then the number of 1's would result as 0, which contradicts the 1 in position 9. I did a similar reasoning for 8's and 7's. Then when I reached 6 I realized there could be a 6, and 6 obviously had to appear in position 0. Cool riddle.
Video: Blindly trying different combinations would take forever.
Me: Wait, what? But I just solved it that way.
oh wow lol
6210001000
I started to think about how many times the 0 must occur.
If you make the number small, you a few kinda high numbers which have to occur at least once. Wherever you place these high numbers, you need to place a high amount of digits, so you wont have enough space to place all these digits.
After you try a 6 at the first spot the solution becomes very obvious.
I think that most of these autobiographic numbers will have lots of 0 in general.
Max Mustermann I took the second number 3211000, added 3 0s to the end to make it 10 digits, changed the 3 to 6, moved the 1 from the 3rd spot onto the 6th spot and I ended up with 6210001000.
Vohasiiv
Nice idea
Jau Jo
Doesnt really matter how much digits you try. The principle is always the same.
I dont know if there is alsways only 1 possible solution (like for the 10 digit number), but the principle X210000000...00000001000 for an X+4 digit number always works, up to infinity.
At first I wasvery excited about the complexity of an autobiographic number, but as I realized that, it got kinda lame.
Jau Jo
Well, this should be proven first...
Green shirt looks as confused as I am
My solution:
1. Figure out that the digits add up to 10
2. Consider any combinations of non-zero digits adding to 10 (there are 39 total combinations, but many you can eliminate right off the bat)
3. Count the number of zeros that would be in each combo (for example, for combo 5, 3, 2 there would have to be be 7 zeros)
4. Eliminate combos where number of zeros you found from #3 is not already in your number combo (left with 7 possibilities)
5. Of the 7 possible combos it is pretty easy to see that it's 6 zeros, 2 ones, and 1 two.
You lost me at 00:01
Unnecessary attention-seeking comment 👎👎
@@hasnain9654 unnecessary comment that doesnt understand a joke.
@@youstolemyhandleyoutwat I know what joke looks like. It wasn't joke and if you still think it is so think it I don't care and for God's sake don't explain me joke 😑😑💆
@@hasnain9654 ok, so jokes go like "knock, knock."
Then you ask
"Who's there?"
And i answer
"sugma."
And you ask "Sugma who?"
And i answer
"your neighbor. Half of your house is on fire. Maybe we should call the fire department."
And that, is a joke.
@@youstolemyhandleyoutwat was that joke or verbal irony? 😑😑 well can be said as joke
6210001000
It took like 5 min
Yes got it correct
I got 6210001000 too
Just type 0000000000 and find errors
0000000000
9000000000
9000000001
8000000001
8000000010
8100000010
7100000010
7200000010
7210000010
6210000010
6210001000
Regele Iaurtului Exactly the process I took.
Wow just wow how did you guys do that
@@neve6759 I can't tell if that is sarcastic
It’s not sarcastic, I’m genuinely impressed
*_*stares at puzzle*_*
who needs art anyways?
Just started with 9000000000 and worked backward. After that I put a one in the nine spot and changed the 9 to 8. Then I moved the one to the 8 spot. Then put a one in the one spot and changed the 0 spot to 7. Then had to change the # of ones to 2 and put a one in the 2 spot. Then had to changed the # of zeros to 6 and moved the one in the 7 spot to the 6 spot. Done. Gets you 6210001000
others:
hmm i know! *puts in code*
me:
*smashes the door*
😂😂😂😂😂🤣🤣🤣🤣🤣
Video: "If you're not sure where to start..."
Video: "Our partners at Brilliant.org..."
Me: "I can just go to google for help."
This explanation was pretty long and overwhelming, but I just started at 9000000000 and then used common sense for the rest.
I started at 9, then placed a 1 at the end, then a 2 for the second digit, then a 1 for the third digit. Then I changed the 9 to 6 and moved the 1 at the end to the "Six slot." It took about 10 seconds.
A 1 digit autobiographical number = a paradox
The first number represents the number of 0s but if it is zero then it means there are no 0s but there has to be a zero to represent that etc.
That's not a paradox, it just means there isn't a 1 digit autobiographical number
And I don't think a 2 digit or 3 digit autobiographical number exists either.
0 now 1 but then 11
Infact, I learned that there cha only be 9 such no. Because there isn't a greater digit to tell the no. of zeroes
4:29 the feeling of disappointment in his eyes XD
what happened to those quotes at the begging of the videos? I miss them
I first wrote 5210010000, and then i noticed it wasnt correct so i put a 6 and put the 1 one forward and it worked. One of the easiest ted talk riddles ever! Nice
I solved it but by using a completely different style of logic.
- I did the same thing knowing that all the numbers would have to add up to 10 as my starting point.
- I knew that the 9 HAD to be a 0. If it was a 1 (or higher), then that would mean another number would have to be a 9, which would be impossible without having to add other numbers. One digit solved!
- Following the logic, I next checked the 8. If the 8 was 1, then that means that the 0s could be an 8. That then leaves a paradox, though, where the 1 can't say 1 without making it a 2, which would cause the 2 to be a one, which pushes it up to 12. Therefore, 8 also HAD to be a 0.
- The above then made me realise that the sweet spot was X 0s, Two 1s, One 2, and One X's.
- 2 + 1 + 1 + X = 10, so X must be 6.
Double checked my work to make sure I was right, and that it all made sense, and got
6 2 1 0 0 0 1 0 0 0
TED-Ed: “Blindly trying different combinations would take forever.”
Me: That is exactly what I do all the time ( seriously ).
3:20 The man is just standing there...
Its a women
@@valkyriesimmonds7012 shut mouth
I actually figured this out in a way that was weird even to me. I started by looking at 9000000000 (it was obvious that zeroes would be most of the answer). From there, I just changed values to reflect what was showing. The steps I followed looked something like this:
9000000001
8100000001
7200000100
7210000100
6210001000
[SPOILERS] I also solved the bonus puzzle, but not using the ridiculously complicated method from the video. I'd say the typical answers (with 10 digits being considered the longest) are these seven numbers: 1210, 2020, 21200, 3211000, 42101000, 521001000, 6210001000. However, I'd also say that this could extend even farther (where digit places past the tenth would obviously be 0 automatically because multiple digits can't fit in one place). The extended ones follow the exact same pattern (where n is the number of digits and is greater than 6, the only autobiographical number of that length will consist of the digits [n-4], 2, 1, an [n-7] number of 0s, 1, and three 0s) and still, technically, describe themselves: 72100001000, 821000001000, 9210000001000. So that gives us ten autobiographical numbers in base 10, but going into higher bases can make even more: A2100000001000, B21000000001000, etc. So I'd say that, even though I can't possibly type them all, I truly found all autobiographical numbers.
Okay I just added 0s to the second number until it had 10 digits and then corrected the errors (just two anyway).
Same here bruh got the no under 1 min
Okay, so, a piece of paper later I think the number is 6210001000. Cuz there are 6 zeroes, two ones, one two and one six. I hope I'm right
Edit: first riddle I manage to get right xD I'm so happy :D
Edit 2: also the way you get to the solution is way too complicated. Someone called Luca Perju explained an easier way. I did something like that.
I'm so surprised i got it. All i did was put ten blanks, started testing a few combinations until i came up with one that had me marking 6 in the zero spot... also, your explanation was waaay more confusing than it needed to be. Like i said you know you need a big digit in the zeros place so you just test out with 5, since that is the minimum possible, see if you can add numbers to equal the sum of 10, and if not move on
I'm so proud I actually managed to solve this
Guess you could say, crack the da Vinci code
1:16 I did it more or less blindly (just a bit of logic), and it took me a lot less time than the length of this video (Didn't time it, but under a minute).
For obvious reasons, this number has a lot of 0. Thus the first number has to be large (don't know how large, that's not important yet). There can't be 2 numbers that large, for the same reason the video stated, it would take too many digits. So that first digit (whatever it is) is listed only once, therefore it has a "1".
Now, can that be the only 1? Of course not. If it was, there would be a "1" under 1, and that'd be the second "1". So there's at least 2 "1". If there's more than 2 (say, 3) then there's a "1" under three, but only 2 ones are used to there needs to be another 1 somewhere, say "4", which means there needs to be 4 of something, and so on, we quickly see that it doesn't work.
So there can only be 2 "1". So we know, at this point, that the number is 'large digit _ 2 _ 1 _ some zeros _ a 1 under the large digit _ more zeroes".
So there has to be 6 zeroes, which tells us what the large digit it, as well as the full number, 6, 2, 1, (three zeroes), 1, (three zeroes).
ok.
good stream of thinking
eh it wasn't too hard, i just started with 9 as the first digit and the rest zeroes and then checked it. if it was incorrect, i'd adjust the list to make it correct. Thought process:
9000000000 - there isn't a 1 to count the number of nines
8000000100 - made it so that there is a 1 to count the number of nines but i realized that would bring the 9 to an 8 since there is one less 0
7100001000 - made it so that there is a 1 to count that 1 that counts the leading digit
6100010000 - made it so that there is a 1 to count the number of ones
6110010000 - made it so that there is a 1 to count the number of ones again
6210010000 - changed it to a 2 because changing that to a 2 would result in 2 1's and 1 2 balancing it out
Dammit, I thought I was so clever with 9 000 000 000 then realized there's one 9 in there. So it can't be 9 000 000 001. 8 000 000 010 is wrong too.
Bedtime.
@@r5t6y7u8 ouch ahaha
I started with 9000000001 Went back realized I need 2 1's 1 for 2 and 1 for the number of 0's added them to 4 realized I only needed 6 at that point and just moved the 1 from 9 to 6 and Came to the answer
And you got it wrong. The answer is 6210001000
NEERRRRRRDDDDDDDDDD
Yaayyyyy.....for the first time in my life I felt genius...I solved this riddle 😍😍😍😍
The guys face tho
Thats look like a girl to me :/
This riddle is more important than my tomorrow's exam ....
Yesss i solved it in just 5 minutes. The first riddle of ted ed i solved this fast..😎😎😎
Yeah, so via the conditions you established, we know that for any case such that t_0>2 any sequences must look like k,2,1,0,...,1,...,0. For lower values this doesn't work cuz k can be included in the 1/2 count. Some case checking shows that n6 the above sequence works, with K=n-4. 3211000,42101000,521001000. The first digit increments, you slide a 0 into the middle and it works.
Basically what you've given is a necessary condition, (and it confines us to only one answer per n), and some case checking can give us enough sufficient conditions to have the full answer
Didn't understood a sh*t, still feeling myself brilliant cause I successfully watched this whole video without letting my brain run out of my 💀.
I mean if the door is wooden then wouldn't fire be the answer?
You not gonna ask how some dude from the 15th century somehow has combinational locks that lasted 600 years without failing for his safe?
Say goodbye to me before I get lost in the sea of comments
EDIT: HOLY MAC 127 LIKES THATS INSANE THANK U SO MUCH
Goodbye
Demon Elmo goodbye
Bye
Bye
Goodbye mate 😫
I can't believe I solved it without the analysis he had to go through!!!! Anyone like me?
I got the right answer but I didn't used the above method tbh! 🙄
I actually didn't get the method explained above! Lol.
Who cares about the method they use?
We each think differently (even though most, if not all, in the comment section used the same method).
What matters is that the puzzle is solved.
@@sharif47 Yh! Absolutely. 😇.
I didn't even bother listening to their solution. It was pretty much what my brain did unconsciously. '^'
"Fortunately, you have 3 codes" proceeds to show you only have 2
Figuring out the ten digit one is easy. Finding a six digit one, using the same process, I found is extremely difficult.
Edit: there are no six digit autobiographical numbers.
got it after 30 mins. Worth it.
WOOHOO I FINALLY SOLVED ONE!!
Now i can go eat. I'm almost dying.
My 11 year old figured it out in 3 minutes
It took me 30 minutes
I have a Maths degree and have no clue what he's talking about.
So for the confusing part, let’s say you have a sum that we’ll call S. It has exactly n terms, and S = n + 1.
All terms are positive, meaning they’re all at least 1. If they were all equal to 1, then S = n, which is a contradiction. This means at least one term is >= 2.
If two or more terms are at least 2, then S >= n + 2, which is also a contradiction. This means at most one term is >= 2.
So there is exactly one term that is at least 2. All the rest have to be 1. If that one term is greater than 2, then S > n + 1, which is a contradiction, so it has to be 2.
Finally, we have that S has exactly one 2, and all the rest are 1’s.