Yeah, when I was trying to solve the riddle I made an early assumption that the 2 numbers must be the same because otherwise Peet and Franky couldn't have figured out the order. However, I couldn't find any solutions given this assumption. Turns out my assumption was wrong, but only because the riddle isn't stated clearly.
Tor Stokkan question isn't 13 a prime number or was he saying that the sum and the product can't be a prime number? I answered my own question didn't I?
Why does it say there are only 25 possible sums? I found 73 numbers from 4(the smallest possible sum) to 198(the largest possible sum) that can not be written as the sum of 2 prime numbers.
4 is not one of the 73 sums that can not be the sum of 2 primes, I just mentioned the parameters of the sum. The 73 sums are: 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97, 101 (that's the 25th) and every uneven number from 103 to 197.
I like some of the other riddles on this channel, but this is a stupid riddle. The riddle on the wall is interesting, but ultimately this riddle just comes down to going through a lot of possible combinations. And the explanation at the end is confusing. The solution breaks down like this: Let A = the product, B = the sum, and the actual values be x and y. So A = x*y, B = x+y. The value of A is not enough to determine x and y, and so x and y are not both prime. However apparently the value of B was enough to know this. So therefore B can not be obtained by summing two prime numbers. So the first step is to go through the numbers from 4 to 100 and check which ones can not be obtained by summing two prime numbers. This takes a while (there is a nice maths short cut) but the list you get for possible values of B is: 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 81, 83, 87, 89, 93, 95 or 97. Even deriving this list is a bit of a chore, but I was still with the puzzle at this point. So now the guy who knows A can work out this list of possible values for B, and we are told this leads him to know what the values of x and y are. How? Well this guy obviously had a lot of time, because for each possible value of B he went through every possible value of x and y and checked what their product was. So for example, for the B=11 case, he considered the x,y combinations (2,9), (3,8), (4,7), (5,6) and worked out the corresponding products (which are 18, 24, 28, 30). He then did the same for EVERY OTHER value of B. Just to be clear, there are a total of 697 combinations. I know - I used excel to iterate them. The table starts like this: A x y B A unique? 18 2 9 11 TRUE 24 3 8 11 TRUE 28 4 7 11 TRUE 30 5 6 11 FALSE 30 2 15 17 FALSE 42 3 14 17 FALSE 52 4 13 17 TRUE And so on, for 697 lines. Now whatever the value of A is, it must be unique, because the guy who knew A could solve it. So we need to check for uniqueness down the A column. And this can only be done by iterating the whole blinking table. (By this point, I had concluded this riddle was stupid). You might think that uniqueness on this column is quite rare, but actually its the reverse - of the 697 combinations, 494 of them have unique products. So we now know that A is one of these 494 possible values. However the thing that makes this whole puzzle solvable, is that the guy who knows B can now also derive the values of x and y. How can he do this? Well in this huge table for each possible value of B there are a number of possible values of A. BUT the guy knows that the value of A is unique in the table, and this is enough to give him the answer. This must be because for the true value of B there is only one corresponding value of A that is unique. For example, B can not be 11, because A=18, A=24 and A=28 are ALL unique values of A in the table. So if B was 11, the guy who knows B would not be able to derive the value of A (and hence x and y). So we now have to trawl through the table and see for which value of B there is only one unique possible value for A. And then that is the answer. Ironically, the actual answer is in the first few lines - its A=52, B=17. All the other values of A when B=17 are not unique. This actually makes is more freaking annoying, because you have to go through the whole table to check uniqueness of A, and then again to make sure that the whole puzzle is well founded and there is only one solution. Final thought - a good logic riddle DOES NOT REQUIRE mindless iterations of possible combinations. Good logic puzzles are elegant, and when you see the solution you admire their beauty. I don't admire this riddle solution - its messy and ugly. I accept it lives up to the title of being difficult, but difficult in a bad way.
Great explanation. It's also bad because he didn't specify that the two numbers had to be different, and because finding the numbers doesn't tell you which order they go on the locks.
Brock savage I agree. Both intelligent men could have had the answer 4 as their known product/sum respectively. Therefore 2 could be the number for both combinations on the lock. Dumb riddle, it offers too many variables while also being over complex.
I actually like this problem even less than you. The solution is inconsistent with its internal logic. Why stopping counting when numbers add up to 100? If both can be as large as 99 shouldn't one check until 198? And as far as I am concerned there are other potential reasons that would have allowed Frankie to know the numbers in the first turn. I can name two: A. Any x^3 with x being a prime number could only be the product of x and x^2. B. Upper bound constraints. The fact that numbers may only be up to 99 leads to an additional constraint. Intuitively 99^2 could only be the product of 99 and 99 given our constraints in spite of none being prime numbers. On the one hand this leads to a nice simplification (if any of the two numbers were a prime number greater than 99/2 the possible decomposition would also be unique. Take 53 and x. Regardless of x the decomposition of 53*x must be 53 and x given that no number may be greater than 100. Therefore on the one hand the list of possible solutions simplifies since numbers can only add up to 54 (if they added more a number could be 53, but then if one was 53 the decomposition given the constraints would be unique). On the other hand however, this upper bound constraint would also selectively affect other products (41*3*7, it could only be decomposed as 41 and 21 for instance ). Hence there is a need to further screen all such possible cases without a self evident efficent method other than brute forcing since the elimination of them in the first piece of information will most likely imply a lack of uniqueness in the third piece of information. There may be or not an answer but the reasoning from the video is wrong and most likely there will be multiple answers. Funny how the so called answer relies on missing important information...
How can i solve this riddle? Ok, if i was in that "house", i would do this: -Wait, how did i even get in here?!?! -Say "Hello???" -Look around -See the riddle -Read the riddle -Don't get the riddle -Panic -Try those numbers: 85 55 (I counted the words) -Oh crap, wrong code -Sees behind -Theres a window -Bye bye "house" -Escape
The "number' is irrelevant. The 'lock' is a round smooth bar around a round handle. It doesn't matter if it is 'solved' or not to unlock it. You can still turn the door knob.
Uziel Perez Am I not to be allowed to make jokes. I thought I was making a joke that was appropriate with the rest of the human population. But I guess not, as I've always said, you can't impress everyone.
Imagine the first guy knows the product is 52. The second guy knows the sum is 17. The first guy is basically telling the second guy the two numbers aren't prime. Then the second guy is saying that he knew that. Which means that his number is not the sum of two primes. So the first guy looks at 2 and 26. They sum to 28. But so do 11 and 17. Two primes. So the sum isn't 28. So he looks at the sum of 4 and 13, which is 17. Can two primes sum to 17? Nope. So he says he knows the numbers. The second guy looks at the numbers which sum to 17. 2 and 15 for a start. The product is 30. But 30 is 2 times 15, 3 times 10, 5 times 6. If the product was 30, the first guy would have to rule out two of these when told the second guy knew he didn't know the numbers. To do that, he would have to see what primes sum to 17, 13 and 11. None sum to 17, none sum to 11, 2 and 11 sum to 13. The second guy would be thinking the numbers could be 2 and 15 or maybe 5 and 6. So he wouldnt have said he knew the numbers. can rule out the first guy having 30, because then he wouldn't have been able to say that he knew the numbers. So the second guy rules out 30 as the product and therefore 2 and 15 as the numbers. He then does the whole thing again with 3 and 14. Same result. He hits gold with 4 and 13. But not quite. He has to rule out 5 and 12 and so on. 4 and 13 is the only possibility. I really do feel that the explanation in the video is a bit lacking in detail. Now make sure that no other combination will work ✌
And thanks for posting! It's a good one. I thought the answer was 2 and 9 to start but then I watched the video. When you know the answer I think there's a good bit of justification required. When you don't there are lots of cases to go through. Making it 25 times as long. Would you agree it takes more work than the video seems to suggest?
Phaionix Suppose 2 and 9 worked. Then the product is 18 and the sum is 11. So suppose that is what our two guys started with. Pete says he doesn't know the numbers, meaning they are not prime. Frankie says he knew that, meaning his sum cannot be two primes. While this is true of 11, that is not all that is required. Frankie DOESN'T KNOW Pete's number, so he can't assume the numbers are 2 and 9. All he knows is that Pete now knows the numbers. The question is, can Frankie find out the numbers now? Is 18 the only possible product? Well, take 4 x 7 = 28. There i one other way to write 28 as a product, 2 × 14 = 28. Imagine Pete had 28, and Frankie says he knew Pete didn't know the numbers. Pete would then be able to rule out 2 and 14 because they sum to 16 and 3 + 13 = 16, which means Frankie could not say he knew Pete did not know the numbers. So Frankie knows Pete could say he knew the numbers if he had 18 as a product (check this) or 28. And maybe more. So he could not say that 2 and 9 are the numbers. I hope that helps.
Phaionix I never was that good with names! That's a nice solution. You are using primes when you say 2+12=14=3+11 and so on. But that's a small point. Your point about what the first two statements mean is a new take. I wonder in the long run if it makes things easier?
I successfully worked through the logic up until the point where I had to start writing down all the possible product/sums of acceptable numbers. That was the point I decided I didn't care THAT much about getting the final answer.
why cant i just bust the door down... i mean, if i was going into a broken down shed boarded up and covered in blood, i would at least bring a weapon with blunt force, like a club
I'm sorry... Can someone explain something to me? I don't see how it is possible for Peet to single out any such number based on simply Frank saying "I know my number". It would seem there would still be too many possibilities for Peet at that point. Please Explain!!!
@ Michael - I explained it in some other comment but will paste it in here as well so you can understand the solution: "You probably approach this riddle incorrectly. You treat Frank's and Pete's statements that they know the numbers as something to be verified and not as another clue. You are also putting yourself in Franky / Peet shoes and subconsiously assume that they are working with the same amount of information as you have which is not the case. You don't know neither sum nor product. You need to figure it out but Franky knows product and Peet knows sum already. And with that additional information that you don't possess they are able to figure out the numbers. Because of that your goal is to determine what numbers make that scenario possible and not whether Peet or Franky can claim that they know the numbers. If some combination of 2 numbers does not allow both Frank and Peet to know what those numbers are then this is wrong combination and not correct numbers. I think it is easiest to understand if you do a kind of reverse engineering and follow the scenario knowing the correct numbers from the start (as if you were the person who programmed the lock): Two numbers are 4 and 13. Frank then is being told that the product is 52 and Peet that sum is 17. Frank knows then that possible combination of numbers is either 4 and 13 or 2 and 26. Therefore he knows that Peet was given a sum of either 17 or 28. Frank says that he doesn't know the answer. Peet knows all possible products that Frank might have (30 if correct numbers are 2 and 15, 42 if numbers are 3 and 14, 52 if numbers are 4 and 13 and so on) Peet says that he knew that Frank doesn't know the answer. For Frank that means that sum must be 17 because if it was 28 than one of possible options would be 5 and 23 in which case Frank would be able to figure out numbers on his own and Peet couldn't have claimed that he knew that Frank doesn't know. So Frank says that he knows the answer because only 4 and 13 gives sum of 17 and product of 52. Peet then also knows the answer because only product of 52 would allow Frank to say he knows the answer. Other products wouldn't allow that. For example: If correct numbers were 2 and 15 than Frank would have product = 30, that would have possible combinations 2 and 15, 3 and 10, 5 and 6. Information that Peet knew that Frank doesn't know the answer would allow Frank to eliminate 3 and 10 but would leave other two combinations as valid options in which case Frank would not have said that he knows the answer. Therefore 2 and 15 cannot be correct numbers. You can go through all other potential products (42, 60, 66, 70 and so on) and you will see that only in case of product =52 Frank could have made that claim. Therefore Peet knows the answer as well. I hope you see now that 4 and 13 allow both Frank and Peet to indeed know what the numbers are and make this scenario possible. And since these are the only numbers that allow for this scenario to be possible then they are the correct numbers. BTW - this solution assumes that order of numbers doesn't matter. This should have been said in the riddle because without this assumption this riddle cannot be solved."
The solution is so simple! Ignore all the mathematical shit and work on that boarded up window. If you kick those planks hard enough you'll get free in no time. Besides, unlocking the door won't help much because it's boarded up outside. It was a trick riddle and I outsmarted it!
illdemoteu I get what you're saying. Personally I'd rather kick the boards off the window on the left than risk getting cut by glass but either window is an easy escape route.
You said there are 25 possible sums. Here you are assuming sum to be less than 100. This information is missing in the video. Ther correct problem and the correct solution is here: www.cs.rug.nl/~grl/ds05/sumproduct.pdf
faizal somani THANK YOU! I wasted so much time! Yes, the riddle is told wrong. Sorry but anyone who got it right made the same falacious assumption that the video maker did.
Looking at your resolution, I have one question.. how did you get only 25 sums left after knowing that it can't be a sum of prime numbers? I got way more than 25 options left. After removing the sum of high numbers, (since 98x95 will give a unique answer that no other 2 2-digit numbers multiplied together will give), I was left with 63 sums left.
Analyzing a little further I was able to remove a bunch of sums from my list, I made a spreadsheet with all possible outcomes from the mutiplication of 2 2-digit integers, excluded the mirror values (3&5 = 5&3), and selected to highlight the duplicate values. Doing that, I scanned quickly the spreadsheet to see which out of the possible 63 sums that I had found out previously, would not be generated by a pair of numbers which would have a unique product. Ex: The sum 51 can't be made by adding 2 prime numbers, so technically it should be analyzed. However: 34 + 17 = 51, and 34 x 17 = 578, which is a value that no other 2 2-digit numbers could make, hence, if the sum was in fact 51, there would be a chance that Frank would already know the answer. So 51 is no longer a possibility. By doing that analysis I reduced from 63 possible sums down to 10!
great work! And if you keep going, using the other information, you will reduce it to 1 and that's your answer! But seriously, this "list of 25" is ridiculous. I think he just counted the table rows on the answer sheet
After trying to solve this riddle and not being able to reduce my possible answers to a single solution, I went ahead and watched the solution. At first I could not understand why their solution had a unique answer but mine did not. After reviewing their explanation, I see that they have more rows in the table than they should. That is, they have rows in the table that contain products where Franky could not claim that he knew the 2 numbers. For example, the very 1st row has a product of 18. It claims Franky knows that the numbers are 2 and 9. In fact they could also be 6 and 3. Therefore Franky could not claim to know the numbers. In fact as soon as Pete says he knew Franky didn't know the 2 numbers, Franky learned quite a bit about more about the sum of the 2 numbers. Basically Pete told Franky that the sum of the 2 numbers was odd (look up Goldbach's Conjecture) and that some odd sums were not allowed, eg. 33 because it is the sum of 2 primes (31+2). Now that Franky knows that 1 of the numbers must be even and the other odd, he knows more about how the factors of the product must be partitioned between the two numbers. When Franky says he knows the numbers, he is telling Pete that there is only one possible way to partition the factors between the 2 numbers that meet all of the criteria outlined above. Some examples are: 24=3*8, 3+8=11 (lowest product and sum) 28=4*7, 4+7=11 208=13*16, 13+16=29 610=10*61, 10+61=71 960=15*64, 15+64=79 9306=94*99, 94+99=193 9506=97*98, 97+98=195 (highest product and sum) If you can understand why these are the only allowable partitions then you've got it. Hint: for the lower value products the only rule is one number is of the form 2^n, n>1 and the other number is a prime. For larger products, the constraint that each number must be < 100 allows for exceptions to the previous rule. Finally, Pete sees that his sum can only correspond to one solution so he and us know the 2 numbers. Unfortunately this isn't the case for numbers larger than 15. If the riddle specified numbers 2-15 instead of 2-99, then the solution given would be the correct and unique solution. If we look at the table in the video at the rows with the sum of 29, we see three rows that should not be there. All three rows contain products that do not have unique partitions, therefore they (and other possible cases) would not have been up for consideration by Pete. The rows that should not be there are listed below. 54=2*27, 2+7=29 - other possibilities: 3*18, 6*9 100=4*25, 4+25=29 - other possibilities: 5*20 138=6*23, 6+23=29 - other possibilities: 2*69 The row that should be there (and presumably is but the video cut it off) is: 208=13*16, 13+16=29 You can see that there is only 1 legal partition for the product 208 that yields an odd number and an even number in turn yielding an odd sum. Therefore this is also a valid answer. Other valid answers include: 148=4*37, 4+37=41 592=16*37, 16+37=53 688=16*43, 16+43=59 And of course not all sums are allowable. Returning to the video, for the sum of 23 there are 3 rows. One of them shouldn't be there but the other 2 should so Pete's sum can't be 23, but it can be 17 (the answer given in the video) or 29, 41, 53, 59... So there is no unique solution to this problem. Others have also added that there is actually no solution at all since there is no way to know which number to enter first. This is also true. While I enjoyed thinking about this problem, I'm going to have to give it a thumbs down for its lack of a unique solution and a poorly stated problem.
But 11 can also be made by 3 and 8, which gives 24. 24 is a product of 12, 2 and 8,3 and 6,4 of which only the 8 and 3 pair gives a sum that is the right one ( 10 and 12 are not correct since 10 is achievable by summing 5 and 5, 12 by, 7 and 5) so Frank would be sure what the numbers are if he got 24 or 18, but! Pete could never be sure what the numbers are if he got 11, which means you have to try the next possible sum - 17 after which you quickly get to 4 and 13. Who thought of this riddle and got it to work? :) amazing
oh god that guy in the barracks must be one of the 3% who can solve it. (i am being sarcastic, why would he not be able to know becuz there is an answer)
This riddle is mistaken. It is quite impossible for Peet to say he knows the numbers after Frank states he is confident in his own. Frank understands that Peet can only be confident if his sum is never possibly of two primes. Call that type of number Tasty. Such numbers are 11, 17, 23... Therefore all Frank must do to be confident in the numbers himself is to see which ones add up to one of those Tasty numbers. Therefore from Peet's perspective, any number which is the product of a Tasty number's possible parts could hypothetically allow Frank to be confident in his number. 18 (9,2) and 24 (8,3), for example, both satisfy Frank and leave Peet clueless. Simply giving us that "Peet knows his number" doesn't imply anything because it has no conceivable conceptual manifestation due to being logically impossible. I don't understand how this solution makes any sense. Please Explain.
the second guy knows the number too in the end. 9,2, 8,3 7,4 would not satisfy because they all add up to 11. 3 different products are possible, but the second guy can't know which pair works, then. Because he knows too, it must be that he can figure out what the product is, as there is only one possible product. There is no other product that has a pair that add up to 17.
memed bengul: Just NO! There is a logical flaw about Frank's Statement! He just cannot say, "-Now I know the numbers!" before Pete says "-I know too". At that point there are many possibilities that satisfy the clues so far. This riddle is just fraud! :D
@ George, @Alper - you guys approach this riddle incorrectly. You treat Frank's and Pete's statements that they know the numbers as something to be verified and not as another clue. You are also putting yourself in Franky / Peet shoes and subconsiously assume that they are working with the same amount of information as you have which is not the case. You don't know neither sum nor product. You need to figure it out but Franky knows product and Peet knows sum already. And with that additional information that you don't possess they are able to figure out the numbers. Because of that your goal is to determine what numbers make that scenario possible and not whether Peet or Franky can claim that they know the numbers. If some combination of 2 numbers does not allow both Frank and Peet to know what those numbers are then this is wrong combination and not correct numbers. I think it is easiest to understand if you do a kind of reverse engineering and follow the scenario knowing the correct numbers from the start (as if you were the person who programmed the lock): Two numbers are 4 and 13. Frank then is being told that the product is 52 and Peet that sum is 17. Frank knows then that possible combination of numbers is either 4 and 13 or 2 and 26. Therefore he knows that Peet was given a sum of either 17 or 28. Frank says that he doesn't know the answer. Peet knows all possible products that Frank might have (30 if correct numbers are 2 and 15, 42 if numbers are 3 and 14, 52 if numbers are 4 and 13 and so on) Peet says that he knew that Frank doesn't know the answer. For Frank that means that sum must be 17 because if it was 28 than one of possible options would be 5 and 23 in which case Frank would be able to figure out numbers on his own and Peet couldn't have claimed that he knew that Frank doesn't know. So Frank says that he knows the answer because only 4 and 13 gives sum of 17 and product of 52. Peet then also knows the answer because only product of 52 would allow Frank to say he knows the answer. Other products wouldn't allow that. For example: If correct numbers were 2 and 15 than Frank would have product = 30, that would have possible combinations 2 and 15, 3 and 10, 5 and 6. Information that Peet knew that Frank doesn't know the answer would allow Frank to eliminate 3 and 10 but would leave other two combinations as valid options in which case Frank would not have said that he knows the answer. Therefore 2 and 15 cannot be correct numbers. You can go through all other potential products (42, 60, 66, 70 and so on) and you will see that only in case of product =52 Frank could have made that claim. Therefore Peet knows the answer as well. I hope you see now that 4 and 13 allow both Frank and Peet to indeed know what the numbers are and make this scenario possible. And since these are the only numbers that allow for this scenario to be possible then they are the correct numbers. BTW - this solution assumes that order of numbers doesn't matter. This should have been said in the riddle because without this assumption this riddle cannot be solved.
George, your logic is pretty sound but you draw wrong conclusions. From the fact that if sum is 11 then Frank would know his numbers but Peet was still clueless even though he said he also now knows them you conclude that this means that Peet can't make that claim instead of concluding that this must be wrong sum. There are plenty of potential sums and products that these guys could have been given that would allow Frank to know the numbers but would leave Peet clueless. But there is only one combination of sum (17) and product (52) that allows both of them to claim that they know the numbers. And since they both claimed that then this is correct answer
I disagree with the 25 possible outcomes left. The two numbers can not be two prime numbers because else Frankie would know inmediately. However there are other products that would allow Frankie to know the numbers. For instance if the product was x^3 with x being a prime number, x^(2a+1) where x is prime and x^(a+2)> 99 (like 2^11 or 3^7, where only solutions are 64/32 and 81/27), x^(2a) where x is prime and x^(a+1)> 99 (like 5^4, only solution being 25/25) x*a where x is a prime number larger than 49, and in general any x*y where x is a prime number larger than 99/c and c is smaller than any prime number that composes y, like 53*a, where only solutions are 53/a, or 41*3*5, where only solution is 49/15, 29*5*7, where only solution is 29/35. This means that the sum can not be larger than 54. If it was 55 or higher one of the numbers could be 53. If one of the numbers was 53 the product would have a single solution and Frankie would have known at first. Taking this into account i only get 11 possible sums, which are 11, 17, 23, 27, 29, 35, 37, 41, 47, 51 and 53. I guess your 25 included 57, 59, 65, 67, 71, 77, 79, 81, 83, 87, 89, 93, 95 and 97. I think it is wrong but following your logic I believe you should have kept checking numbers until they summed 198.
*riddle channel* : bla bla bla i likemath * as me...* : in the corner of my room twitching of to much math. ya i shouldn't have watched this.....ehhhhhh
He asked it wrong as he says if they got the numbers wrong than the door will lock forever suggesting that it is not already locked so you can just walk out
No, he's meaning that it will stay locked permanently even if you subsequently enter the right combination, and so the problem is constrained in that they can't test multiple numbers as a strategy to rule out wrong ones and must reach a definitive answer to enter right the first time.
These intelligent men are intelligent like having the power of telepathy? IT'S IMPOSSIBLE TO SOLVE ONLY BY KNOWING THAT OTHER GUY DON'T KNOW THE ANSWER!!! Look take the numbers 4 and 13 ONLY Franky know that sum is 17 ONLY Peet know that product is 52 Franky know that possible answers could be: 2,15; 3,14; 4,13; 5,12; 6,11; 7;10; 8,9... Peet know that possible answers could be: 2,26; 4,13; Franky tell that he can't realize what is correct answer, but Peet already knew that (it's ok there is to many combinations for every his answer) And with this information Franky can't do anything!!! It's just impossible Unless if he tell him his sum or product... but then that information that he don't know the answer is not necessarily And you still have just 50% chance of solving it because there two operation is commutative 4*13 = 13*4 4+13 = 14+4 Sorry about my bad English, it's not my main language :)
you mixed the characters up. it is the one with the sum who already realizes that the other one can't narrow down to one. after he states this, the guy that knows only the product realizes that the sum must be such a number that it's can't ever be derived from two primes.
It is 100% unsolvable, Once you turn a dial on the number lock, you already put in an incorrect code, so you are already screwed. And who in the world have a lock that permently locks if a incorrect code is entered when thee is no "enter button" Plus It is the woods, so just find coin or whatever and unscrew the barracks and get out.
in this case m gonna break the wall ,its look like pretty old and in another case i couldn't be able to enter in that room lol the door is blocked already
You totally stole TED-ED's intro. If you look at the beginning of TED's bridge riddle, the intro is formatted the exact same way, and since this video was released after TED's, #EXPOSED!
1. The lock isnt even locking the door. Its just sitting on the door knob. 2. That house is completely destroyed. U could ceawl out. 3. 3% is made up. Good try for the click bait lol. 4. You could pull the "lock" off the door knob.(also 5. At 1:09 there is no lock)
But he still only has a 50/50 chance of escaping since the answer could be either 4 and 13 or 13 and 4. Screw you riddle channel, I spent three days trying to figure this out under the assumption that the answer had to be two identical numbers
The First i though it was 4,9,6 Cause there was Written In The Wall DIF Then I Count D A,B,C,D Sooo 4 A,B,C,D,E,F,G,H,I Sooo 9 A,B,C,D,E,F Soo 6 After The Answer Reveal I Though i was correct but i was wrong....
First of all there is a broken window that I can just go through, Second of all there is another window just barricaded with a few wooden planks that I can break And third of all the damn walls are made out of wood I could just break through the house
Sorry I'm a little slow on this. Could someone explain why the numbers couldn't be 3 and 33? Ergo first guy is baffled because the product of 99 could have two solutions and the second guy is also puzzled because the sum is 36 which has multiple factors.
This is what I would do 1 see metal door with lock 2 look around the room 3 see riddle 4 not get riddle 5 keep looking around 6 find window 7 break outa there!!! (If there was no window) 8 SCREAM HELPPPPPPPPP 9 try to remove the lock 10 break the tiny window on metal door (you saw it) 11 grab a glass shard 12 start trying to dig dig and DIG 13 crawl out
There may be an error. The solution can be 13 and 4 as well, rather than 4 and 13. Therefore better not to include the sum, but the difference: 9. This mean 13 and 4. Or -9, meaning 4 and 13
Fuck the riddle, bust the door down. Wait... Considering, the door was boarded from the outside, why in the world is there a lock inside? How did the guy who put a lock outside? I know, pictures can be inaccurate, but wtf?
Here's my solution. If Franky only knows the product and Peet only knows the sum and they both know the numbers at the end, another solution could be that the product and sum are the same. In this case, 02 + 02 and 02 * 02 are the same, so the numbers are 02 and 02. Neither of them know what the numbers are initially, so the product could have also been from 01 and 04 and the sum could have been from 01 and 03. This does not violate the rules stated here because it would be of the same prime number, not two different ones.
but how would you know which order the two numbers go in the passcode? I did all the math, then declared it impossible for this reason. This riddle is flawed, very interesting, but flawed.
Well SHIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIT! I got it half right!
In the 'i already knew you couldnt' the T looks a lot like a 4. Giving a clue the numbers are disguised as letters. The S in sum also looks like an 8. I cant solve this but i guess 84?
not really a solution for he still has a 50% chance to miss if the numbers are used in the wrong order
That's a valid point. The solution is not complete.
my point exactly, not that i have solved the riddle but i was like, yeah which one goes in which lock now ?
Chris S. The only part I was correct was the prime number....
How was I supposed to know who the fuck Peet and Frank were without the damn narrator? I could've mixed them up, and I did.
Yeah, when I was trying to solve the riddle I made an early assumption that the 2 numbers must be the same because otherwise Peet and Franky couldn't have figured out the order. However, I couldn't find any solutions given this assumption.
Turns out my assumption was wrong, but only because the riddle isn't stated clearly.
You still end up with two possible solutions:
1. 4-13
2. 13-4
So you have a 50/50 chance...
50 likes!
Tor Stokkan question isn't 13 a prime number or was he saying that the sum and the product can't be a prime number? I answered my own question didn't I?
Bhintuna Maharjan no, it wasnt that both hat to he not prime, it was that they both couldnt be prime, so one of them being prime is fine
0413
Tor Stokkan - Exactly! Didn't realise that earlier!
But how do you know which number comes first???
hey jodie, i guess it works both ways ;) And if it doesnt, a 50% chance to survive is way better than a certain death i think
also.. to be fair the riddle never said they had to be in a certain order.. just that you needed the 2 correct numbers :P technically speaking ofc
Why does it say there are only 25 possible sums? I found 73 numbers from 4(the smallest possible sum) to 198(the largest possible sum) that can not be written as the sum of 2 prime numbers.
radu443 1 may not be used. So when the sum is 4, u would already know it was 2*2.
4 is not one of the 73 sums that can not be the sum of 2 primes, I just mentioned the parameters of the sum. The 73 sums are: 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97, 101 (that's the 25th) and every uneven number from 103 to 197.
I like some of the other riddles on this channel, but this is a stupid riddle. The riddle on the wall is interesting, but ultimately this riddle just comes down to going through a lot of possible combinations. And the explanation at the end is confusing. The solution breaks down like this:
Let A = the product, B = the sum, and the actual values be x and y. So A = x*y, B = x+y.
The value of A is not enough to determine x and y, and so x and y are not both prime. However apparently the value of B was enough to know this. So therefore B can not be obtained by summing two prime numbers. So the first step is to go through the numbers from 4 to 100 and check which ones can not be obtained by summing two prime numbers. This takes a while (there is a nice maths short cut) but the list you get for possible values of B is: 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 81, 83, 87, 89, 93, 95 or 97. Even deriving this list is a bit of a chore, but I was still with the puzzle at this point.
So now the guy who knows A can work out this list of possible values for B, and we are told this leads him to know what the values of x and y are. How? Well this guy obviously had a lot of time, because for each possible value of B he went through every possible value of x and y and checked what their product was. So for example, for the B=11 case, he considered the x,y combinations (2,9), (3,8), (4,7), (5,6) and worked out the corresponding products (which are 18, 24, 28, 30). He then did the same for EVERY OTHER value of B. Just to be clear, there are a total of 697 combinations. I know - I used excel to iterate them. The table starts like this:
A x y B A unique?
18 2 9 11 TRUE
24 3 8 11 TRUE
28 4 7 11 TRUE
30 5 6 11 FALSE
30 2 15 17 FALSE
42 3 14 17 FALSE
52 4 13 17 TRUE
And so on, for 697 lines.
Now whatever the value of A is, it must be unique, because the guy who knew A could solve it. So we need to check for uniqueness down the A column. And this can only be done by iterating the whole blinking table. (By this point, I had concluded this riddle was stupid). You might think that uniqueness on this column is quite rare, but actually its the reverse - of the 697 combinations, 494 of them have unique products. So we now know that A is one of these 494 possible values.
However the thing that makes this whole puzzle solvable, is that the guy who knows B can now also derive the values of x and y. How can he do this? Well in this huge table for each possible value of B there are a number of possible values of A. BUT the guy knows that the value of A is unique in the table, and this is enough to give him the answer. This must be because for the true value of B there is only one corresponding value of A that is unique. For example, B can not be 11, because A=18, A=24 and A=28 are ALL unique values of A in the table. So if B was 11, the guy who knows B would not be able to derive the value of A (and hence x and y). So we now have to trawl through the table and see for which value of B there is only one unique possible value for A. And then that is the answer.
Ironically, the actual answer is in the first few lines - its A=52, B=17. All the other values of A when B=17 are not unique. This actually makes is more freaking annoying, because you have to go through the whole table to check uniqueness of A, and then again to make sure that the whole puzzle is well founded and there is only one solution.
Final thought - a good logic riddle DOES NOT REQUIRE mindless iterations of possible combinations. Good logic puzzles are elegant, and when you see the solution you admire their beauty. I don't admire this riddle solution - its messy and ugly. I accept it lives up to the title of being difficult, but difficult in a bad way.
Great explanation. It's also bad because he didn't specify that the two numbers had to be different, and because finding the numbers doesn't tell you which order they go on the locks.
Brock savage
I agree. Both intelligent men could have had the answer 4 as their known product/sum respectively. Therefore 2 could be the number for both combinations on the lock. Dumb riddle, it offers too many variables while also being over complex.
Stephen Wokes did he say which person is which/who?
I actually like this problem even less than you.
The solution is inconsistent with its internal logic. Why stopping counting when numbers add up to 100? If both can be as large as 99 shouldn't one check until 198?
And as far as I am concerned there are other potential reasons that would have allowed Frankie to know the numbers in the first turn. I can name two:
A. Any x^3 with x being a prime number could only be the product of x and x^2.
B. Upper bound constraints. The fact that numbers may only be up to 99 leads to an additional constraint. Intuitively 99^2 could only be the product of 99 and 99 given our constraints in spite of none being prime numbers. On the one hand this leads to a nice simplification (if any of the two numbers were a prime number greater than 99/2 the possible decomposition would also be unique. Take 53 and x. Regardless of x the decomposition of 53*x must be 53 and x given that no number may be greater than 100.
Therefore on the one hand the list of possible solutions simplifies since numbers can only add up to 54 (if they added more a number could be 53, but then if one was 53 the decomposition given the constraints would be unique). On the other hand however, this upper bound constraint would also selectively affect other products (41*3*7, it could only be decomposed as 41 and 21 for instance ). Hence there is a need to further screen all such possible cases without a self evident efficent method other than brute forcing since the elimination of them in the first piece of information will most likely imply a lack of uniqueness in the third piece of information. There may be or not an answer but the reasoning from the video is wrong and most likely there will be multiple answers.
Funny how the so called answer relies on missing important information...
The riddle was just dumb No way could 3% of people could solve it
I would climb out of the broken glass window ...
+xhai zerna fun fact . you can actually break the glass :o
Panda Gaming So smart!
Panda Gaming me too 😐
same
Can u solve the toughest ghost riddle
th-cam.com/video/EfnhlZOrXmU/w-d-xo.html
Subscribe if u like it
can someone bring a math professor, i didnt even understand the solution
You really don't need to be a maths professor to get it...
Red Knight same
not enough information to solve it lol. there's also too many numbers to choose from that are different products and sums that are not 1 and 100
i understood nothing tbh
dont worry, nobody did
How can i solve this riddle? Ok, if i was in that "house", i would do this:
-Wait, how did i even get in here?!?!
-Say "Hello???"
-Look around
-See the riddle
-Read the riddle
-Don't get the riddle
-Panic
-Try those numbers: 85 55 (I counted the words)
-Oh crap, wrong code
-Sees behind
-Theres a window
-Bye bye "house"
-Escape
Ciao Ciao perfect something I would do
lmao thats the same thing I got
Ciao Ciao Why does everyone say 'hello' to see if anyone is there in horror movies?! Do you rly think someone will answer?
Also if they did answer wouldn't you just try to run I mean it's common sense.
P.S. would tottaly just escape through a window too lol!
The "number' is irrelevant. The 'lock' is a round smooth bar around a round handle. It doesn't matter if it is 'solved' or not to unlock it. You can still turn the door knob.
lol
lol
First Last lmao!
First Last 1:12 there's no lock
First Last the door probably opens inwards, but then the bar would stop it and the bar cant be removed until the right code is put in the lock
Why do math when you can jump out the window...
Mr. Creepy Cuz you will pie
I never knew we could turn into a pie
Tru
Mr. Creepy uhhh is that logo moon chasers it just looks the same
What's moon chasers? I've made this logo myself in paint.net a few months ago
I dint even understand what he was saying lol
It says they all each have a green badge, blue badge, and red badge so they each have 1 of each ?
dariso06 same
same
Can u solve the toughest ghost riddle
th-cam.com/video/EfnhlZOrXmU/w-d-xo.html
Subscribe if u like it
Blue has green, Red has blue, Green has red
Who the fuck can solve this riddle
haha maybe some crazy math professor :D
I can solve it :-)
yeah right Max Müller
Max müller yea when u know the answer dumbass
+Max müller you say that in every video lol
So.... where did the get this 3% claim? Did you do a survey or something like that?
69% of statistics are made up bro
Jayo Caine I am 420% sure you made up those statistics.
I am 100% sure that I want to die because of these jokes
Uziel Perez and on what do you base that?
Uziel Perez Am I not to be allowed to make jokes. I thought I was making a joke that was appropriate with the rest of the human population. But I guess not, as I've always said, you can't impress everyone.
It says all strangers will die. Just make friends with the person killing strangers.
TheHolySC No.It says all strangers will pie.
TheHolySC Look closely!
Introduce yourself. Stranger no more...
Who said anyone is Killin anyone ?
TheHolySC or you can just intruduce yourself to the person and you won’t be a stranger anymore
How do you know now the answer is 0413 not 1304 ? Both numbers can be interchanged I think now that is my genius mind
Shubham Goyal I agree. I think this riddle is wrong. You can't know if it's 0413 or 1304.
yup
I don't think the order of input matters
Well done! You are probably right.
If you listen carefully....3:58. The order of digits don't matter
I can't understand the riddle, even after he explained. I AM So CONFUSED!!! 😕
minecraft guy me too😧😧
Imagine the first guy knows the product is 52. The second guy knows the sum is 17. The first guy is basically telling the second guy the two numbers aren't prime. Then the second guy is saying that he knew that. Which means that his number is not the sum of two primes. So the first guy looks at 2 and 26. They sum to 28. But so do 11 and 17. Two primes. So the sum isn't 28. So he looks at the sum of 4 and 13, which is 17. Can two primes sum to 17? Nope. So he says he knows the numbers. The second guy looks at the numbers which sum to 17. 2 and 15 for a start. The product is 30. But 30 is 2 times 15, 3 times 10, 5 times 6. If the product was 30, the first guy would have to rule out two of these when told the second guy knew he didn't know the numbers. To do that, he would have to see what primes sum to 17, 13 and 11. None sum to 17, none sum to 11, 2 and 11 sum to 13. The second guy would be thinking the numbers could be 2 and 15 or maybe 5 and 6. So he wouldnt have said he knew the numbers. can rule out the first guy having 30, because then he wouldn't have been able to say that he knew the numbers. So the second guy rules out 30 as the product and therefore 2 and 15 as the numbers. He then does the whole thing again with 3 and 14. Same result. He hits gold with 4 and 13. But not quite. He has to rule out 5 and 12 and so on. 4 and 13 is the only possibility. I really do feel that the explanation in the video is a bit lacking in detail. Now make sure that no other combination will work ✌
Good job, thanks for your effort :)
I didnt want to make it too long in the video.
And thanks for posting! It's a good one. I thought the answer was 2 and 9 to start but then I watched the video. When you know the answer I think there's a good bit of justification required. When you don't there are lots of cases to go through. Making it 25 times as long. Would you agree it takes more work than the video seems to suggest?
u da real mvp
Phaionix Suppose 2 and 9 worked. Then the product is 18 and the sum is 11. So suppose that is what our two guys started with. Pete says he doesn't know the numbers, meaning they are not prime. Frankie says he knew that, meaning his sum cannot be two primes. While this is true of 11, that is not all that is required. Frankie DOESN'T KNOW Pete's number, so he can't assume the numbers are 2 and 9. All he knows is that Pete now knows the numbers. The question is, can Frankie find out the numbers now? Is 18 the only possible product? Well, take 4 x 7 = 28. There i one other way to write 28 as a product, 2 × 14 = 28. Imagine Pete had 28, and Frankie says he knew Pete didn't know the numbers. Pete would then be able to rule out 2 and 14 because they sum to 16 and 3 + 13 = 16, which means Frankie could not say he knew Pete did not know the numbers. So Frankie knows Pete could say he knew the numbers if he had 18 as a product (check this) or 28. And maybe more. So he could not say that 2 and 9 are the numbers. I hope that helps.
Phaionix I never was that good with names! That's a nice solution. You are using primes when you say 2+12=14=3+11 and so on. But that's a small point. Your point about what the first two statements mean is a new take. I wonder in the long run if it makes things easier?
Break the window or crawl through it! It's fragile! 300th comment!
I successfully worked through the logic up until the point where I had to start writing down all the possible product/sums of acceptable numbers. That was the point I decided I didn't care THAT much about getting the final answer.
when do you get stuck in a scary house with no escape?!??!?!!??!?!?!?!?!
why cant i just bust the door down... i mean, if i was going into a broken down shed boarded up and covered in blood, i would at least bring a weapon with blunt force, like a club
I'm sorry... Can someone explain something to me? I don't see how it is possible for Peet to single out any such number based on simply Frank saying "I know my number". It would seem there would still be too many possibilities for Peet at that point. Please Explain!!!
I agree. What if the numbers were 49 & 48? It seems it would work for that combo also.
@ Michael - I explained it in some other comment but will paste it in here as well so you can understand the solution:
"You probably approach this riddle incorrectly.
You treat Frank's and Pete's statements that they know the numbers as something to be verified and not as another clue.
You are also putting yourself in Franky / Peet shoes and subconsiously assume that they are working with the same amount of information as you have which is not the case. You don't know neither sum nor product. You need to figure it out but Franky knows product and Peet knows sum already. And with that additional information that you don't possess they are able to figure out the numbers.
Because of that your goal is to determine what numbers make that scenario possible and not whether Peet or Franky can claim that they know the numbers. If some combination of 2 numbers does not allow both Frank and Peet to know what those numbers are then this is wrong combination and not correct numbers.
I think it is easiest to understand if you do a kind of reverse engineering and follow the scenario knowing the correct numbers from the start (as if you were the person who programmed the lock):
Two numbers are 4 and 13. Frank then is being told that the product is 52 and Peet that sum is 17.
Frank knows then that possible combination of numbers is either 4 and 13 or 2 and 26. Therefore he knows that Peet was given a sum of either 17 or 28.
Frank says that he doesn't know the answer.
Peet knows all possible products that Frank might have (30 if correct numbers are 2 and 15, 42 if numbers are 3 and 14, 52 if numbers are 4 and 13 and so on)
Peet says that he knew that Frank doesn't know the answer.
For Frank that means that sum must be 17 because if it was 28 than one of possible options would be 5 and 23 in which case Frank would be able to figure out numbers on his own and Peet couldn't have claimed that he knew that Frank doesn't know.
So Frank says that he knows the answer because only 4 and 13 gives sum of 17 and product of 52.
Peet then also knows the answer because only product of 52 would allow Frank to say he knows the answer. Other products wouldn't allow that.
For example: If correct numbers were 2 and 15 than Frank would have product = 30, that would have possible combinations 2 and 15, 3 and 10, 5 and 6. Information that Peet knew that Frank doesn't know the answer would allow Frank to eliminate 3 and 10 but would leave other two combinations as valid options in which case Frank would not have said that he knows the answer. Therefore 2 and 15 cannot be correct numbers.
You can go through all other potential products (42, 60, 66, 70 and so on) and you will see that only in case of product =52 Frank could have made that claim. Therefore Peet knows the answer as well.
I hope you see now that 4 and 13 allow both Frank and Peet to indeed know what the numbers are and make this scenario possible.
And since these are the only numbers that allow for this scenario to be possible then they are the correct numbers.
BTW - this solution assumes that order of numbers doesn't matter. This should have been said in the riddle because without this assumption this riddle cannot be solved."
The solution is so simple! Ignore all the mathematical shit and work on that boarded up window. If you kick those planks hard enough you'll get free in no time. Besides, unlocking the door won't help much because it's boarded up outside.
It was a trick riddle and I outsmarted it!
gg
Read back my original post then admit you're the idiot.
illdemoteu That doesn't mean shit. You are still pretty dumb.
illdemoteu I get what you're saying. Personally I'd rather kick the boards off the window on the left than risk getting cut by glass but either window is an easy escape route.
Like I say either window works better than dealing with that maths bullshit.
You said there are 25 possible sums. Here you are assuming sum to be less than 100. This information is missing in the video.
Ther correct problem and the correct solution is here: www.cs.rug.nl/~grl/ds05/sumproduct.pdf
faizal somani THANK YOU! I wasted so much time! Yes, the riddle is told wrong. Sorry but anyone who got it right made the same falacious assumption that the video maker did.
Matthew Acuff even i thought about it for a long time. I also made a javascript for it. Still i was not getting what the video maker got.
Hard to solve them all. but really fun Riddle channel!
very hard
WTF
Call a locksmith there's my solution
Even if u can solve this u still need to explain the HR why your appointment is crucial for the organisation.
1:15 i can go out now :)))))))
รัตนาภรณ์ บุญเกตุ Strangers will pie beware.
รัตนาภรณ์ บุญเกตุ iii
The broken window is of another room. Just in case you didn't know. That's the second riddle.
click bait
: (
Yeah because you solved that puzzle correctly
the window is broken...
Looking at your resolution, I have one question.. how did you get only 25 sums left after knowing that it can't be a sum of prime numbers? I got way more than 25 options left. After removing the sum of high numbers, (since 98x95 will give a unique answer that no other 2 2-digit numbers multiplied together will give), I was left with 63 sums left.
Analyzing a little further I was able to remove a bunch of sums from my list, I made a spreadsheet with all possible outcomes from the mutiplication of 2 2-digit integers, excluded the mirror values (3&5 = 5&3), and selected to highlight the duplicate values. Doing that, I scanned quickly the spreadsheet to see which out of the possible 63 sums that I had found out previously, would not be generated by a pair of numbers which would have a unique product.
Ex: The sum 51 can't be made by adding 2 prime numbers, so technically it should be analyzed. However:
34 + 17 = 51, and
34 x 17 = 578, which is a value that no other 2 2-digit numbers could make, hence, if the sum was in fact 51, there would be a chance that Frank would already know the answer. So 51 is no longer a possibility.
By doing that analysis I reduced from 63 possible sums down to 10!
great work! And if you keep going, using the other information, you will reduce it to 1 and that's your answer!
But seriously, this "list of 25" is ridiculous. I think he just counted the table rows on the answer sheet
Claudio Antunes Joffe he said only numbers between 1 - 100 n not 1 or 100
Who else thought that the 4 numbers were in the sentences that the two guys were saying 😂
I 😂😂
ManjuDragon samee
Same Lol
Meeeeeeeeee bra
After trying to solve this riddle and not being able to reduce my possible answers to a single solution, I went ahead and watched the solution. At first I could not understand why their solution had a unique answer but mine did not.
After reviewing their explanation, I see that they have more rows in the table than they should. That is, they have rows in the table that contain products where Franky could not claim that he knew the 2 numbers. For example, the very 1st row has a product of 18. It claims Franky knows that the numbers are 2 and 9. In fact they could also be 6 and 3. Therefore Franky could not claim to know the numbers.
In fact as soon as Pete says he knew Franky didn't know the 2 numbers, Franky learned quite a bit about more about the sum of the 2 numbers. Basically Pete told Franky that the sum of the 2 numbers was odd (look up Goldbach's Conjecture) and that some odd sums were not allowed, eg. 33 because it is the sum of 2 primes (31+2). Now that Franky knows that 1 of the numbers must be even and the other odd, he knows more about how the factors of the product must be partitioned between the two numbers.
When Franky says he knows the numbers, he is telling Pete that there is only one possible way to partition the factors between the 2 numbers that meet all of the criteria outlined above. Some examples are:
24=3*8, 3+8=11 (lowest product and sum)
28=4*7, 4+7=11
208=13*16, 13+16=29
610=10*61, 10+61=71
960=15*64, 15+64=79
9306=94*99, 94+99=193
9506=97*98, 97+98=195 (highest product and sum)
If you can understand why these are the only allowable partitions then you've got it. Hint: for the lower value products the only rule is one number is of the form 2^n, n>1 and the other number is a prime. For larger products, the constraint that each number must be < 100 allows for exceptions to the previous rule.
Finally, Pete sees that his sum can only correspond to one solution so he and us know the 2 numbers. Unfortunately this isn't the case for numbers larger than 15. If the riddle specified numbers 2-15 instead of 2-99, then the solution given would be the correct and unique solution.
If we look at the table in the video at the rows with the sum of 29, we see three rows that should not be there. All three rows contain products that do not have unique partitions, therefore they (and other possible cases) would not have been up for consideration by Pete. The rows that should not be there are listed below.
54=2*27, 2+7=29 - other possibilities: 3*18, 6*9
100=4*25, 4+25=29 - other possibilities: 5*20
138=6*23, 6+23=29 - other possibilities: 2*69
The row that should be there (and presumably is but the video cut it off) is:
208=13*16, 13+16=29
You can see that there is only 1 legal partition for the product 208 that yields an odd number and an even number in turn yielding an odd sum. Therefore this is also a valid answer. Other valid answers include:
148=4*37, 4+37=41
592=16*37, 16+37=53
688=16*43, 16+43=59
And of course not all sums are allowable. Returning to the video, for the sum of 23 there are 3 rows. One of them shouldn't be there but the other 2 should so Pete's sum can't be 23, but it can be 17 (the answer given in the video) or 29, 41, 53, 59...
So there is no unique solution to this problem. Others have also added that there is actually no solution at all since there is no way to know which number to enter first. This is also true. While I enjoyed thinking about this problem, I'm going to have to give it a thumbs down for its lack of a unique solution and a poorly stated problem.
If
The lock was made in China it was way more easier to get out
🎷🎷🎷🎷🍾🍾🍾😼😼😼👍🏻
I would just go out where I entered
please please explain this answer more this is very important for me.
Khushi Taneja You dont
I got 2 and 9 lol
Sum 11
Product 18
But 11 can also be made by 3 and 8, which gives 24. 24 is a product of 12, 2 and 8,3 and 6,4 of which only the 8 and 3 pair gives a sum that is the right one ( 10 and 12 are not correct since 10 is achievable by summing 5 and 5, 12 by, 7 and 5) so Frank would be sure what the numbers are if he got 24 or 18, but! Pete could never be sure what the numbers are if he got 11, which means you have to try the next possible sum - 17 after which you quickly get to 4 and 13. Who thought of this riddle and got it to work? :) amazing
What If I didnt know english?
This was an old house. I would just
blow away the house with 1 blow.
One Punch
Dude...
I
oh god that guy in the barracks must be one of the 3% who can solve it. (i am being sarcastic, why would he not be able to know becuz there is an answer)
Why can't you just bust a wall? It looks pretty old to me.
This riddle is mistaken.
It is quite impossible for Peet to say he knows the numbers after Frank states he is confident in his own.
Frank understands that Peet can only be confident if his sum is never possibly of two primes. Call that type of number Tasty.
Such numbers are 11, 17, 23...
Therefore all Frank must do to be confident in the numbers himself is to see which ones add up to one of those Tasty numbers.
Therefore from Peet's perspective, any number which is the product of a Tasty number's possible parts could hypothetically allow Frank to be confident in his number. 18 (9,2) and 24 (8,3), for example, both satisfy Frank and leave Peet clueless.
Simply giving us that "Peet knows his number" doesn't imply anything because it has no conceivable conceptual manifestation due to being logically impossible.
I don't understand how this solution makes any sense. Please Explain.
the second guy knows the number too in the end. 9,2, 8,3 7,4 would not satisfy because they all add up to 11. 3 different products are possible, but the second guy can't know which pair works, then. Because he knows too, it must be that he can figure out what the product is, as there is only one possible product. There is no other product that has a pair that add up to 17.
the correct answer is 4 4
memed bengul: Just NO!
There is a logical flaw about Frank's Statement!
He just cannot say, "-Now I know the numbers!" before Pete says "-I know too".
At that point there are many possibilities that satisfy the clues so far.
This riddle is just fraud! :D
@ George, @Alper - you guys approach this riddle incorrectly.
You treat Frank's and Pete's statements that they know the numbers as something to be verified and not as another clue.
You are also putting yourself in Franky / Peet shoes and subconsiously assume that they are working with the same amount of information as you have which is not the case. You don't know neither sum nor product. You need to figure it out but Franky knows product and Peet knows sum already. And with that additional information that you don't possess they are able to figure out the numbers.
Because of that your goal is to determine what numbers make that scenario possible and not whether Peet or Franky can claim that they know the numbers. If some combination of 2 numbers does not allow both Frank and Peet to know what those numbers are then this is wrong combination and not correct numbers.
I think it is easiest to understand if you do a kind of reverse engineering and follow the scenario knowing the correct numbers from the start (as if you were the person who programmed the lock):
Two numbers are 4 and 13. Frank then is being told that the product is 52 and Peet that sum is 17.
Frank knows then that possible combination of numbers is either 4 and 13 or 2 and 26. Therefore he knows that Peet was given a sum of either 17 or 28.
Frank says that he doesn't know the answer.
Peet knows all possible products that Frank might have (30 if correct numbers are 2 and 15, 42 if numbers are 3 and 14, 52 if numbers are 4 and 13 and so on)
Peet says that he knew that Frank doesn't know the answer.
For Frank that means that sum must be 17 because if it was 28 than one of possible options would be 5 and 23 in which case Frank would be able to figure out numbers on his own and Peet couldn't have claimed that he knew that Frank doesn't know.
So Frank says that he knows the answer because only 4 and 13 gives sum of 17 and product of 52.
Peet then also knows the answer because only product of 52 would allow Frank to say he knows the answer. Other products wouldn't allow that.
For example: If correct numbers were 2 and 15 than Frank would have product = 30, that would have possible combinations 2 and 15, 3 and 10, 5 and 6. Information that Peet knew that Frank doesn't know the answer would allow Frank to eliminate 3 and 10 but would leave other two combinations as valid options in which case Frank would not have said that he knows the answer. Therefore 2 and 15 cannot be correct numbers.
You can go through all other potential products (42, 60, 66, 70 and so on) and you will see that only in case of product =52 Frank could have made that claim. Therefore Peet knows the answer as well.
I hope you see now that 4 and 13 allow both Frank and Peet to indeed know what the numbers are and make this scenario possible.
And since these are the only numbers that allow for this scenario to be possible then they are the correct numbers.
BTW - this solution assumes that order of numbers doesn't matter. This should have been said in the riddle because without this assumption this riddle cannot be solved.
George, your logic is pretty sound but you draw wrong conclusions. From the fact that if sum is 11 then Frank would know his numbers but Peet was still clueless even though he said he also now knows them you conclude that this means that Peet can't make that claim instead of concluding that this must be wrong sum. There are plenty of potential sums and products that these guys could have been given that would allow Frank to know the numbers but would leave Peet clueless. But there is only one combination of sum (17) and product (52) that allows both of them to claim that they know the numbers. And since they both claimed that then this is correct answer
How do you get in theres a peace of Woody on the door
the door opens inwards
I disagree with the 25 possible outcomes left. The two numbers can not be two prime numbers because else Frankie would know inmediately. However there are other products that would allow Frankie to know the numbers. For instance if the product was x^3 with x being a prime number, x^(2a+1) where x is prime and x^(a+2)> 99 (like 2^11 or 3^7, where only solutions are 64/32 and 81/27), x^(2a) where x is prime and x^(a+1)> 99 (like 5^4, only solution being 25/25) x*a where x is a prime number larger than 49, and in general any x*y where x is a prime number larger than 99/c and c is smaller than any prime number that composes y, like 53*a, where only solutions are 53/a, or 41*3*5, where only solution is 49/15, 29*5*7, where only solution is 29/35.
This means that the sum can not be larger than 54. If it was 55 or higher one of the numbers could be 53. If one of the numbers was 53 the product would have a single solution and Frankie would have known at first. Taking this into account i only get 11 possible sums, which are 11, 17, 23, 27, 29, 35, 37, 41, 47, 51 and 53. I guess your 25 included 57, 59, 65, 67, 71, 77, 79, 81, 83, 87, 89, 93, 95 and 97. I think it is wrong but following your logic I believe you should have kept checking numbers until they summed 198.
oh thank you for that riddle I´ll try to solve it tomorrow in scool
good luck! If you can solve this you dont need to go to school anymore ;)
no scool any more didn´t solved it today but I´llmaybe solve it tomorrow ;-)
now 2 1/2 days later I have a solution I´m gonna check it out :-)
13 & 4
haha sick! Great commitment.
congratulation ;)
yeah im really proud of me but where do you found the riddle?
smh bruh that padlock is just hung over a doorknob it won't do anything
You can just scream and the people next door would hear it since it's an old house!
1:34 just climb out the window!!!!!🤦🏽♂️
*riddle channel* : bla bla bla i likemath
* as me...* : in the corner of my room twitching of to much math. ya i shouldn't have watched this.....ehhhhhh
He asked it wrong as he says if they got the numbers wrong than the door will lock forever suggesting that it is not already locked so you can just walk out
No, he's meaning that it will stay locked permanently even if you subsequently enter the right combination, and so the problem is constrained in that they can't test multiple numbers as a strategy to rule out wrong ones and must reach a definitive answer to enter right the first time.
thats f'ing brilliant!! not only that but I cant think of any lock (like one pictured) that would stop a doorknob from turning.
Kevin Aye it's not brilliant. He meant that if you answer incorrectly, it will be locked forever. That means it was locked before.
Kiwi45679765 Maybe he meant that the door was already locked and when you set a wrong password another lock will triggered that makes it permanent.
Ugh. I'm having a brain fart. My head hurts
Minh Pham 😂😂😂😂😂😂
Minh Pham brain poop 💩
What that f**k solve this riddle
*CAUTION*
Medical students don't watch it!!
I have also one that no one can solve, which thing can't move, can't see, can't broke? Please like if you want to know answer
I mean 2 and 2 worked
Is everyone just going to ignore the fact that you don't know who's Peet and who's Franky? That made the riddle just waaay harder.
The narrator said which was which.
@@alexandergilbert1023 Yes , he did , but he changed their names again.
This whole solution is a mess
GET OUT OF THE DAMN WINDOW!
Theres litteraly a FRICKEN WINDOW
I'm kinda glad I didn't get this one. Welcome back to the 97%.
These intelligent men are intelligent like having the power of telepathy?
IT'S IMPOSSIBLE TO SOLVE ONLY BY KNOWING THAT OTHER GUY DON'T KNOW THE ANSWER!!!
Look take the numbers 4 and 13
ONLY Franky know that sum is 17
ONLY Peet know that product is 52
Franky know that possible answers could be: 2,15; 3,14; 4,13; 5,12; 6,11; 7;10; 8,9...
Peet know that possible answers could be: 2,26; 4,13;
Franky tell that he can't realize what is correct answer, but Peet already knew that (it's ok there is to many combinations for every his answer)
And with this information Franky can't do anything!!! It's just impossible
Unless if he tell him his sum or product... but then that information that he don't know the answer is not necessarily
And you still have just 50% chance of solving it because there two operation is commutative
4*13 = 13*4
4+13 = 14+4
Sorry about my bad English, it's not my main language :)
you mixed the characters up. it is the one with the sum who already realizes that the other one can't narrow down to one. after he states this, the guy that knows only the product realizes that the sum must be such a number that it's can't ever be derived from two primes.
Miloš Popović iii
It is 100% unsolvable, Once you turn a dial on the number lock, you already put in an incorrect code, so you are already screwed. And who in the world have a lock that permently locks if a incorrect code is entered when thee is no "enter button" Plus It is the woods, so just find coin or whatever and unscrew the barracks and get out.
I can' t understand this, because I'm German. 😰
in this case m gonna break the wall ,its look like pretty old and in another case i couldn't be able to enter in that room lol the door is blocked already
There are only two people who can solve this. One is the guy who made it and the other is a professional detective
You totally stole TED-ED's intro. If you look at the beginning of TED's bridge riddle, the intro is formatted the exact same way, and since this video was released after TED's, #EXPOSED!
good eye
He can stay inside the room without solving this
He would die from lack of oxygen ._.
This riddle was easy all it needed was
Common sense
What about 22?
1. The lock isnt even locking the door. Its just sitting on the door knob. 2. That house is completely destroyed. U could ceawl out. 3. 3% is made up. Good try for the click bait lol. 4. You could pull the "lock" off the door knob.(also 5. At 1:09 there is no lock)
Who would right a riddle on that wall anyone
But he still only has a 50/50 chance of escaping since the answer could be either 4 and 13 or 13 and 4.
Screw you riddle channel, I spent three days trying to figure this out under the assumption that the answer had to be two identical numbers
Yes it is 2 and 2
The First i though it was 4,9,6 Cause there was Written In The Wall DIF Then I Count D A,B,C,D Sooo 4 A,B,C,D,E,F,G,H,I Sooo 9 A,B,C,D,E,F Soo 6
After The Answer Reveal I Though i was correct but i was wrong....
um how are u trapped exactly?
Why dont u just whip out ur phone n call for help
i can solve that
i can do that too
i do
Wow this is was good... It took me some time to understand even after the solution was given 😅
*kanye shrug*
This made me look dumb :p
You lost me at 0:30 when you said “whole numbers”
First of all there is a broken window that I can just go through,
Second of all there is another window just barricaded with a few wooden planks that I can break
And third of all the damn walls are made out of wood I could just break through the house
2+2=4
2x2=4
*Opens console* "TCL" *Hits enter* *walks through wall* *opens console* "TCL" *Hits enter* *walks away*
I was able to solve all😂
Sorry I'm a little slow on this. Could someone explain why the numbers couldn't be 3 and 33? Ergo first guy is baffled because the product of 99 could have two solutions and the second guy is also puzzled because the sum is 36 which has multiple factors.
I would just make friends with the person who says Stranger will die
This is what I would do
1 see metal door with lock
2 look around the room
3 see riddle
4 not get riddle
5 keep looking around
6 find window
7 break outa there!!!
(If there was no window)
8 SCREAM HELPPPPPPPPP
9 try to remove the lock
10 break the tiny window on metal door (you saw it)
11 grab a glass shard
12 start trying to dig dig and DIG
13 crawl out
Aparate datorii si 2 masini belea
If this was real life i would instantly kill myself because I already know I would NEVER be able to figure this out
There may be an error. The solution can be 13 and 4 as well, rather than 4 and 13. Therefore better not to include the sum, but the difference: 9. This mean 13 and 4.
Or -9, meaning 4 and 13
UGH WHY ARE ALL THESE RIDDLES LIKE, RELATED TO MATH?! ITS NOT GONNA BE FUN ANYMORE!
no offence for shouting
Fuck the riddle, bust the door down.
Wait... Considering, the door was boarded from the outside, why in the world is there a lock inside?
How did the guy who put a lock outside?
I know, pictures can be inaccurate, but wtf?
Here's my solution. If Franky only knows the product and Peet only knows the sum and they both know the numbers at the end, another solution could be that the product and sum are the same. In this case, 02 + 02 and 02 * 02 are the same, so the numbers are 02 and 02. Neither of them know what the numbers are initially, so the product could have also been from 01 and 04 and the sum could have been from 01 and 03. This does not violate the rules stated here because it would be of the same prime number, not two different ones.
The riddle is very easy but you need to look at all corners of the room at different times. What can you see at 1:11?
I did IT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
but how would you know which order the two numbers go in the passcode? I did all the math, then declared it impossible for this reason. This riddle is flawed, very interesting, but flawed.
Well
SHIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII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I got it half right!
What he said in the beginning of the video reminds me of the bridge riddle on TED-Ed’s channel.
Anyone needing to be told that neither of the two digit numbers was 100 - may as well have given up there and then. 0:26 ☺☺☺
In the 'i already knew you couldnt' the T looks a lot like a 4. Giving a clue the numbers are disguised as letters. The S in sum also looks like an 8.
I cant solve this but i guess 84?
I was able to get information from the first two statements but from the last two, I got very confused.