The Most DIFFICULT Riddle EVER! 

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

You still end up with two possible solutions: 1. 4-13 2. 13-4 So you have a 50/50 chance...

not really a solution for he still has a 50% chance to miss if the numbers are used in the wrong order

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.

I would climb out of the broken glass window ...

can someone bring a math professor, i didnt even understand the solution

But how do you know which number comes first???

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.

Why do math when you can jump out the window...

It says all strangers will die. Just make friends with the person killing strangers.

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 ✌

So.... where did the get this 3% claim? Did you do a survey or something like that?

I dint even understand what he was saying lol

How do you know now the answer is 0413 not 1304 ? Both numbers can be interchanged I think now that is my genius mind

Who the fuck can solve this riddle

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!

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 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.

Even if u can solve this u still need to explain the HR why your appointment is crucial for the organisation.

I can't understand the riddle, even after he explained. I AM So CONFUSED!!! 😕

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!!!

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.

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

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.

Break the window or crawl through it! It's fragile! 300th comment!

Call a locksmith there's my solution

when do you get stuck in a scary house with no escape?!??!?!!??!?!?!?!?!

Hard to solve them all. but really fun Riddle channel!

1:15 i can go out now :)))))))

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

This was an old house. I would just blow away the house with 1 blow.

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.

17 of following combinations gives a multiplication value that can be derieved from 2 or more whole number combo. 1) 2+15 multiplication of which is 30, which can be a combo of 10x3 or 5x6 or 2x15 2) 3+14 multiplication of which is 42, which can be derieved from 6x7, 21x2, 3x14 3) 4+13, multiplication of which is 52, which can be derived from 26x2, 4x13 4) 5+12, multiplication of which is 60, which can be derived from 10x6, 30*2, 5x12 5) 6+11, multiplication of which is 66, which can be derived from 33x2 or 6x11 or 22x3 6) 7+10, multiplication of which is 70, which can be derived from 35*2 or 7x10 7) 8+9, multiplication of which is 72, which can be derived from 36x2 or 18x4 or 8x9 or 3x24. so when Frankie can be confused with 7 combinations, how did this guy with a very funny accent concluded that the number combo is 4 and 13 ??

the window is broken...

Why can't you just bust a wall? It looks pretty old to me.

oh thank you for that riddle I´ll try to solve it tomorrow in scool

The numbers are 4 and 7. The Sum is 11 and Product is 28. From the clues we gather 1. Frankie "I cannot tell what the numbers are" --> There is more than 2 possible factors Product is at Least 3 primes multiplied together IE: (2,2,3)=2x6=4x3 2. Pete "I already knew you couldn't" --> Pete knows the sum's combinations all have more than 2 factors All combinations of sum also have at least 3 prime factors 3. Frankie "I know the numbers" --> Frankie now knows Pete's Sum and there is only one possible answer There is one unique prime factor and the are only 3 prime factors in the product IE (2,2,13) 4. Pete "Oh I know too" --> Pete knows there is a prime and only sees 1 solution There is only one unique prime factor in all combinations of the sum (where the produce has 3 prime factors) For 4, 7 Sum 11 Product 28 9, 2 (3,3,2) 14, 2 (7,2,2) 16 Not Possible [13,3] 8, 3 (2,2,2,3) 7, 4* (7,2,2) 11 Possible 7, 4* (7,2,2) 6, 5 (2,3,5) 1. Frankie: Can be 14, 2 or 7, 4 2. Pete: All combinations have more than 3 prime 3. Frankie: Cannot be 16 4. Pete: Only 7, 4 is unique For 4, 13 Sum 17 Product 54 15, 2 (3,5,2) 27, 2 (3,3,3,2) 29 Possible 14, 3 (2,7,3) 18, 3 (3,3,2,3) 21 Not Possible [19,2] 13, 4* (13,2,2) 9, 6 (3,3,2,3) 15 Not Possible [13,2] 12, 5 (3,2,2,5) 13, 4* (13,2,2) 17 Possible 11, 6* (11,2,3) 10, 7 (2,5,7) 9, 8 (3,3,2,2,2) 1. Frankie: Can be any of 4 possibilities 2. Pete: All combinations have more than 3 prime 3. Frankie: Can be either 29 or 17 // Stop 4. Pete: Would not know if its 13, 4 or 11, 6 // In General, if the sum > 11 it opens up the possibility for more potential combinations leaving both Frankie and Pete unsure, however there may be another solution.

There are only two people who can solve this. One is the guy who made it and the other is a professional detective

I would just go out where I entered

I got 2 and 9 lol Sum 11 Product 18

please please explain this answer more this is very important for me.

Alternative simpler solution for the lock is (2,6) or (6,2) although no information on which order to apply. QU 1. Frankie actually has the product as 12, so there are two possible answers (2,6) or (3,4) so initially says he doesn’t know the answer. QU 2. Pete has the sum of 8, so knows Frankie couldn’t know the answer given possible solutions could be (2,6) or (4,4). Pete therefore still doesn’t know but knows the product must be 12 or 16. QU.3 Frankie now knows the answer as he knows Pete has a sum of 7 or 8. If the sum was 7, then the possible solutions are (2,5) or (3,4). If it was (2,5) then Pete couldn’t have said definitively that he knew Frankie didn’t know the answer as for a product of 10, (2,5) is the only solution and Frankie would have been able to say he knew the answer at the start. Therefore Frankie now knows the sum must be 8. Therefore the solution (2,6) is the correct solution rather than the alternative (3,4). QU.4 Pete now knows the answer. If the solution was (4,4) and so Frankie’s product was 16, he would still not know the answer as he would still have 2 possible solutions (2,8) or (4,4). The solution must therefore have a product of 12 and so by the same logic as Frankie employs, Pete also now knows the answer. However he still only has a 50/50 chance of getting out given the duplicate solutions of (2,6) or (4,13).

*Opens console* "TCL" *Hits enter* *walks through wall* *opens console* "TCL" *Hits enter* *walks away*

very hard

click bait : (

Who else thought that the 4 numbers were in the sentences that the two guys were saying 😂

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

I figured it out, but unless there was water in that room, I probably would have died from dehydration first.

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.

You can just scream and the people next door would hear it since it's an old house!

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)

What If I didnt know english?

I'm kinda glad I didn't get this one. Welcome back to the 97%.

THAT IS ABSOLUTELY UNREAL.I WOULD NEVER HAVE GOT THAT IN TEN LIFETIMES!THAT IS MIND BOGGLING DEDUCTION!

smh bruh that padlock is just hung over a doorknob it won't do anything

Ugh. I'm having a brain fart. My head hurts

How do you get in theres a peace of Woody on the door

Here is what i will do: 1. go on youtube 2. search IT'S EVERYDAY BRO but louder 3. play a other song called BABE but also louder 4. play a other song again called LOVE YOURSELF and it's also louder 5. wait until someone wakes up 6. they kidnap me 7. i do the same thing over again

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

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.

*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

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 :)

Wait. What if: The sum of the numbers can be either odd or even. If it's even, the numbers can be: (even + even) or (odd + odd), if the two numbers are odd, there is a chance that they are both primes, so Peet can't be sure about whether Franky knows the numbers or not. If the sum is odd, numbers are (odd + even) which means Franky doesn't know the numbers. Now: Franky knows one of the numbers is odd and one is even. He has split his product and now he knows the factors. He can be sure about the numbers if one of them is a power of 2 and other is a prime>2 Now Peet knows that the sum is in the form of 2^n + p. He starts taking powers of two from his sum and looks for a prime remaining. Two numbers that satisfy this are: 8 and 17 (or 17 and 8) And btw, neither the sum or the product of two numbers changes when changing the order, so it's all dumb

I didn't quite get the explanation behind the answer to be honest, but this does not change the fact that the riddle has a glaring flaw: if the answer is anything but a single number duplicated (like 11 and 11), than knowing the two numbers DOESN'T allow you to achieve your goal fully - if you are allowed just ONE try than knowing which numbers you have to imput still leaves you with guess which of the two goes were, because if it's only TWO numbers, multiplitation and addition can interchange the numbers used in the equasation (4 times 5 gives the same result as 5 times 4 for example). So, in this version, even KNOWING the numbers still gives you only 50% chance of escape :P

WTF

GET OUT OF THE DAMN WINDOW!

I can' t understand this, because I'm German. 😰

There are still two possible solutions: 4,13 and 13,4. so you would need another hint, e.-g. "the first number is lower" or "the first number is prime" to reduce it to 1 solution.

Disagree. Let's say the sum is 11. The options are 2+9, 3+8, 4+7 or 5+6. For all these options the product can be divided by two different numbers (2*9=3*6, 3*8=2*12, 4*7=2*14, 5*6=3*10). Therefore, when Pete says, "I already knew you couldn’t", it is not necessarily because of that solution.

What that f**k solve this riddle

I mean 2 and 2 worked

He can stay inside the room without solving this

- The outside you see windows and a door nailed with "plank". - The inside you see no window and some accessories on the door = a lock(but somehow disappear after a few second) now the question how to get out? - twist the handle door - the door can't be pushed because there is a plank blocked it - pull it (duh!) - if the door can't be pulled try to crash(start run to the door) it with your body

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.

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!

This riddle was easy all it needed was Common sense

1. I'm bored 2. Watch this 3. My brain cells is broken. I feel so dumb watching this lol

I still think 2, 9 works. The sum is 11, therefore Frankie is only considering 2 numbers, 18 or 24, and Pete knows this initially, thus his confidence that he knows Frankie doesn't know. 18 has 2 factors (6*3, 2*9). Had the sum been 9, then Pete wouldn't make the statement that he knew that Frankie doesn't know because 9 = 2+7 (two primes) so there'd be a chance that Frankie did know, which means Pete couldn't proclaim that he knew that Frankie didn't know. This bit of information is what excludes 24 too because 24 has 3 factors and Frankie wouldn't be able to deduce the number simply by learning that Pete knew that Frankie didn't know. Only 18 could be the product if 11 were the given sum. And therefore it still seems like the numbers 2, 9 would satisfy.

I do not understand why after second statement we have only 25 possible sums? I think there are 54 possible sums: 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, 107, 113, 117, 119, 121, 123, 125, 127, 131, 135, 137, 143, 145, 147, 149, 155, 157, 161, 163, 167, 171, 173, 177, 179, 185, 187, 189, 191, 197, which can't be splitted as a sum of two prime numbers.

@Riddle Channel Jokes beside, why should there only be 25 possible sums? The Numbers have to be between 2 and 99 => Min Sum is 4 and Max Sum is 198 => 195 Numbers Now taking Goldbachs Conjecture into account, all even numbers can be ignored => 97 Numbers For every Prime between 3 and 195, an additional Number can be ignored (Prime + 2). There are 43 Primes in this Range => 54 Number. So how do you get 25?

This is where Common CORE math would actually be useful.

I solved this by writing a haskell program : ) import Data.List(groupBy, sortOn) import qualified Data.Map as M -- The possibilities for the two numbers are expressed as a list of tuples of integers between 2 and 99 allPossibilities :: [(Int,Int)] allPossibilities = [(x,y) | x [(2,9),(3,6)] -- ... -- is the initial possible state of a person knowing the product of the two numbers. -- Filter out any entry where there is only one possibile pair of numbers because then the person would know the answer already personKnowsOpButNotTwoNums :: Ord a => [(Int,Int)] -> (Int -> Int -> a) -> M.Map a [(Int,Int)] personKnowsOpButNotTwoNums possiblities op = M.filter ((>1) . length) $ groupNumbersByOp allPossibilities op groupNumbersByOp :: Ord a => [(Int,Int)] -> (Int -> Int -> a) -> M.Map a [(Int,Int)] groupNumbersByOp possibilities op = M.fromList $ map (\ps@((x,y):_) -> (op x y,ps)) groupedByOp where groupedByOp = groupBy (\(x0,y0) (x1,y1) -> op x0 y0 == op x1 y1) sortedByOp sortedByOp = sortOn (\(x,y) -> op x y) possibilities -- Gradually restrict the possible states the two people could be in. I will refer to the product-knowing person as "ProdPerson" and the -- sum-knowing person as "SumPerson'" -- ProdPerson does not know the two numbers step0 = personKnowsOpButNotTwoNums allPossibilities (*) -- SumPerson' does not know the two numbers step0' = personKnowsOpButNotTwoNums allPossibilities (+) -- SumPerson' already knew that ProdPerson did not know, so every pair of numbers SumPerson' thinks are possible must be in step0 -- (if any pair of numbers SumPerson' thought were possible were not in the possibilities where ProdPerson didn't know the answer, -- then SumPerson' couldn't know for sure that ProdPerson didn't know the answer). step1' = M.filter (all multInStep0) step0' where multInStep0 = \(x,y) -> (M.member (x*y) step0) -- ProdPerson knows that SumPerson' is limited to the options in step1' (because they were told) and then ProdPerson knows the answer, -- so the only options that remain are those where the list of numbers ProdPerson thinks are possible have size 1. step2 = M.filter ((==1) . length) (filterSumInStep1' step0) where filterSumInStep1' = filter (\(x,y) -> M.member (x+y) step1') -- SumPerson' knows that ProdPerson is limited to the options in step2 (because they were told) and then SumPerson' knows the answer, -- so the only options that remain are those where the list of numbers SumPerson' thinks are possible have size 1. step3' = M.filter ((==1) . length) (filterProdInStep2 step1') where filterProdInStep2 = filter (\(x,y) -> M.member (x*y) step2)

Everybody who solved this please read. There is a mistake, when I was solving this riddle I have found that it has more solutions, if there is only your solution, then the final buble must come from Franky (said somethink like:I know you can know the numbers). You just forget that Peet knows the sum! So every combination of prime number and 4, which gives the sum from "not sums of prime numbers" is correct answer.

To escape :WHY DON'T YOU JUST GET OUT OF THAT SHATTERED WINDOW.This is so unforeseen in my life.

"Two perfectly intelligent men are talking to each other." *Pete spelled his damn name wrong* 🤦🏾‍♂️😂😂

Frankie would also know the two numbers if they weren't both prime, but the product was a cube of a prime (8, 27, 125, 343, the rest are too large as 11^2=121). All of these will lead to even sums (and therefore can be formed with a pair of primes, allowing Frankie potentially know the numbers). The sum must not be possible through addition of two primes, so it must be 2 more than a composite odd number, for Peet to know the first person couldn't know the two numbers. I'm still kind of confused on which pairs are listed for the allowable sums (I understand why these sums are allowable (e.g. what's wrong with 12, 5 for 17?)

Actually the number of cases can be drastically reduced. There is no need to check those many iterations. I did it on pen and paper and I was able to solve it. It think there is no need of excel sheet for this. All we need is a few extra constraints to reduce the number further. Yes, the solution in the video is not great. Because it is trying to solve it in a brute force manner after a point. So instead of 25 different possible sums, we can reduce to only 10 cases by knowing the fact that "any one of the number cannot be a prime number greater than 50". If it is 20 = 4*5 and 2*10 are possible. But if the number is (53*2*2) then (53,4) and (106,2) are not possible. The only way is (53,4). So, we ruled out the possibility of both primes and also one prime greater than 50. Later in those 10 cases we need to prove that there are atleast two possible solutions in 9 cases and only one solution in one case. There is no need to check everything. But only the numbers with the power of 2. (2,4,8,16...) And a prime number. And in this process of checking we encounter 4,13 (the only pair of power of 2 and a prime in the case of sum 17). In all other cases there are atleast 2. [ If we prove the there are atleast two cases then we can proceed to next case. And in order to check this power of 2 and the prime pair gives the fastest result. Because if there is a power of 2...then any combination where the 2 is splitted gives 2 even numbers. And the sum of even numbers is always even. So we can rule out that as all the sums are odd.] In total I checked 19 cases (2 each for 9 cases+1 for the answer case) by hand... But that's enough to pin point the answer. "By finding patterns we can reduce the grunt work. One has to be clever enough to find it. We cannot say the problem is stupid always. We can say that the problem is stupid as far as I thought." Anyways it is good problem. It feels stupid for sometime. But after finding more and more patterns it starts to become beautiful.

I disagree with the given solution...I believe that this is the correct solution. Please correct me if I am wrong. I agree about the product not being the result of 2 prime numbers and as a result gives Frankie at least 2 combinations that give the product value. I also agree that Pete can tell that the sum is not the result of 2 prime numbers. This is where my logic strays from the given solution. Frankie now says "I know the numbers". This means that the 2 options that give Frankie's product number result in one sum that is allowed and one sum that is not allowed. Since Pete already told him that he definitely knew that he couldn't solve it, that lead Frankie to realize that the other solution is the correct value. For Frankie: the product is 18. This has two options 2x9 and 3x6. So Frankie knows that those are the only 2 combinations that would satisfy his product value. However, he doesn't know which one is correct. Consider the 2 cases separately: If the answer is 2 and 9 the sum is 11. This is a sum that does not come from 2 prime numbers and therefore any combination of numbers that sum to 11 will result in multiple possible product combinations. Therefore Pete would KNOW that Frankie couldn't figure it out alone. However, if the answer was 3 and 6, the sum of the numbers would be 9 and one of the options of the sum could be 2 + 7 which would be 2 prime numbers. Therefore there would have been one option that Frankie could have solved by himself with a sum of 9. Since Pete was SURE Frankie couldn't solve it himself, 3 and 6 can't be a possible solution. When Franke hears what Pete says, he realizes that only one of his possible solutions will work, 2 and 9. AFTERWARDS, Pete then comes to the same conclusion.

Answers: A) The lock is on a knob, so I can open the door anyway. B) I'm in a wooden broken down shack, I snash through a wall. C) Oh look, after I read the riddle the lock disappeared! D) The window us shattered, why not climb out??? E) I don't get trapped in the first place, I'm not an idiot!!!!

I just love watching overly hard riddles to confuse myself

Part 1: Assuming Frankie's product is 52 and Pete's sum is 17. Frankie: "I don't know what the numbers are" ( it could be 2 and 26, 4 and 13...) Pete: "I already KNEW you couldn't" ( No two primes add to 17, if two did, Frank COULD have potentially known) *This means there is a manageable list of sums for which no two addends are primes. 17 is on this list. *Here is some of that list from a short program I made: [11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97] Frankie: "I know what the numbers are now" (he knows about this list, since Pete effectively said his number is not made from two primes. Now Frankie knows the numbers are his factors 4 and 13 since they are the only ones that add to a number on that list) Part 2: Pete "I know what the numbers are too now" It gets more tricky here: other sets of two numbers could have replaced those in part 1, but with the information in part 2, only 52 and 17 will fit. Still assuming Pete is given sum 17: Lets test for potential addends and products from Pete's perspective to see if he could know 2 and 15 (30) -> 2, 3, 5, 6, 10, 15 then Frankie couldn't have known the numbers since two sets of factors here add to a number on the list of possible sums. (5 + 6 is 11 and 2 + 15 is 17) So Pete knows it's not this 3 and 14 (42) -> 2, 3, 6, 7, 14, 21 -> 21 and 2 add to a number on the list, so does 3 and 14 So Pete knows it's not this 5 and 12 (60) "" So Pete knows it's not this 6 and 11 (66) "" So Pete knows it's not this 7 and 10 (70) "" So Pete knows it's not this 8 and 9 (72) "" So Pete knows it's not this 4 and 13 (52) - > 1, 2, 4, 13, 26, 52 -> ONLY 4 and 13 add to a number on the list, so Frankie would know the two numbers if his product was 52, since it can't be any other factors AND it is the only set of numbers that add up to Pete's sum that has this property. (If many of these had this property, we must be looking at the wrong set of numbers, because Pete couldn't know which Frankie had) For ex: Say it was 18 and sum is 11 2 and 9 (18) -> 2, 3, 6, 9 -> Only 1 set of factors would get a number on the list, so Frankie would know the numbers... 3 and 8 (24) -> 2, 3, 4, 6, 8, 12 Frankie would also know here 4 and 7 (28) - > 2, 4, 7, 14 And here. These issues exist for al the combinations of values that add to the sum of all the other numbers on the list except 4 and 13 with 52. Pete can't know which addends are Frankie's factors here since all 3 could be 4 and 13 with 52 are the only numbers that satisfy these constraints 1. The product is not composed of two primes AND 2. Only one set of factors add to a number on the list (so Frankie knows the number) AND 3. There is only one instance where the addends of Pete's number multiply to a number with only one set of factors on the list. So Pete knows it must be that product and those addends.

It would be nice if you labeled which character was who. It's somewhat arbitrary and unfair to assume the character on the left is the first one listed in the riddle. I had no idea who said what until you gave the solution.

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

Another thing is that the sums factor of 17 besides 4 & 13 are 3 & 14 which multiplies to 42 that is also a possible product, then there’s 2 & 15 that multiplies to 30 in which is also another possible product. So mathematically I say this riddle may be broken.

I solved it right off the bat because you can't get yourself locked in that room without unlocking the door in the first place

I solved it but threw it away. The drawing showed 2x2 numbers to enter. So either there should be a 0 in front, or there should be 3 open spaces.

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

Well, I would never go exploring without my phone and a knife, so I would just cut my way out of the house, and if I can't I call the police

Riddle:A guy is trapped in a room with his leg chained to the wall. He found a saw in the room. He tries to saw through the chain, but the chain is too hard to be sawed through. What do you think the saw is for?