taeto 93 Posted February 21, 2019 1 hour ago, Ghideon said: Can we see the original question? (Even if it's in Turkish it might help at this point) Excellent points, I agree! 0 Share this post Link to post Share on other sites

Strange 4227 Posted February 22, 2019 17 hours ago, Ghideon said: Can we see the original question? (Even if it's in Turkish it might help at this point) It would be interesting to see how a Google translation differed from the OP 0 Share this post Link to post Share on other sites

taeto 93 Posted February 22, 2019 Whatever the original question might be, it does seem like an interesting and original attempt at creating an attractive mathematical/logical puzzle. However, it appears a failed attempt. It is really hard to come up with interpretations of the different parts of the question which make things fall into place and provide a solution. I have been thinking about it a bit, and I have not yet discovered a way to reasonably interpret the question which allows for just a single solution, let alone a unique one. 0 Share this post Link to post Share on other sites

md65536 340 Posted June 18, 2019 (edited) On 2/21/2019 at 6:58 AM, taeto said: End of analysis of the case 4333. I do not see how John can rule it out as the solution if Jack knows R. But it is tricky, and I can definitely have overlooked something. Did I make a mistake? R=(8,1). Yes. If the range is known to be [3, 4], then Jack cannot know that James doesn't know the answer, as he states. If the range is [3, 4] then Jack knows that James must know the answer. The only way that Jack can assert that James doesn't know the answer is if the range includes factors that can multiply to the same product in different ways, eg. 9*1 or 3*3 (requires a range of at least [1, 9]) or 8*2 or 4*4 (range at least [2, 8]). First ruling out 0, the smallest range for which this works is [1, 4]. *We* know this much after Jack's statement, so everyone else (John, James) are able to know this after his statement too, so John can rule it out. No wait... my line of reasoning is wrong and might not go anywhere useful... Edited June 18, 2019 by md65536 0 Share this post Link to post Share on other sites

md65536 340 Posted June 18, 2019 (edited) On 2/21/2019 at 6:58 AM, taeto said: So far John seems unable to eliminate 4333 as a possibility. But we have not yet considered James's statement. The product 108 cannot occur in many ways: 9322, 9431 are the only two alternatives. As above, 9431 can be dismissed just from the info that Jack gave. James observes that the possible sums are 13, 16. Neither of these allows John to determine a solution, certainly this is true of the sum 13, which is shared by 8221, the (supposed) actual solution. And sum 16 has enough candidates that no exact identification is possible, it seems. What about 6332? Its product is also 108. It's the only such one that adds up to 14. If John supposes the answer is 4333, then James would have considered 6332 as a possible solution, and wouldn't have been certain that John wouldn't know the answer??? Is that enough to rule out John thinking it's 4333? Or would we have to also show that every other possibility that James might suppose that John might suppose, doesn't have 2 or more combinations that add up to 14? But wait... 8411 adds up to 14... So John considering 4333 as an answer might have considered that James knew both 6332 and 8411 add up to 14, so could say that John didn't know the answer, meaning that 4333 is still a possible answer to John? Edited June 18, 2019 by md65536 0 Share this post Link to post Share on other sites

md65536 340 Posted June 18, 2019 On 2/21/2019 at 6:58 AM, taeto said: Now there is a complication: if the actual solution were 9431, then Jack would have R=(9,1) and maybe Jack would have to admit that James might possibly deduce the solution from knowing p. The solution could be 9751 in that case.R=(8,1). If the solution were 9xy1, the product would be the same as xy33, so Jack could claim that James doesn't know the digits. (Then, 9621 adding up to 18, and 9431 adding up to 17, both with products 108, might be possible reasons why John can eliminate 4333.) 0 Share this post Link to post Share on other sites

md65536 340 Posted June 19, 2019 (edited) I think I've solved it and the answer surprised me. I'll restate the puzzle as I solved it: There's some number with digits abcd, with 0<=d<=c<=b<=a<=9. Person P1 knows a and d. P2 knows a*b*c*d. P3 knows a+b+c+d. P1 says "A) I don't know the digits but B) neither does P2". P2 says "C) I don't know the digits but D) neither does P3." P2 says "D) I know that P3 doesn't know the digits, and C) I don't know the digits either." --- I believe this order is important because P3 can know the answer while the statement is being given. P3 says "E) I didn't know the digits but F) now I do." It does not seem to need to say more but it could say now P1 knows or now everyone knows. Solution: A) a and d must be different or it would know. B) The range must contain some number with ambiguous factorization. For example, 9*1=3*3. If the range contains 1 and 9, or just 3, then it is still good. Do this for all possibilities and you find the range must contain 0, 2, 3, or 4. If it contains 1 we already know it must also contain either 0 or 2. So, eliminate all possibilities where d>=5. C) Eliminate all remaining possibilities that have a unique product. I think that a key element here is the concept of common knowledge. At each step, everyone can eliminate any possibilities that are based on what was said, but not necessarily the possibilities that the others could eliminate. Anyway, that might not be important yet... D) P2 can eliminate all possibilities that add up to a unique sum. E,F) Here is the key to the puzzle and where most of the possibilities are eliminated. P3 can also eliminate all possibilities that add up to a unique sum. However, since it now knows what the answer is, it must be that it was possible for P3 to assume that P2 knew the digits. The answer must be a combination whose sum is not unique sum, but where all other combinations that add up to the same sum have a unique product. Edit: No wait, that's not what I implemented! I looked for combinations that P2 knew had a unique sum *among* only possibilities that have no unique product, but which P3 did not know had a unique sum. Very confusing! I'm not sure I got that right or implemented it right, but I end up with only one possibility left and that is Spoiler 9994 A) P1 can't guess the middle digits. B) every product 9xy4 can be factored multiple ways, eg. C) 9*9*9*4 = 9*9*6*6 D) 9994 sums to 31. So do 8887, 9877, 9886, 9976, 9985, but these others all have unique factorizations. E) P3 figured 9994 is the only possibility that does not have a unique sum, but it has a unique sum among possibilities without unique factorizations. If it was any of the other possibilities that add up to 31, then P2 would have guessed the digits from the product. Edited June 19, 2019 by md65536 0 Share this post Link to post Share on other sites

md65536 340 Posted June 19, 2019 1 hour ago, md65536 said: However, since it now knows what the answer is, it must be that it was possible for P3 to assume that P2 knew the digits. The answer must be a combination whose sum is not unique sum, but where all other combinations that add up to the same sum have a unique product. On second thought I think what I crossed out is true, and possibly equivalent to what I implemented. 0 Share this post Link to post Share on other sites

md65536 340 Posted June 20, 2019 (edited) The solution also works if Jack/P1 doesn't know a and d but just d-a, so the puzzle could be interpreted either way. The less information Jack is given, the harder it is for us to reason about it, but the easier it is to eliminate possibilities once we figure it out, because for Jack to be certain that James/P2 doesn't know the answer implies that the info Jack actually has must be substantial. If d<=4, then we know that d-a>4, eliminating possibilities like 3222 earlier. Edited June 20, 2019 by md65536 0 Share this post Link to post Share on other sites

md65536 340 Posted June 20, 2019 (edited) 17 hours ago, md65536 said: The solution also works if Jack/P1 doesn't know a and d but just d-a, so the puzzle could be interpreted either way. Changed my mind! Spoiler If Jack only knows a-d, then 9994 doesn't work, because 7772 also has a difference of 5 but it has a unique product, so Jack could not know given a-d=5, that James doesn't know the answer. On 2/20/2019 at 1:56 PM, taeto said: You really have to give up on your idea of "range". If Jack only knows that the difference a-d is equal to 7, he most definitely cannot ascertain that James will not figure out the precise solution once James knows the product. No, I think the interpretation of "James knows only a minus d" is the right one. There are no numbers with a-d=7 that have a unique product. Therefore he knows that James doesn't know the answer. I think 8221 might work in the end? There was the idea that 4421 or 4222 would then also work. However, a-d=3 can be eliminated by 5552 (a unique product), and a-d=2 by 3111. If 0's are not allowed, then 8211 fits all the criteria. However I end up with 3 working solutions: 8211, 9881, and 9981 all seem to work. Edited June 20, 2019 by md65536 0 Share this post Link to post Share on other sites

Commander 35 Posted October 15, 2019 (edited) I think this puzzle needs to be restated by the proposer again so that any ambiguity is removed ! Also I think the Puzzle is that a,b,c,d are all from digits 1 to 9 [no zero] each unique and you can have a > b> c> d ! The question is to find the smallest 4 digit number which will satisfy this as well as the Conditions stated for Jack, John and James ! >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> As an aside please solve this related Puzzle : Some children were playing in the Garden. Mr. Johnson asked Mrs. Albert if they were all hers. “No” she replied. “ My children are playing with friends from 3 other Families “. “Our Family happens to be the largest, The Berry Family have a smaller Number of Children, The Charles Family have an even smaller number and the Dickens have the smallest of all”. “How many children altogether ?” asked Mr. Johnson. Mrs. Albert replied , “There are fewer than 18 and the product of the Numbers in the 4 Families happens to be my House number which you already know”. Mr. Johnson asked if there was more than one child in the Dickens Family. Mrs Albert replied. How many children are there in each family ? Edited October 15, 2019 by Commander clarity 0 Share this post Link to post Share on other sites

swansont 7232 Posted October 15, 2019 4 hours ago, Commander said: As an aside please solve this related Puzzle : ! Moderator Note A new puzzle should go in a new thread. 0 Share this post Link to post Share on other sites

Commander 35 Posted October 15, 2019 As advised by the Moderator the related Puzzle is posted in a separate topic Plz reply to that Puzzle in the new Thread. 0 Share this post Link to post Share on other sites