rookie1ja 14 Posted March 30, 2007 Report Share Posted March 30, 2007 Complex Deduction - Back to the Number Puzzles This is definitely one of the harder number puzzles on this site. A teacher says: I'm thinking of two natural numbers greater than 1. Try to guess what they are. The first student knows their product and the other one knows their sum. First: I do not know the sum. Second: I knew that. The sum is less than 14. First: I knew that. However, now I know the numbers. Second: And so do I. What were the numbers? This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Savoury - solution The numbers were 2 and 9. And here comes the entire solution. There shall be two natural numbers bigger than 1. First student knows their product and the other one knows their sum. The sum is smaller than 14 (for natural numbers bigger than 1), so the following combinations are possible: 2 2 ... NO - the first student would have known the sum as well 2 3 ... NO - the first student would have known the sum as well 2 4 ... NO - the first student would have known the sum as well 2 5 ... NO - the first student would have known the sum as well 2 6 2 7 ... NO - the first student would have known the sum as well 2 8 2 9 2 10 2 11 ... NO - the first student would have known the sum as well 3 3 ... NO - the first student would have known the sum as well 3 4 3 5 ... NO - the first student would have known the sum as well 3 6 3 7 ... NO - the first student would have known the sum as well 3 8 ... NO - the product does not have all possible sums smaller than 14 (eg. 2 + 12) 3 9 ... NO - the first student would have known the sum as well 3 10 ... NO - the product does not have all possible sums smaller than 14 4 4 4 5 4 6 ... NO - the product does not have all possible sums smaller than 14 4 7 ... NO - the product does not have all possible sums smaller than 14 4 8 ... NO - the product does not have all possible sums smaller than 14 4 9 ... NO - the product does not have all possible sums smaller than 14 5 5 ... NO - the first student would have known the sum as well 5 6 ... NO - the product does not have all possible sums smaller than 14 5 7 ... NO - the first student would have known the sum as well 5 8 ... NO - the product does not have all possible sums smaller than 14 6 6 ... NO - the product does not have all possible sums smaller than 14 6 7 ... NO - the product does not have all possible sums smaller than 14 So there are the following combinations left: 2 6 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 (it is impossible to create a pair of numbers from sum 8, so that the product would have an alternative sum bigger than 14 ... eg. if 4 and 4, then there is no sum – created from their product 16 – bigger than 14 – eg. 2 + 8 = only 10) 2 8 2 9 2 10 3 4 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 3 6 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 4 4 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 4 5 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 The second student (knowing the sum) knew, that the first student (knowing the product) does not know the sum and he thought that the first student does not know that the sum is smaller than 14. Only 3 combinations left: 2 8 ... product = 16, sum = 10 2 9 ... product = 18, sum = 11 2 10 ... product = 20, sum = 12 Let’s eliminate the sums, which can be created using a unique combination of numbers – if the sum is clear when knowing the product (this could have been done earlier, but it wouldn’t be so exciting) - because the second student knew, that his sum is not created with such a pair of numbers. And so the sum can not be 10 (because 7 and 3) – the second student knew, that the first student does not know the sum – but if the sum was 10, then the first student could have known the sum if the pair was 7 and 3. The same reasoning is used for eliminating sum 12 (because 5 and 7). So we have just one possibility – the only solution – 2 and 9. And that’s it. This is definitely one of the hardest number puzzles on this site. A teacher says: I'm thinking of two natural numbers bigger than 1. Try to guess what they are. The first student knows their product and the other one knows their sum. First: I do not know the sum. Second: I knew that. The sum is less than 14. First: I knew that. However, now I know the numbers. Second: And so do I. What were the numbers? Link to post Share on other sites

Guest Posted April 28, 2007 Report Share Posted April 28, 2007 lol, this one brings back memories and I figured this out in like 10 min when I was a sophmore in HS (teacher was preaty impressed), I actually found this one preaty eazy as its a process of elimination. The 2nd stament "I knew that, and the sum is less than 14" forces the sum to be 11 because it is the only number where all the sets of the sums have have to have 1 non-prime number and have more than one possible set of sums. 2-9 3-8 4-7 5-6 this means the product has to be 18,24,28, or 30 and of those numbers the only one where all possible sums of two numbers gives a product less than 14 is 18. 2+9 < 14 2+12 = 14 2+14 > 14 2+15 > 14 so the product is 18 and the sum is 11, the two numbers are 2 and 9. I think my solution is slightly more elegant, ...ok my ego got the best of me again. *if two natural numbers x*y = K and x<y then (x+1)(y-1) >= K because if multiplied out it would be xy+y-x-1 so if x<y then y-x-1 >= 0. I just figured this on the spot so I thing it's right but I'm not 2 sure if there are exceptions. Keep in mind the initial conditions. Link to post Share on other sites

Guest Posted April 28, 2007 Report Share Posted April 28, 2007 accidental repeat Link to post Share on other sites

Guest Posted June 28, 2007 Report Share Posted June 28, 2007 was teacher preaty impressed with your spelling? Link to post Share on other sites

Guest Posted July 6, 2007 Report Share Posted July 6, 2007 I came up with the same solution as you did. Your equation with K was interesting, but trivial. The largest the product of any two numbers that sum to X can be is if each of the numbers is x/2. Geometrically this is saying that a square has more area per outer edge than a non-square rectangle. Your statement is a corollary to this much more powerful theorem. Link to post Share on other sites

Guest Posted August 1, 2007 Report Share Posted August 1, 2007 hahaha... right; one of the hardest on the site, please. Link to post Share on other sites

Guest Posted August 8, 2007 Report Share Posted August 8, 2007 Read the question again and assume that each students responds to the other instantly as I thought. They are geniuses.. Link to post Share on other sites

Guest Posted August 26, 2007 Report Share Posted August 26, 2007 "3 10 ... NO - the product does not have all possible sums smaller than 14" ++++++++++++++++++++++++++++++++++++++++++++++ Why the the product should have all possible sums smaller than 14? and what's the meaning of "First: I knew that. However, now I know the numbers." what does the First "knew"? I can't understand from the text, maybe that's because i'm not a English native people~ Link to post Share on other sites

Guest Posted September 25, 2007 Report Share Posted September 25, 2007 You don't have to take all combinations to work this one out. It is based on the following: If a number is the product of two primes, then you know their sum. Clues: A: I don't know the sum. So the number is not the product of two primes. One of the numbers must be 4, 6, 8, 9 or 10. B: I knew that. This is the biggest clue! If the sum can also be a sum of two primes, then B would not be absolutely sure that A does not know the numbers eg if B's sum was 6, then it could be that A's product is 9=3x3, from which A could immediately deduce the two numbers. So the fact that B knew that A did not know the sum limits the sum to be 11. Every other number under 14 is the sum of two primes: 4=2+2, 5=2+3, 6=3+3, 7=3+4, 8=3+5, 9=2+7, 10=3+7, 12=5+7, 13=2+11. So the possibilities are 2 9, 3 8, 4 7, or 5 6. B: The sum is less than 14 A: I knew that. This implies that the product cannot be factored out into two numbers whose sum is larger or equal to 14. Now the largest sum that the factors can add to is when we take a large and a small factor. This eliminates 3x8=24=2x12, which allows 2+12=14;v or 4x7=28=2x14 with 2+14>14; or 5x6=30=2x15, 2+15>14; leaving only 2x9=18=3x6. So the numbers are 2 and 9. Link to post Share on other sites

Guest Posted December 9, 2007 Report Share Posted December 9, 2007 Can someone explain to me why it is not 3 and 4? Link to post Share on other sites

Guest Posted March 3, 2008 Report Share Posted March 3, 2008 Can someone explain to me why it is not 3 and 4? Even without following the "rules", a sum of 7 would have given B the answer immediately upon hearing that A doesn't know the sum. because the product of 7's alternative pair (2,5) would have given the answer to A immediately. Link to post Share on other sites

Guest Posted March 13, 2008 Report Share Posted March 13, 2008 I disagree with answer 2 and 9. Because: first person know the product is 18. He don't know the answer now. He just could guess the set of the answer: 1x18 imposible, coz 1 not allowed 2x9 could be the answer 3x6 could be the answer By know the sum is less than 14, he still don't know the answer coz 2+9 < 14 3+6 < 14 So he don't know the answer and couldn't say wether he know the answer Link to post Share on other sites

rookie1ja 14 Posted March 13, 2008 Author Report Share Posted March 13, 2008 I disagree with answer 2 and 9. Because: first person know the product is 18. He don't know the answer now. He just could guess the set of the answer: 1x18 imposible, coz 1 not allowed 2x9 could be the answer 3x6 could be the answer By know the sum is less than 14, he still don't know the answer coz 2+9 < 14 3+6 < 14 So he don't know the answer and couldn't say wether he know the answer as mentioned in the solution ... 3 6 ... NO - it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 if 3 and 6 then they would think the following ... A teacher says: I'm thinking of two natural numbers bigger than 1. Try to guess what they are. The first student knows their product (18) and the other one knows their sum (9). First: I do not know the sum. Second: I knew that. (pairs of numbers that have sum of 9 can have various products) The sum is less than 14. (here is the problem, I thought that there would be some other combination for which sum would be bigger than 14 and thus I would help you by eliminating that combination ... but for sum of 9 there is no such combination ... whereas for sum 11 there is - 7 and 4 - which would make the product of 28 - that could be 14 and 2 as well - so 16 is bigger than 14 ... the same for 8 and 3, 5 and 6) First: I knew that. (there is no way to have sum of 2 numbers 14 or more, when their product is 18 ... but you must have thought that I do not know it - and that's the problem - you did not know that) However, now I know the numbers. Second: And so do I. What were the numbers? Link to post Share on other sites

Guest Posted March 16, 2008 Report Share Posted March 16, 2008 can sum1 tell y not 4,6.its product is 24 and v can even have 8,3 which have the same product and their sum is less than 14 Link to post Share on other sites

Guest Posted April 9, 2008 Report Share Posted April 9, 2008 Can someone explain to me why it is not 3 and 4? he ment to put 2+5, this problem has made me pull my hair out for 2 days and i was convinced it was 2 and 6, but anyway, 1sponer or watever his name is explains the prosses the best Link to post Share on other sites

Guest Posted April 23, 2008 Report Share Posted April 23, 2008 (edited) can sum1 tell y not 4,6.its product is 24 and v can even have 8,3 which have the same product and their sum is less than 14 Here's why it can't be 4 & 6. Follow the process: First student doesn't know the sum: this is fine, he knows the product is 24 but the sum could be 10 (4+6), 11 (3+8) or 14 (2+12). It fails right here. He also knows, per the 3rd line, that the sum is less than 14, so it can't be 2 & 12. However, if the product, which he knows, is 24, there is a possibility that the numbers are 2 & 12. Since there is this possibility he cannot possibly know that the sum is LESS (not equal) to 14. It is quite possible that the sum would be equal to 14, so this is a contradiction. Edited April 23, 2008 by puzzlmaster Link to post Share on other sites

Guest Posted April 24, 2008 Report Share Posted April 24, 2008 (edited) That was pretty hard since we have to think about all possible combinations. But I don't understand why the solution should be 9 and 2. Edited April 24, 2008 by Vishmi Link to post Share on other sites

Guest Posted May 28, 2008 Report Share Posted May 28, 2008 Can someone explain to me why it is not 3 and 4? It the numbers were 3 and 4 that means that one kid knows that the sum is 7. The teacher said that the numbers were larger than 1. So the combinations that give you 7 and the nums are larger than 1 are: 2:5 3:4 The sum kid(kid that knows the sum) is aware that the product kid knows the product. So if the other kid knew that the product was 10 then they wouldn't have the discussion because only 2 numbers give you a product of 10 and those are 2 and 5. Since the product kid did not speak saying he knows the numbers, the sum kid is safe to assume that the numbers are NOT 2 and 5. That leaves only one combination for a sum of 7 and that is the combination 3:4. The sum kid now knows this and would have spoken about what the numbers were. Since none of the kids spoke about their numbers and had that discussion above, this means that the numbers were not 2:5 or 3:4 Link to post Share on other sites

Guest Posted May 28, 2008 Report Share Posted May 28, 2008 I disagree with answer 2 and 9. Because: first person know the product is 18. He don't know the answer now. He just could guess the set of the answer: 1x18 imposible, coz 1 not allowed 2x9 could be the answer 3x6 could be the answer By know the sum is less than 14, he still don't know the answer coz 2+9 < 14 3+6 < 14 So he don't know the answer and couldn't say wether he know the answer Ok assume the numbers were 3 and 6. State 1: Sum kid knows the sum is 9. Possible combinations: 2:7 Product 14 3:6 product 18 4:5 product 20 State 2: The 2:7 gives a product of 14. A product of 14 can only derive from 2 and 7. Product kid would need no help to figure out the numbers. 4:5 gives a product of 20 which has the combinations of State 3 3:6 which has a product of 18 and the combinations of State 4 State 3: 2+10=12 In this case the product boy assumes that the sumn boy knows the sum is 12. What could be going in the sum boys mind right now? 2+10=12 Product 20 Could be the answer 3+9=12 Nope product is 27 giving away the answer 4+8=12 Product 36 Could be the answer 5+7=12 Nope product is 35 giving away the answer 6+6=12 Product 36 Could be the answer State 4: 3+6 = 9 Could be the answer 2+9 = 11 Could be the answer Both have a sum less than 14 This gets too confusing to continue, but all you need to understand is that none of the boys can eliminate enough combinations to have only one combination left to speak the answer. So the boys have to talk to each other to get some more info. The sum boys says the sum is less than 14. After this statement, both boys know the numbers. At this point the product boy knows the numbers because he states so. The sum boy eliminates the combination he had that would have left the product boy in doubt, thus he figures out the numbes too. BUT in the above example, the product boy doesn't figure out the numbers. Check state 3 and state 4. Both of them have combination that sum up to less that 14. So the product boy cant eliminate any combination. The sum at this situation doesn't help him. But in our example, the product boy had figured out the numbers after he know the sum was less than 14. So 3 and 6 are not the combination since the leave the boys still struggling to figure out the solution. Link to post Share on other sites

Guest Posted June 12, 2008 Report Share Posted June 12, 2008 You don't have to take all combinations to work this one out. It is based on the following: If a number is the product of two primes, then you know their sum. Clues: A: I don't know the sum. So the number is not the product of two primes. One of the numbers must be 4, 6, 8, 9 or 10. B: I knew that. This is the biggest clue! If the sum can also be a sum of two primes, then B would not be absolutely sure that A does not know the numbers eg if B's sum was 6, then it could be that A's product is 9=3x3, from which A could immediately deduce the two numbers. So the fact that B knew that A did not know the sum limits the sum to be 11. Every other number under 14 is the sum of two primes: 4=2+2, 5=2+3, 6=3+3, 7=3+4, 8=3+5, 9=2+7, 10=3+7, 12=5+7, 13=2+11. So the possibilities are 2 9, 3 8, 4 7, or 5 6. B: The sum is less than 14 A: I knew that. This implies that the product cannot be factored out into two numbers whose sum is larger or equal to 14. Now the largest sum that the factors can add to is when we take a large and a small factor. This eliminates 3x8=24=2x12, which allows 2+12=14;v or 4x7=28=2x14 with 2+14>14; or 5x6=30=2x15, 2+15>14; leaving only 2x9=18=3x6. So the numbers are 2 and 9. This explanation is pretty neat and simple, thanks Link to post Share on other sites

Guest Posted August 19, 2008 Report Share Posted August 19, 2008 But the statement "Every other number under 14 is the sum of 2 primes" is not true. 10 = 6+4 and 8 = 4+4 or 6+2. I think there is a leap of logic here, b/c there are exceptions. For example, if Product Guy knows the Product is 16, the sum could be 8 (4+4) or 10 (8+2). Let's say the sum is 10 and Sum Guy knows it. Sum Guy says "the sum is less than 14" - which is still true (true), but that doesn't help Product Guy solve anything. We are relying on Product Guy saying "now I know the numbers" as meaning he DOES KNOW THE CORRECT NUMBERS... but he may not! So, why can't the numbers be 8 and 2 ? Link to post Share on other sites

Guest Posted September 16, 2008 Report Share Posted September 16, 2008 Here is how I see it: The first one says he doesnt know the sum, so he knows the product. Another thing we can derive from this that the product has more than 2 divisors, since if the number had only 2 divisors it would be pretty easy to make the product as follows: x is the product and x = a*b is the only way to write x as a product of smaller numbers then the sum is a + b. So x can be produced by at least 2 different product like x = a * b = c * d, where a is different from both c and d. The second 1 says that the sum is smaller than 14. From this the first 1 knows the numbers. So if x can be written as x = a1 * b1 = a2 * b2 = .... = an * bn, there can be only 1 i in {1,n} for which ai + bi < 14 since the first couldnt guess the numbers if there were more pairs like that. But from the information that the "product guy" can guess the numbers if he knows that the sum is less than 14 he can guess the numbers too. This means that if his number(the sum) y can be written as y = c1 + d1 = ... = cm + dm there should be only 1 pair say cj,dj such that cj * dj = x is the only way to write x as a product such that the sum is smaller than 14. With such a pair both can decide as the conversation goes in the excercise. Lets see the solution: I try for y-s from 13 13 = 7+6 where with product reordering it can be written as 14 + 3 and 21 + 2 so its a good candidate. 13 = 8 + 5 with reordering 10 + 4 , 20 + 2 so it is a good candidate too since there are at least 2 good candidates the "sum guy" cant decide which is good so the sum cant be 13 12 = 6 + 6 with reordering 12 + 3, 18 + 2 so it is a good candidate 12 = 7 + 5 which is not good becouse it is the only way we can write 35 as a product is 5 * 7 so the "product guy" would know the sum from start 12 = 8 + 4 with reordering 16 + 2 so it makes it a good candidate and 12 fails the same way as 13 with 2 candidates. (*) 11 = 6 + 5 with reordering 10 + 3 which makes it a bad candidate since the product guy has 30 but he doesnt know if the numbers are 3,10 or 6,5 since they are both below 14. 11 = 7 + 4 with reordering 14 + 2 so its a good candidate 11 = 8 + 3 with reordering 6 + 4 so the problem is the same as with 6 + 5 11 = 9 + 2 with reordering 6 + 3 so the problem is the same as above. 11 can be a sum and 4,7 is a solution pair for the exercise since its a lonely candidate. Lets see how the conversation goes as a proof of (4,7) solution: -The first knows 42, but he doesnt know the sum which can be 4 + 7 = 11 or 14 + 2 = 16 -The second tells its below 14. -Now the first knows the numbers. -The second can guess the numbers using the discussion from paragraph (*) There can be more solution pairs which have a sum of less than 11 I didnt try those since 1 solution was enough for me. Link to post Share on other sites

Guest Posted September 17, 2008 Report Share Posted September 17, 2008 I missed the "I knew thats" and my solution is wrong therefore Link to post Share on other sites

Guest Posted September 27, 2008 Report Share Posted September 27, 2008 The problem is that the students didn't have anything better to do...They could play video games,but,no,they have to do the task teacher made for them...Sheesh,the teacher should do something more fun with his life... Link to post Share on other sites

Guest Posted October 26, 2008 Report Share Posted October 26, 2008 Okay, after reading all the posts very carefully, I have the DEFINITIVE ANSWER: The sum is...............the parts of the giraffes left over after the crocs ate. The product is..........what came out in the end. I'd give my right arm to be ambidextrous. Bows Link to post Share on other sites

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