rookie1ja 11 Report post Posted March 30, 2007 Easy Deduction - Back to the Number Puzzles A teacher thinks of two consecutive numbers between 1 and 10 (1 and 10 included). The first student knows one number and the second student knows the second number. The following exchange takes place: First: I do not know your number. Second: Neither do I know your number. First: Now I know. What are the 4 solutions of this easy number puzzle? 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. Easy Savoury - solution None of the students can have numbers 1 or 10, since they would guess the other one’s number with no problems. I will describe solutions at one end of the interval of numbers 1-10 (the same can be done on the other end). Information that the second student does not know must be important for the first student. So the first one must expect that the second one has 1 or 3 (if the first one has 2). And as the second student does not know, then he has certainly not 1. So the first pair is 2 and 3. If the first one had 3, then he would expect the other one to have either 2 or 4. But if the second one had 2 (and the second one would have known that the first one does not have 1), then he would know the number of the first student. However, neither the second student knows the answer – so he has 4. The second pair of numbers is 3 and 4. Solutions at the other end of interval are 9 and 8 or 8 and 7. A teacher thinks of two consecutive numbers between 1 and 10 (Edit: 1 and 10 included). The first student knows one number and the second student knows the second number. The conversation of the students is as follows: First: I do not know your number. Second: Neither do I know your number. First: Now I know. Will you find all 4 solutions? Share this post Link to post Share on other sites

Guest Report post Posted June 14, 2007 What happened if the student A has the number five? Student B can have either the number 4 or 6, in which case he could guess that the other student has 3, 5, or 7. Share this post Link to post Share on other sites

Guest Report post Posted June 27, 2007 Honestly, when I tried doing this one, I never included 1 or 10 in the first place since the original problem says BETWEEN 1 and 10, meaning the numbers in question are 2 through 9. That might seem nit picky but these ARE brainteasers and it'd be nice if this one was clarified a little bit more. Also, I agree with the other person, what happens in the case of a 5? Why has that one been eliminated? Share this post Link to post Share on other sites

Guest Report post Posted July 5, 2007 If it was a 5, then the first one would not be able to say "Now I know". Instead, he would have said "I still don't know". Then, the 2nd person would know that the 1st person has either 4,5, or 6. If the 2nd person had a 3, 4, 6 or a 7, he would be able to say "I know now". But if he had a 5, he would still not be sure. And then the 1st person would know that the 2nd person has a 5. Share this post Link to post Share on other sites

Guest Report post Posted August 2, 2007 I can think of two situations but not four. If A has the number two, he knows B is either one or three. If B is one, B would KNOW A is two. Since B doesn't know, B must be three. The same would go for eight and nine I'm getting addicted to these puzzles. I totally forgot to watch TV tonight. Anyone know where to find lateral thinking puzzles? I guess I could look a little harder but I just can't put off watvching TV any longer Share this post Link to post Share on other sites

Guest Report post Posted August 31, 2007 The questions asks "Will you find all 4 solutions?" The answer, then, should be "yes" or "no." There are a lot of riddles on here that ask that sort of question. The question SHOULD state, "what are the 4 solutions?" Share this post Link to post Share on other sites

rookie1ja 11 Report post Posted August 31, 2007 The questions asks "Will you find all 4 solutions?" The answer, then, should be "yes" or "no." There are a lot of riddles on here that ask that sort of question. The question SHOULD state, "what are the 4 solutions?" right ... so will you find all 4 solutions? yes or no? there are a lot of people writing the same as you, but just a few of them really answer it (and prove it) Share this post Link to post Share on other sites

Guest Report post Posted September 26, 2007 isn't this assuming alot? Share this post Link to post Share on other sites

Guest Report post Posted September 26, 2007 I think I CAN see four solutions... I'll call 1st person A, 2nd person B A = 2, B = 3: A doesn't know at first. If B had 1, he would know. Since he doesn't know, he must have 3. A solves it at second round A = 9, B = 8: analogous to the above. A = 3, B = 4: A doesn't know at first. If B had 2, he would see that A could have only 1 or 3, but with 1, A would have known immediately. Therefor, since B doesn't know, he cannot have 2. A solves it at second round... A = 8, B = 7: analogous to the above. I'm still in the process of finding out what happens if A and B were reversed... Maybe there are even more than 4 solutions then BoilingOil Share this post Link to post Share on other sites

Guest Report post Posted November 9, 2007 I do not understand why 1 and/or 10 are 'easy' to guess. As this is pretty fundamental, I think this is why I missed this one. Share this post Link to post Share on other sites

Guest Report post Posted November 17, 2007 This is very, very confusing. Share this post Link to post Share on other sites

Guest Report post Posted December 1, 2007 What if either student lies? Share this post Link to post Share on other sites

Guest Report post Posted December 4, 2007 I'd say there are 7 solutions: 2-3, 3-4, 4-5, 5-6, 6-7, 7-8, 8-9 Share this post Link to post Share on other sites

Guest Report post Posted December 17, 2007 There are not seven solutions, there are only four: The only possible numbers A could have and be certain of B's number are 2, 3, 8, and 9. If A had any of 4, 5, 6, or 7, then he would not be certain of B's number. Share this post Link to post Share on other sites

Guest Report post Posted December 17, 2007 Surely student B knows after student A's first statement only that A does not have have 1 or 10. Neither does he have 1 or 10 or he would have known the numbers and not responded that neither did he. At this stage either of them could still have 2 or 9 or anything between so when the first student then says he now knows the numbers it would seem that he must have 2 or 9 with the second student having 3 or 8. Unless I completely misunderstand the question this means there are only 2 possibilities when he could know the numbers at that stage - 2 and 3 or 8 and 9. Share this post Link to post Share on other sites

Guest Report post Posted December 20, 2007 Having come across this one again, and thought about it again, I now see that after the second 'don't know' the first student clearly does not have no.1 and the second does not have 1 or 2. Equally first does not have 10 and second does not have 9 or 10. First student however then says he knows the numbers so he must have 2 or 3 or 8 or 9 with second student having 3 or 4 or 7 or 8 to make the pair. Phew!!! seems easy when the mist clears. Share this post Link to post Share on other sites

Guest Report post Posted February 18, 2008 Easy Savoury - Back to the Number Puzzles A teacher thinks of two consecutive numbers between 1 and 10 (Edit: 1 and 10 included). The first student knows one number and the second student knows the second number. The conversation of the students is as follows: First: I do not know your number. Second: Neither do I know your number. First: Now I know. Will you find all 4 solutions? First: 3 Second: 4 First: 2 Second: 3 First: 8 Second: 7 First: 9 Second: 8 Share this post Link to post Share on other sites

Guest Report post Posted October 30, 2008 There is a lot of confusion in this thread, let me clear it up: Since the first child does not know the other's number, he does not have a 1 or a 10. (Or he would know the other student's number would be 2 or 9 respectively. Now the second child knows that the first child does not have a 1 or a 10, he knows that if he has a 2 or a 9 then the other child must have a 3 or an 8, respectively. He says he does not know, so therefore he does not have a 1, 2, 9 or 10. The first child now knows that the second child does not have a 1, 2, 9 or 10, and we know that the first doesn't have 1 or 10. The only way the first child would know the answer at this point is if he has a 2, a 3, an 8 or a 9. Explanation: If child 1 has a 2: Child 2 does not have 1(discovered earlier) or 2(I have it), the only other consecutive number is 3. If child 1 has a 3: Child 2 does not have a 2(discovered earlier) or a 3(I have it), the only other consecutive number is 4. If child 1 has an 8: Child 2 does not have a 9(discovered earlier) or an 8(I have it), the only other consecutive number is 7. If child 1 has a 9: Child 2 does not have a 10(discovered earlier) or a 9(I have it), the only other consecutive number is 8. Therefore the only possible answers are: 2 and 3 3 and 4 7 and 8 8 and 9 Share this post Link to post Share on other sites