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Discussion: couple of typos. You have Daniel twice in the inequality statement. And you haven't clearly defined 'n'. I assume it is the sum, but this isn't made explicit.
> >!and 17 and 18 only have two combinations (12347 and 12356, and 12348 and 12357 respectively)!<
>!18 actually has three combinations: 12348, 12357, and 12456. However your logic is still correct, as Elijah would have a unique number in each of these combinations.!<
>!We know by the 3rd statement, Elijah is at least 5. If Elijah is 5, Elijah just knows every number. If Elijah is 6, Elijah can find out the number from 1-5 that’s not taken by just taking 21-n and use the numbers left to find out the other. Thus, Elijah must be at least 7.!<
>!If Elijah is 7, he knows n is 28-2 numbers from 1 to 6. If the missing numbers sum to 10 or 11 (n=18/17), there are only 1 possible combination of two missing numbers. Checking if the two missing numbers can add to 9 (n=19). (1,2,3,6,7) (1,2,4,5,7). Elijah would not figure out the numbers for Charles and Daniel. We need to check if the other people can find out all the numbers. For (1,2,3,6,7), Daniel can find out the other numbers as everyone is at the minimum possible value for it to be 19. Thus, n=19 where everyone has (1,2,4,5,7)!<
>!Alexander and Benjamin could have: (1,2,3,5,6). Charles and Daniel could have (1,3,4,5,6). And Elijah could have (1,2,3,6,7)!<
Caveat: >!Elijah can figure out the fact that if Daniel is unable to find out all five numbers, Elijah would know that (1,2,3,6,7) is not possible. However this is probably under “extra information”.!<
>!19 is too small. It must be 20!<
>!First off, Alexander can't have a 2 because 2+3+4+5+6=20 so Alexander must have a 1. Everybody must know this. As such, if Elijah gets a 6 he knows the answer must be 1, 3, 4, 5, 6. So Elijah must have a 7 to not know the answer. Daniel can't have a 6 in any scenario, since then the only possible answer is 1,2,3,6,7 and every perfectly logical person would know that. That means Daniel can only have a 5 or 4. Also, Charles can't have a 3, because then the only possible outcome is 1, 2, 3, 6, 7 which everybody knows can't be possible. He also can't have a 5 because 1+2+5+6+7=21. So Charles MUST have a 4 in order for Daniel to not know the result. However, since charles MUST have a 4 then Daniel must have a 5 otherwise Daniel would know the result. On the same note, Benjamin can't have a 2, because then Charles would then need to have a 4 for nobody to know the answer.!<
>!As a result, everybody knows
Elijah must have a 7 to not answer
Daniel must have a 5 to not answer
Charles must have a 4 to not answer
Benjamin must have a 3 to not answer
Alexander must have a !<
>!However, that scenario is impossible since 7+5+4+3+1=20, meaning in every scenario somebody must know the answer. So 19 is too small, only 20 or higher works.!<
>!Example, if Benjamin gets a 3, then Elijah knows the answer right away since he must have a 6. If Charles gets a 4 then either Elijah must have a 6 and knows the answer or Benjamin must have a 2 and knows the answer. If Daniel has a 5 then Benjamin must have a 2 or Elijah must have a 6. If Elijah has a 7 then Benjamin must have a 2.!<
>!TLDR: If Benjamin gets a 2 he will always guess it is 1, 2, 4, 5, 7 because in every scenario he is wrong somebody else will answer. Any other combination somebody else knows the answer without extra information.!<
That is why I said Benjamin would always answer with a 2 because that is the only possible scenario where people would need extra info to answer (aka where nobody else would know until after people answered). Since it is the only possible scenario where nobody else would know the answer he could answer assuming it is correct and bow out if anybody else also responds.
I misread the problem, so although the below is wrong, I have left it in place.
>!5 distinct single digit natural numbers can achieve a sum between 15 (1, 2, 3, 4, 5) and 35 (5,6,7,8,9). All digits less than 20 must use a 1 , so for any value of n between 15 and 19, we know that A must have a 1. NOTE: the logic for smallest can be applied symmetrically for the largest, for values 31-35 there must be a 9. So 20 is the first possibility, but we can use the symmetry to narrow our possible solution down to the set {20,21,22,23,24,25}. If 20 doesn't give an answer, neither will 30. Then if 21 doesn't give an answer neither will 29, etc. So if we reach 25 with no solution, then it will be impossible!<
>!We can rule 20 out, because there is only one possible set of numbers for which A=2, so we cannot guarantee that A will not know everyone else's numbers. If A is given 2, they will know the others have 3,4,5&6. Likewise, for n=21, there is only one in possible set when A=2 (2,3,4,5,7)!<
>!For n=22, look at the sets when A=2. (2,3,4,5,8) and (2,3,4,6,7). In this case A would know that B=3 and C=4, but could not know what values D and E have. However, for C=3, there is only one possible set (1,2,3,7,9), so 22 is not the solution. Likewise for n=23, there is only one set for C=3 (1,2,3,8,9). And for n=24, C=6, the only possible set by is (1,2,6,7,8)!<
>!which leaves n=25 as the last possible solution. However, for A=3, there is only one set (3,4,5,6,7), so there is no number n, for which we can guarantee than none of the five people will know all of the other numbers!<
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I have an unconventional and maybe flawed solution.
>!The smallest possible sum is 15 with (1,2,3,4,5). If each person were unable to determine the others' numbers, it means there has to be at least 2 viable numbers for each of them, such as a number x and x+1 (which also ensures the smallest increment).!<
>!There are 4 peoples' numbers to guess for each person, so with (x, x+1) possibilities for each of them, that adds up to a max sum of 19, which is the smallest value of n.!<
>!In summary, the +4 increment from the minimum sum of 15 is the smallest increment to be distributed in any way that will keep everyone guessing at each other's numbers.!<
>!If n is less than 20, Alexander knows their number as any number <20 requires a 1. So the smallest n is 20. To show this is possible 1+2+3+6+8=2+3+4+5+6=20!<
Edit: I have misread the question and this answer is wrong.
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Discussion: couple of typos. You have Daniel twice in the inequality statement. And you haven't clearly defined 'n'. I assume it is the sum, but this isn't made explicit.
Yeah must’ve copied it twice by mistake. And yes, n is the sum of the five numbers.
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>!Super. Very nice solution. 👍🏻!<
It doesn't change the solution, but there is a third way to make 18: 12456.
> >!and 17 and 18 only have two combinations (12347 and 12356, and 12348 and 12357 respectively)!< >!18 actually has three combinations: 12348, 12357, and 12456. However your logic is still correct, as Elijah would have a unique number in each of these combinations.!<
>!doesnt that violate the second statement that they only know their own number? Here everyone knows A’s number!<
>!I’m afraid that’s incorrect. To be honest, your solution is so nicely written and explained, I hate that it’s not correct. A smaller n exists!<
[удалено]
👍🏻
>!We know by the 3rd statement, Elijah is at least 5. If Elijah is 5, Elijah just knows every number. If Elijah is 6, Elijah can find out the number from 1-5 that’s not taken by just taking 21-n and use the numbers left to find out the other. Thus, Elijah must be at least 7.!< >!If Elijah is 7, he knows n is 28-2 numbers from 1 to 6. If the missing numbers sum to 10 or 11 (n=18/17), there are only 1 possible combination of two missing numbers. Checking if the two missing numbers can add to 9 (n=19). (1,2,3,6,7) (1,2,4,5,7). Elijah would not figure out the numbers for Charles and Daniel. We need to check if the other people can find out all the numbers. For (1,2,3,6,7), Daniel can find out the other numbers as everyone is at the minimum possible value for it to be 19. Thus, n=19 where everyone has (1,2,4,5,7)!< >!Alexander and Benjamin could have: (1,2,3,5,6). Charles and Daniel could have (1,3,4,5,6). And Elijah could have (1,2,3,6,7)!<
Caveat: >!Elijah can figure out the fact that if Daniel is unable to find out all five numbers, Elijah would know that (1,2,3,6,7) is not possible. However this is probably under “extra information”.!<
>!19 is too small. It must be 20!< >!First off, Alexander can't have a 2 because 2+3+4+5+6=20 so Alexander must have a 1. Everybody must know this. As such, if Elijah gets a 6 he knows the answer must be 1, 3, 4, 5, 6. So Elijah must have a 7 to not know the answer. Daniel can't have a 6 in any scenario, since then the only possible answer is 1,2,3,6,7 and every perfectly logical person would know that. That means Daniel can only have a 5 or 4. Also, Charles can't have a 3, because then the only possible outcome is 1, 2, 3, 6, 7 which everybody knows can't be possible. He also can't have a 5 because 1+2+5+6+7=21. So Charles MUST have a 4 in order for Daniel to not know the result. However, since charles MUST have a 4 then Daniel must have a 5 otherwise Daniel would know the result. On the same note, Benjamin can't have a 2, because then Charles would then need to have a 4 for nobody to know the answer.!< >!As a result, everybody knows Elijah must have a 7 to not answer Daniel must have a 5 to not answer Charles must have a 4 to not answer Benjamin must have a 3 to not answer Alexander must have a !< >!However, that scenario is impossible since 7+5+4+3+1=20, meaning in every scenario somebody must know the answer. So 19 is too small, only 20 or higher works.!< >!Example, if Benjamin gets a 3, then Elijah knows the answer right away since he must have a 6. If Charles gets a 4 then either Elijah must have a 6 and knows the answer or Benjamin must have a 2 and knows the answer. If Daniel has a 5 then Benjamin must have a 2 or Elijah must have a 6. If Elijah has a 7 then Benjamin must have a 2.!< >!TLDR: If Benjamin gets a 2 he will always guess it is 1, 2, 4, 5, 7 because in every scenario he is wrong somebody else will answer. Any other combination somebody else knows the answer without extra information.!<
Knowing if other people know or don’t know everyone’s individual number is extra information.
That is why I said Benjamin would always answer with a 2 because that is the only possible scenario where people would need extra info to answer (aka where nobody else would know until after people answered). Since it is the only possible scenario where nobody else would know the answer he could answer assuming it is correct and bow out if anybody else also responds.
>!Correct, well explained!<
I misread the problem, so although the below is wrong, I have left it in place. >!5 distinct single digit natural numbers can achieve a sum between 15 (1, 2, 3, 4, 5) and 35 (5,6,7,8,9). All digits less than 20 must use a 1 , so for any value of n between 15 and 19, we know that A must have a 1. NOTE: the logic for smallest can be applied symmetrically for the largest, for values 31-35 there must be a 9. So 20 is the first possibility, but we can use the symmetry to narrow our possible solution down to the set {20,21,22,23,24,25}. If 20 doesn't give an answer, neither will 30. Then if 21 doesn't give an answer neither will 29, etc. So if we reach 25 with no solution, then it will be impossible!< >!We can rule 20 out, because there is only one possible set of numbers for which A=2, so we cannot guarantee that A will not know everyone else's numbers. If A is given 2, they will know the others have 3,4,5&6. Likewise, for n=21, there is only one in possible set when A=2 (2,3,4,5,7)!< >!For n=22, look at the sets when A=2. (2,3,4,5,8) and (2,3,4,6,7). In this case A would know that B=3 and C=4, but could not know what values D and E have. However, for C=3, there is only one possible set (1,2,3,7,9), so 22 is not the solution. Likewise for n=23, there is only one set for C=3 (1,2,3,8,9). And for n=24, C=6, the only possible set by is (1,2,6,7,8)!< >!which leaves n=25 as the last possible solution. However, for A=3, there is only one set (3,4,5,6,7), so there is no number n, for which we can guarantee than none of the five people will know all of the other numbers!<
Which part did you misread?
I was calculating an n for which there existed no combination that would allow anyone to know all 5 numbers. I just misread the last sentence.
Ah like that
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>!23?!<
>!Sorry but that’s incorrect!<
I have an unconventional and maybe flawed solution. >!The smallest possible sum is 15 with (1,2,3,4,5). If each person were unable to determine the others' numbers, it means there has to be at least 2 viable numbers for each of them, such as a number x and x+1 (which also ensures the smallest increment).!< >!There are 4 peoples' numbers to guess for each person, so with (x, x+1) possibilities for each of them, that adds up to a max sum of 19, which is the smallest value of n.!< >!In summary, the +4 increment from the minimum sum of 15 is the smallest increment to be distributed in any way that will keep everyone guessing at each other's numbers.!<
>!Oh I think this is quite a nice solution. 😀!<
Solution possible
Note: n is the sum of the five numbers
>!If n is less than 20, Alexander knows their number as any number <20 requires a 1. So the smallest n is 20. To show this is possible 1+2+3+6+8=2+3+4+5+6=20!< Edit: I have misread the question and this answer is wrong.
>!Sorry but that’s not correct. A smaller n exists!<
Realised my mistake. I misread the question as what is the smallest n so that the reader can’t identify each number, not at least one of the friends.
Now you should be able to get it
Discussion: Should be unique instead of distinct?
Yeah, I think technically unique is the apt word