Logic problem on sum of possible numbers a given person can have if they had a conversation with another.
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Larry tells Marry and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Marry one of the numbers and then tells Jerry the other number. Then occurs a conversation between Marry and Jerry:
Marry: I don't know your number.
Jerry: I don't know your number either.
Marry: Ah, I now know your number.
Assuming both of them use correct logic, what is the sum of all possible numbers Marry could have?
What I have tried:
Marry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Jerry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Since Marry doesn't know Jerry's number, Marry's number could not have been 1 or 10.
Jerry's number then could not have been either 2 or 9 because then since he already knows Marry's # is not 1, then Marry's number would have been three. Same logic for 9.
So I am left with the possibilities as follows:
Marry's #s: 2, 3, 4, 5, 6, 7, 8, 9
Jerry's #s: 1, 3, 4, 5, 6, 7, 8, 10
I'm stuck here!
Help would be appreciated!
Also, it would also be nice if you would help me on this question(Transferring bases of numbers.) too!
Thanks!
Max0815
logic
$endgroup$
add a comment |
$begingroup$
Larry tells Marry and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Marry one of the numbers and then tells Jerry the other number. Then occurs a conversation between Marry and Jerry:
Marry: I don't know your number.
Jerry: I don't know your number either.
Marry: Ah, I now know your number.
Assuming both of them use correct logic, what is the sum of all possible numbers Marry could have?
What I have tried:
Marry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Jerry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Since Marry doesn't know Jerry's number, Marry's number could not have been 1 or 10.
Jerry's number then could not have been either 2 or 9 because then since he already knows Marry's # is not 1, then Marry's number would have been three. Same logic for 9.
So I am left with the possibilities as follows:
Marry's #s: 2, 3, 4, 5, 6, 7, 8, 9
Jerry's #s: 1, 3, 4, 5, 6, 7, 8, 10
I'm stuck here!
Help would be appreciated!
Also, it would also be nice if you would help me on this question(Transferring bases of numbers.) too!
Thanks!
Max0815
logic
$endgroup$
add a comment |
$begingroup$
Larry tells Marry and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Marry one of the numbers and then tells Jerry the other number. Then occurs a conversation between Marry and Jerry:
Marry: I don't know your number.
Jerry: I don't know your number either.
Marry: Ah, I now know your number.
Assuming both of them use correct logic, what is the sum of all possible numbers Marry could have?
What I have tried:
Marry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Jerry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Since Marry doesn't know Jerry's number, Marry's number could not have been 1 or 10.
Jerry's number then could not have been either 2 or 9 because then since he already knows Marry's # is not 1, then Marry's number would have been three. Same logic for 9.
So I am left with the possibilities as follows:
Marry's #s: 2, 3, 4, 5, 6, 7, 8, 9
Jerry's #s: 1, 3, 4, 5, 6, 7, 8, 10
I'm stuck here!
Help would be appreciated!
Also, it would also be nice if you would help me on this question(Transferring bases of numbers.) too!
Thanks!
Max0815
logic
$endgroup$
Larry tells Marry and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Marry one of the numbers and then tells Jerry the other number. Then occurs a conversation between Marry and Jerry:
Marry: I don't know your number.
Jerry: I don't know your number either.
Marry: Ah, I now know your number.
Assuming both of them use correct logic, what is the sum of all possible numbers Marry could have?
What I have tried:
Marry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Jerry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Since Marry doesn't know Jerry's number, Marry's number could not have been 1 or 10.
Jerry's number then could not have been either 2 or 9 because then since he already knows Marry's # is not 1, then Marry's number would have been three. Same logic for 9.
So I am left with the possibilities as follows:
Marry's #s: 2, 3, 4, 5, 6, 7, 8, 9
Jerry's #s: 1, 3, 4, 5, 6, 7, 8, 10
I'm stuck here!
Help would be appreciated!
Also, it would also be nice if you would help me on this question(Transferring bases of numbers.) too!
Thanks!
Max0815
logic
logic
asked 50 mins ago
Max0815Max0815
49816
49816
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.
So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).
Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).
Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.
So, Mary's number is 2,3,8, or 9. Sum is 22
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Thanks! I got it!
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– Max0815
29 mins ago
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@Max0815 You're welcome! Fun puzzle, thanks! :)
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– Bram28
27 mins ago
add a comment |
$begingroup$
Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.
If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.
New contributor
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If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
$endgroup$
– Max0815
30 mins ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
7 mins ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.
So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).
Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).
Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.
So, Mary's number is 2,3,8, or 9. Sum is 22
$endgroup$
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
29 mins ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
27 mins ago
add a comment |
$begingroup$
Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.
So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).
Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).
Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.
So, Mary's number is 2,3,8, or 9. Sum is 22
$endgroup$
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
29 mins ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
27 mins ago
add a comment |
$begingroup$
Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.
So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).
Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).
Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.
So, Mary's number is 2,3,8, or 9. Sum is 22
$endgroup$
Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.
So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).
Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).
Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.
So, Mary's number is 2,3,8, or 9. Sum is 22
edited 28 mins ago
answered 32 mins ago
Bram28Bram28
61.7k44793
61.7k44793
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
29 mins ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
27 mins ago
add a comment |
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
29 mins ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
27 mins ago
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
29 mins ago
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
29 mins ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
27 mins ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
27 mins ago
add a comment |
$begingroup$
Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.
If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.
New contributor
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$begingroup$
If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
$endgroup$
– Max0815
30 mins ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
7 mins ago
add a comment |
$begingroup$
Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.
If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.
New contributor
$endgroup$
$begingroup$
If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
$endgroup$
– Max0815
30 mins ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
7 mins ago
add a comment |
$begingroup$
Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.
If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.
New contributor
$endgroup$
Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.
If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.
New contributor
edited 5 mins ago
New contributor
answered 36 mins ago
Robert ShoreRobert Shore
715
715
New contributor
New contributor
$begingroup$
If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
$endgroup$
– Max0815
30 mins ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
7 mins ago
add a comment |
$begingroup$
If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
$endgroup$
– Max0815
30 mins ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
7 mins ago
$begingroup$
If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
$endgroup$
– Max0815
30 mins ago
$begingroup$
If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
$endgroup$
– Max0815
30 mins ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
7 mins ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
7 mins ago
add a comment |
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