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
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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 1 hour ago
Max0815Max0815
51116
51116
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
1 hour ago
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@Max0815 You're welcome! Fun puzzle, thanks! :)
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– Bram28
1 hour ago
add a comment |
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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)
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– Max0815
1 hour ago
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Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
1 hour 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
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$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
1 hour ago
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@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
1 hour 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
1 hour ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
1 hour 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 1 hour ago
answered 1 hour ago
Bram28Bram28
61.7k44793
61.7k44793
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
1 hour ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
1 hour ago
add a comment |
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
1 hour ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
1 hour ago
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
1 hour ago
$begingroup$
Thanks! I got it!
$endgroup$
– Max0815
1 hour ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
1 hour ago
$begingroup$
@Max0815 You're welcome! Fun puzzle, thanks! :)
$endgroup$
– Bram28
1 hour 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
1 hour ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
1 hour 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
1 hour ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
1 hour 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 59 mins ago
New contributor
answered 1 hour 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
1 hour ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
1 hour 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
1 hour ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
1 hour 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
1 hour 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
1 hour ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
1 hour ago
$begingroup$
Yes, that's correct. Sorry for the error.
$endgroup$
– Robert Shore
1 hour ago
add a comment |
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