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Writing differences on a blackboard
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 11:00UTC (8:00pm US/Eastern)Do better than chanceBinary manipulation gameLost In Boston: How do I Get Home?Trying to understand an unusual timestamp encountered on the webThe game of 1036Addition hangmanObtain the “Master of stones” titleMonte Carlo ChessPicking A Number Gamesystematic number removal
$begingroup$
The numbers 25 and 36 are written on a blackboard. At each turn,
a player writes on the blackboard the (positive) difference between two numbers
already on the blackboard, if this number does not already appear on the blackboard. The loser is the player who cannot write a number.
I tried but wasn't able to find any approach to this.
Original source appears to be: Mathematical Circles (Russian Experience), page 58.
logical-deduction calculation-puzzle strategy game
$endgroup$
add a comment |
$begingroup$
The numbers 25 and 36 are written on a blackboard. At each turn,
a player writes on the blackboard the (positive) difference between two numbers
already on the blackboard, if this number does not already appear on the blackboard. The loser is the player who cannot write a number.
I tried but wasn't able to find any approach to this.
Original source appears to be: Mathematical Circles (Russian Experience), page 58.
logical-deduction calculation-puzzle strategy game
$endgroup$
4
$begingroup$
What's the question? "Which player has a winning strategy" maybe?
$endgroup$
– 2012rcampion
Mar 30 at 17:54
1
$begingroup$
Hi and welcome to Puzzling SE! This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted.
$endgroup$
– Akari
Mar 30 at 18:24
2
$begingroup$
P.S. The wording of the above comment are taken from Rubio's answer here This wasn't added in the above comment as it was exceeding the word limit by 42 characters.
$endgroup$
– Akari
Mar 30 at 18:25
1
$begingroup$
(Thanks @Akari. I was able to find what seems to be the original source, and added it. $@!$ all about everything, please be mindful of our attribution requirements here going forward. Thanks for contributing and welcome to Puzzling!)
$endgroup$
– Rubio♦
Mar 30 at 21:28
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
Apr 2 at 18:11
add a comment |
$begingroup$
The numbers 25 and 36 are written on a blackboard. At each turn,
a player writes on the blackboard the (positive) difference between two numbers
already on the blackboard, if this number does not already appear on the blackboard. The loser is the player who cannot write a number.
I tried but wasn't able to find any approach to this.
Original source appears to be: Mathematical Circles (Russian Experience), page 58.
logical-deduction calculation-puzzle strategy game
$endgroup$
The numbers 25 and 36 are written on a blackboard. At each turn,
a player writes on the blackboard the (positive) difference between two numbers
already on the blackboard, if this number does not already appear on the blackboard. The loser is the player who cannot write a number.
I tried but wasn't able to find any approach to this.
Original source appears to be: Mathematical Circles (Russian Experience), page 58.
logical-deduction calculation-puzzle strategy game
logical-deduction calculation-puzzle strategy game
edited Mar 30 at 21:49
Rubio♦
30.5k567188
30.5k567188
asked Mar 30 at 17:03
all about everythingall about everything
361
361
4
$begingroup$
What's the question? "Which player has a winning strategy" maybe?
$endgroup$
– 2012rcampion
Mar 30 at 17:54
1
$begingroup$
Hi and welcome to Puzzling SE! This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted.
$endgroup$
– Akari
Mar 30 at 18:24
2
$begingroup$
P.S. The wording of the above comment are taken from Rubio's answer here This wasn't added in the above comment as it was exceeding the word limit by 42 characters.
$endgroup$
– Akari
Mar 30 at 18:25
1
$begingroup$
(Thanks @Akari. I was able to find what seems to be the original source, and added it. $@!$ all about everything, please be mindful of our attribution requirements here going forward. Thanks for contributing and welcome to Puzzling!)
$endgroup$
– Rubio♦
Mar 30 at 21:28
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
Apr 2 at 18:11
add a comment |
4
$begingroup$
What's the question? "Which player has a winning strategy" maybe?
$endgroup$
– 2012rcampion
Mar 30 at 17:54
1
$begingroup$
Hi and welcome to Puzzling SE! This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted.
$endgroup$
– Akari
Mar 30 at 18:24
2
$begingroup$
P.S. The wording of the above comment are taken from Rubio's answer here This wasn't added in the above comment as it was exceeding the word limit by 42 characters.
$endgroup$
– Akari
Mar 30 at 18:25
1
$begingroup$
(Thanks @Akari. I was able to find what seems to be the original source, and added it. $@!$ all about everything, please be mindful of our attribution requirements here going forward. Thanks for contributing and welcome to Puzzling!)
$endgroup$
– Rubio♦
Mar 30 at 21:28
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
Apr 2 at 18:11
4
4
$begingroup$
What's the question? "Which player has a winning strategy" maybe?
$endgroup$
– 2012rcampion
Mar 30 at 17:54
$begingroup$
What's the question? "Which player has a winning strategy" maybe?
$endgroup$
– 2012rcampion
Mar 30 at 17:54
1
1
$begingroup$
Hi and welcome to Puzzling SE! This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted.
$endgroup$
– Akari
Mar 30 at 18:24
$begingroup$
Hi and welcome to Puzzling SE! This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted.
$endgroup$
– Akari
Mar 30 at 18:24
2
2
$begingroup$
P.S. The wording of the above comment are taken from Rubio's answer here This wasn't added in the above comment as it was exceeding the word limit by 42 characters.
$endgroup$
– Akari
Mar 30 at 18:25
$begingroup$
P.S. The wording of the above comment are taken from Rubio's answer here This wasn't added in the above comment as it was exceeding the word limit by 42 characters.
$endgroup$
– Akari
Mar 30 at 18:25
1
1
$begingroup$
(Thanks @Akari. I was able to find what seems to be the original source, and added it. $@!$ all about everything, please be mindful of our attribution requirements here going forward. Thanks for contributing and welcome to Puzzling!)
$endgroup$
– Rubio♦
Mar 30 at 21:28
$begingroup$
(Thanks @Akari. I was able to find what seems to be the original source, and added it. $@!$ all about everything, please be mindful of our attribution requirements here going forward. Thanks for contributing and welcome to Puzzling!)
$endgroup$
– Rubio♦
Mar 30 at 21:28
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
Apr 2 at 18:11
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
Apr 2 at 18:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The numbers $25$ and $36$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $1$. The sequence is $36$, $25$, $36-25=11$, $25-11=14$, $14-11=3$, $11-3=8$, $8-3=5$, $5-3=2$, $3-2=1$.
Once there is a $1$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $1$ to $36$. This shows that if you have two cooperating players, all number from $1$ to $36$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $1$ must be produced, and then as long as there are missing numbers between $1$ and $36$, there is at least one move available.
This means that regardless of what moves are played, all numbers $1$ to $36$ will appear. We started with $2$ numbers on the board, so the game ends after $34$ moves.
The result is that
The game always ends after $34$ moves, after which the first player cannot move and loses.
$endgroup$
4
$begingroup$
Beat me by a minute, nice one :-)
$endgroup$
– Rand al'Thor
Mar 30 at 18:04
add a comment |
$begingroup$
The key fact is as follows:
The only time the game can end is when the numbers are in the form $k,2k,3k,4k,dots,ak$ for some fixed $a$ and $kgeq1$. This is because, given two numbers $m$ and $n$ on the board, we can always apply repeated subtraction between them to reach their GCD, and then from there to reach every multiple of their GCD up to $textmax(m,n)$.
Therefore this particular game ends when the numbers on the board are
$1,2,3,4,dots,35,36$. This will take a total of 34 moves, since there are two numbers at the start and a new one is written each time.
So the conclusion is
no matter how the game goes, the first player loses, since 34 is even.
$endgroup$
2
$begingroup$
Beat me by about 3 seconds :) Oh, and Jaap beat the two of us as well.
$endgroup$
– Arnaud Mortier
Mar 30 at 18:02
add a comment |
$begingroup$
Answer:
The player who plays first will lose, no matter what choices the player make.
This is because
The Euclidean algorithm tells you that as long as $1$ (the gcd of $25$ and $36$) is not on the board, there are legal ways to continue the process. Now, from the moment when $1$ does appear on the board, no matter how long it took to get there, every positive number between $1$ and $36$ becomes reachable.
Therefore
no matter what path is taken, the amount of steps before the game is over is $34$, one for each integer between $1$ and $36$, excluding the two already on the board at the beginning of the game.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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oldest
votes
$begingroup$
The numbers $25$ and $36$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $1$. The sequence is $36$, $25$, $36-25=11$, $25-11=14$, $14-11=3$, $11-3=8$, $8-3=5$, $5-3=2$, $3-2=1$.
Once there is a $1$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $1$ to $36$. This shows that if you have two cooperating players, all number from $1$ to $36$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $1$ must be produced, and then as long as there are missing numbers between $1$ and $36$, there is at least one move available.
This means that regardless of what moves are played, all numbers $1$ to $36$ will appear. We started with $2$ numbers on the board, so the game ends after $34$ moves.
The result is that
The game always ends after $34$ moves, after which the first player cannot move and loses.
$endgroup$
4
$begingroup$
Beat me by a minute, nice one :-)
$endgroup$
– Rand al'Thor
Mar 30 at 18:04
add a comment |
$begingroup$
The numbers $25$ and $36$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $1$. The sequence is $36$, $25$, $36-25=11$, $25-11=14$, $14-11=3$, $11-3=8$, $8-3=5$, $5-3=2$, $3-2=1$.
Once there is a $1$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $1$ to $36$. This shows that if you have two cooperating players, all number from $1$ to $36$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $1$ must be produced, and then as long as there are missing numbers between $1$ and $36$, there is at least one move available.
This means that regardless of what moves are played, all numbers $1$ to $36$ will appear. We started with $2$ numbers on the board, so the game ends after $34$ moves.
The result is that
The game always ends after $34$ moves, after which the first player cannot move and loses.
$endgroup$
4
$begingroup$
Beat me by a minute, nice one :-)
$endgroup$
– Rand al'Thor
Mar 30 at 18:04
add a comment |
$begingroup$
The numbers $25$ and $36$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $1$. The sequence is $36$, $25$, $36-25=11$, $25-11=14$, $14-11=3$, $11-3=8$, $8-3=5$, $5-3=2$, $3-2=1$.
Once there is a $1$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $1$ to $36$. This shows that if you have two cooperating players, all number from $1$ to $36$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $1$ must be produced, and then as long as there are missing numbers between $1$ and $36$, there is at least one move available.
This means that regardless of what moves are played, all numbers $1$ to $36$ will appear. We started with $2$ numbers on the board, so the game ends after $34$ moves.
The result is that
The game always ends after $34$ moves, after which the first player cannot move and loses.
$endgroup$
The numbers $25$ and $36$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $1$. The sequence is $36$, $25$, $36-25=11$, $25-11=14$, $14-11=3$, $11-3=8$, $8-3=5$, $5-3=2$, $3-2=1$.
Once there is a $1$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $1$ to $36$. This shows that if you have two cooperating players, all number from $1$ to $36$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $1$ must be produced, and then as long as there are missing numbers between $1$ and $36$, there is at least one move available.
This means that regardless of what moves are played, all numbers $1$ to $36$ will appear. We started with $2$ numbers on the board, so the game ends after $34$ moves.
The result is that
The game always ends after $34$ moves, after which the first player cannot move and loses.
edited Mar 30 at 18:06
answered Mar 30 at 17:58
Jaap ScherphuisJaap Scherphuis
16.6k12872
16.6k12872
4
$begingroup$
Beat me by a minute, nice one :-)
$endgroup$
– Rand al'Thor
Mar 30 at 18:04
add a comment |
4
$begingroup$
Beat me by a minute, nice one :-)
$endgroup$
– Rand al'Thor
Mar 30 at 18:04
4
4
$begingroup$
Beat me by a minute, nice one :-)
$endgroup$
– Rand al'Thor
Mar 30 at 18:04
$begingroup$
Beat me by a minute, nice one :-)
$endgroup$
– Rand al'Thor
Mar 30 at 18:04
add a comment |
$begingroup$
The key fact is as follows:
The only time the game can end is when the numbers are in the form $k,2k,3k,4k,dots,ak$ for some fixed $a$ and $kgeq1$. This is because, given two numbers $m$ and $n$ on the board, we can always apply repeated subtraction between them to reach their GCD, and then from there to reach every multiple of their GCD up to $textmax(m,n)$.
Therefore this particular game ends when the numbers on the board are
$1,2,3,4,dots,35,36$. This will take a total of 34 moves, since there are two numbers at the start and a new one is written each time.
So the conclusion is
no matter how the game goes, the first player loses, since 34 is even.
$endgroup$
2
$begingroup$
Beat me by about 3 seconds :) Oh, and Jaap beat the two of us as well.
$endgroup$
– Arnaud Mortier
Mar 30 at 18:02
add a comment |
$begingroup$
The key fact is as follows:
The only time the game can end is when the numbers are in the form $k,2k,3k,4k,dots,ak$ for some fixed $a$ and $kgeq1$. This is because, given two numbers $m$ and $n$ on the board, we can always apply repeated subtraction between them to reach their GCD, and then from there to reach every multiple of their GCD up to $textmax(m,n)$.
Therefore this particular game ends when the numbers on the board are
$1,2,3,4,dots,35,36$. This will take a total of 34 moves, since there are two numbers at the start and a new one is written each time.
So the conclusion is
no matter how the game goes, the first player loses, since 34 is even.
$endgroup$
2
$begingroup$
Beat me by about 3 seconds :) Oh, and Jaap beat the two of us as well.
$endgroup$
– Arnaud Mortier
Mar 30 at 18:02
add a comment |
$begingroup$
The key fact is as follows:
The only time the game can end is when the numbers are in the form $k,2k,3k,4k,dots,ak$ for some fixed $a$ and $kgeq1$. This is because, given two numbers $m$ and $n$ on the board, we can always apply repeated subtraction between them to reach their GCD, and then from there to reach every multiple of their GCD up to $textmax(m,n)$.
Therefore this particular game ends when the numbers on the board are
$1,2,3,4,dots,35,36$. This will take a total of 34 moves, since there are two numbers at the start and a new one is written each time.
So the conclusion is
no matter how the game goes, the first player loses, since 34 is even.
$endgroup$
The key fact is as follows:
The only time the game can end is when the numbers are in the form $k,2k,3k,4k,dots,ak$ for some fixed $a$ and $kgeq1$. This is because, given two numbers $m$ and $n$ on the board, we can always apply repeated subtraction between them to reach their GCD, and then from there to reach every multiple of their GCD up to $textmax(m,n)$.
Therefore this particular game ends when the numbers on the board are
$1,2,3,4,dots,35,36$. This will take a total of 34 moves, since there are two numbers at the start and a new one is written each time.
So the conclusion is
no matter how the game goes, the first player loses, since 34 is even.
answered Mar 30 at 17:59
Rand al'ThorRand al'Thor
71.1k14236472
71.1k14236472
2
$begingroup$
Beat me by about 3 seconds :) Oh, and Jaap beat the two of us as well.
$endgroup$
– Arnaud Mortier
Mar 30 at 18:02
add a comment |
2
$begingroup$
Beat me by about 3 seconds :) Oh, and Jaap beat the two of us as well.
$endgroup$
– Arnaud Mortier
Mar 30 at 18:02
2
2
$begingroup$
Beat me by about 3 seconds :) Oh, and Jaap beat the two of us as well.
$endgroup$
– Arnaud Mortier
Mar 30 at 18:02
$begingroup$
Beat me by about 3 seconds :) Oh, and Jaap beat the two of us as well.
$endgroup$
– Arnaud Mortier
Mar 30 at 18:02
add a comment |
$begingroup$
Answer:
The player who plays first will lose, no matter what choices the player make.
This is because
The Euclidean algorithm tells you that as long as $1$ (the gcd of $25$ and $36$) is not on the board, there are legal ways to continue the process. Now, from the moment when $1$ does appear on the board, no matter how long it took to get there, every positive number between $1$ and $36$ becomes reachable.
Therefore
no matter what path is taken, the amount of steps before the game is over is $34$, one for each integer between $1$ and $36$, excluding the two already on the board at the beginning of the game.
$endgroup$
add a comment |
$begingroup$
Answer:
The player who plays first will lose, no matter what choices the player make.
This is because
The Euclidean algorithm tells you that as long as $1$ (the gcd of $25$ and $36$) is not on the board, there are legal ways to continue the process. Now, from the moment when $1$ does appear on the board, no matter how long it took to get there, every positive number between $1$ and $36$ becomes reachable.
Therefore
no matter what path is taken, the amount of steps before the game is over is $34$, one for each integer between $1$ and $36$, excluding the two already on the board at the beginning of the game.
$endgroup$
add a comment |
$begingroup$
Answer:
The player who plays first will lose, no matter what choices the player make.
This is because
The Euclidean algorithm tells you that as long as $1$ (the gcd of $25$ and $36$) is not on the board, there are legal ways to continue the process. Now, from the moment when $1$ does appear on the board, no matter how long it took to get there, every positive number between $1$ and $36$ becomes reachable.
Therefore
no matter what path is taken, the amount of steps before the game is over is $34$, one for each integer between $1$ and $36$, excluding the two already on the board at the beginning of the game.
$endgroup$
Answer:
The player who plays first will lose, no matter what choices the player make.
This is because
The Euclidean algorithm tells you that as long as $1$ (the gcd of $25$ and $36$) is not on the board, there are legal ways to continue the process. Now, from the moment when $1$ does appear on the board, no matter how long it took to get there, every positive number between $1$ and $36$ becomes reachable.
Therefore
no matter what path is taken, the amount of steps before the game is over is $34$, one for each integer between $1$ and $36$, excluding the two already on the board at the beginning of the game.
answered Mar 30 at 18:01
Arnaud MortierArnaud Mortier
2,176827
2,176827
add a comment |
add a comment |
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4
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What's the question? "Which player has a winning strategy" maybe?
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– 2012rcampion
Mar 30 at 17:54
1
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Hi and welcome to Puzzling SE! This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted.
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– Akari
Mar 30 at 18:24
2
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P.S. The wording of the above comment are taken from Rubio's answer here This wasn't added in the above comment as it was exceeding the word limit by 42 characters.
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– Akari
Mar 30 at 18:25
1
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(Thanks @Akari. I was able to find what seems to be the original source, and added it. $@!$ all about everything, please be mindful of our attribution requirements here going forward. Thanks for contributing and welcome to Puzzling!)
$endgroup$
– Rubio♦
Mar 30 at 21:28
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Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
Apr 2 at 18:11