Mixing
After swapping the positions of the hour and the minute hand, when will a clock still give a valid time?
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At 12 o'clock, the hour hand and minute hand of the clock can be swapped, and the clock still gives the same time, but at 6 o'clock, it can not be swapped. So in what cases when we swap the hour and the minute hand position does a clock still give a valid time?
number-theory puzzle
edited Oct 30 '11 at 22:59
J. M. is not a mathematician
61.3k5152290
asked Aug 24 '11 at 2:48
KevinBuiKevinBui
5641814
$endgroup$
|
show 3 more comments
$begingroup$
At 12 o'clock, the hour hand and minute hand of the clock can be swapped, and the clock still gives the same time, but at 6 o'clock, it can not be swapped. So in what cases when we swap the hour and the minute hand position does a clock still give a valid time?
number-theory puzzle
edited Oct 30 '11 at 22:59
J. M. is not a mathematician
61.3k5152290
asked Aug 24 '11 at 2:48
KevinBuiKevinBui
5641814
$endgroup$
1
$begingroup$
Enumeration works: 12:00, 1:05, 2:10, 3:15... the pattern should be clear. On the other hand, depending on the analog clock's mechanism, the latter times might no longer have the hands to be swappable.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 2:57
2
$begingroup$
At 1:05, the minute hand's on the 1, the hour hand's a little past it; if you swap, the hour hand's exactly on the 1, the minute hand's a little past it, and that's not a valid position.
$endgroup$
– Gerry Myerson
Aug 24 '11 at 3:10
$begingroup$
For the clock I have, that happens on "later times" (e.g. 9:45). I guess it does depend on the clock.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:20
1
$begingroup$
@Gerry Myerson - Hopefully J.M. meant the hands-pointing-in-the-same-direction positions and just rounded (i.e. every 12/11 hours). This is correct but incomplete, as your answer shows (every 13th solution of yours is a same-direction solution).
$endgroup$
– Rex Kerr
Aug 24 '11 at 3:22
$begingroup$
Yes, @Rex has it. I wasn't thinking of the "non-coincident hands" solutions, and I'm not in the mood for the needed arithmetic... hence I left it as a comment.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:38
|
show 3 more comments
$begingroup$
At 12 o'clock, the hour hand and minute hand of the clock can be swapped, and the clock still gives the same time, but at 6 o'clock, it can not be swapped. So in what cases when we swap the hour and the minute hand position does a clock still give a valid time?
number-theory puzzle
edited Oct 30 '11 at 22:59
J. M. is not a mathematician
61.3k5152290
asked Aug 24 '11 at 2:48
KevinBuiKevinBui
5641814
$endgroup$
At 12 o'clock, the hour hand and minute hand of the clock can be swapped, and the clock still gives the same time, but at 6 o'clock, it can not be swapped. So in what cases when we swap the hour and the minute hand position does a clock still give a valid time?
number-theory puzzle
number-theory puzzle
edited Oct 30 '11 at 22:59
J. M. is not a mathematician
61.3k5152290
asked Aug 24 '11 at 2:48
KevinBuiKevinBui
5641814
edited Oct 30 '11 at 22:59
J. M. is not a mathematician
61.3k5152290
asked Aug 24 '11 at 2:48
KevinBuiKevinBui
5641814
edited Oct 30 '11 at 22:59
J. M. is not a mathematician
61.3k5152290
edited Oct 30 '11 at 22:59
J. M. is not a mathematician
61.3k5152290
edited Oct 30 '11 at 22:59
J. M. is not a mathematician
61.3k5152290
61.3k5152290
asked Aug 24 '11 at 2:48
KevinBuiKevinBui
5641814
asked Aug 24 '11 at 2:48
KevinBuiKevinBui
5641814
asked Aug 24 '11 at 2:48
KevinBuiKevinBui
5641814
5641814
1
$begingroup$
Enumeration works: 12:00, 1:05, 2:10, 3:15... the pattern should be clear. On the other hand, depending on the analog clock's mechanism, the latter times might no longer have the hands to be swappable.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 2:57
2
$begingroup$
At 1:05, the minute hand's on the 1, the hour hand's a little past it; if you swap, the hour hand's exactly on the 1, the minute hand's a little past it, and that's not a valid position.
$endgroup$
– Gerry Myerson
Aug 24 '11 at 3:10
$begingroup$
For the clock I have, that happens on "later times" (e.g. 9:45). I guess it does depend on the clock.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:20
1
$begingroup$
@Gerry Myerson - Hopefully J.M. meant the hands-pointing-in-the-same-direction positions and just rounded (i.e. every 12/11 hours). This is correct but incomplete, as your answer shows (every 13th solution of yours is a same-direction solution).
$endgroup$
– Rex Kerr
Aug 24 '11 at 3:22
$begingroup$
Yes, @Rex has it. I wasn't thinking of the "non-coincident hands" solutions, and I'm not in the mood for the needed arithmetic... hence I left it as a comment.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:38
|
show 3 more comments
1
$begingroup$
Enumeration works: 12:00, 1:05, 2:10, 3:15... the pattern should be clear. On the other hand, depending on the analog clock's mechanism, the latter times might no longer have the hands to be swappable.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 2:57
2
$begingroup$
At 1:05, the minute hand's on the 1, the hour hand's a little past it; if you swap, the hour hand's exactly on the 1, the minute hand's a little past it, and that's not a valid position.
$endgroup$
– Gerry Myerson
Aug 24 '11 at 3:10
$begingroup$
For the clock I have, that happens on "later times" (e.g. 9:45). I guess it does depend on the clock.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:20
1
$begingroup$
@Gerry Myerson - Hopefully J.M. meant the hands-pointing-in-the-same-direction positions and just rounded (i.e. every 12/11 hours). This is correct but incomplete, as your answer shows (every 13th solution of yours is a same-direction solution).
$endgroup$
– Rex Kerr
Aug 24 '11 at 3:22
$begingroup$
Yes, @Rex has it. I wasn't thinking of the "non-coincident hands" solutions, and I'm not in the mood for the needed arithmetic... hence I left it as a comment.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:38
1
1
$begingroup$
Enumeration works: 12:00, 1:05, 2:10, 3:15... the pattern should be clear. On the other hand, depending on the analog clock's mechanism, the latter times might no longer have the hands to be swappable.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 2:57
$begingroup$
Enumeration works: 12:00, 1:05, 2:10, 3:15... the pattern should be clear. On the other hand, depending on the analog clock's mechanism, the latter times might no longer have the hands to be swappable.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 2:57
2
2
$begingroup$
At 1:05, the minute hand's on the 1, the hour hand's a little past it; if you swap, the hour hand's exactly on the 1, the minute hand's a little past it, and that's not a valid position.
$endgroup$
– Gerry Myerson
Aug 24 '11 at 3:10
$begingroup$
At 1:05, the minute hand's on the 1, the hour hand's a little past it; if you swap, the hour hand's exactly on the 1, the minute hand's a little past it, and that's not a valid position.
$endgroup$
– Gerry Myerson
Aug 24 '11 at 3:10
$begingroup$
For the clock I have, that happens on "later times" (e.g. 9:45). I guess it does depend on the clock.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:20
$begingroup$
For the clock I have, that happens on "later times" (e.g. 9:45). I guess it does depend on the clock.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:20
1
1
$begingroup$
@Gerry Myerson - Hopefully J.M. meant the hands-pointing-in-the-same-direction positions and just rounded (i.e. every 12/11 hours). This is correct but incomplete, as your answer shows (every 13th solution of yours is a same-direction solution).
$endgroup$
– Rex Kerr
Aug 24 '11 at 3:22
$begingroup$
@Gerry Myerson - Hopefully J.M. meant the hands-pointing-in-the-same-direction positions and just rounded (i.e. every 12/11 hours). This is correct but incomplete, as your answer shows (every 13th solution of yours is a same-direction solution).
$endgroup$
– Rex Kerr
Aug 24 '11 at 3:22
$begingroup$
Yes, @Rex has it. I wasn't thinking of the "non-coincident hands" solutions, and I'm not in the mood for the needed arithmetic... hence I left it as a comment.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:38
$begingroup$
Yes, @Rex has it. I wasn't thinking of the "non-coincident hands" solutions, and I'm not in the mood for the needed arithmetic... hence I left it as a comment.
$endgroup$
– J. M. is not a mathematician
Aug 24 '11 at 3:38
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Let $x$ be the position of the hour hand, as measured in degrees clockwise from 12 o'clock. So, for example, at 1 o'clock, $x=30$. Let $y$ be the position of the minute hand; then $yequiv12xpmod{360}$, because the minute hand spins 12 times as fast as the hour hand. In order for $(y,x)$ to be a valid pair of positions for (hour hand, minute hand), we must also have $xequiv12ypmod{360}$. Putting these together, we get $xequiv144xpmod{360}$, which is $143xequiv0pmod{360}$, which has the solutions $x=0,360/143,720/143,1080/143,dots$.
$x=360/143$ is $12times360/143=30.20979dots$ minutes past 12 o'clock; 30 minutes, 12 and four-sevenths seconds after 12 o'clock. And then any integer multiple of that will do.
EDIT: As Henry points out in a comment, the 2nd paragraph contains an error. $x=360/143$ is $12times360/143=30.20979dots$ degrees past 12 o'clock, but it is $2times360/143$ minutes past 12, which is (as Henry says) 5 minutes, $2{14over143}$ seconds after 12.
answered Aug 24 '11 at 3:08
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
$begingroup$
How many solution in this case?
$endgroup$
– KevinBui
Aug 24 '11 at 3:50
$begingroup$
@DKahnh - 143. 360*143/143 = 360 ≡ 0 (mod 360).
$endgroup$
– Rex Kerr
Aug 24 '11 at 4:14
7
$begingroup$
The situation can be viewed geometrically as follows. The handles' position can be specified giving the angles from 12 o' clock. Thus the clock space is the torus $T=S^1times S^1$ and the valid time positions describe the curve $phi(theta)=(theta, 12theta)$ in $T$. Swapping the handles gives an automorphism of the clock space $T$ which transforms the curve $phi(theta)$ into the curve $psi(theta)=(12theta,theta)$. The solutions of the problem are the intersections of the two curves.
$endgroup$
– Andrea Mori
Aug 24 '11 at 8:46
1
$begingroup$
I don't understand the second paragraph. $frac{12}{143}$ hours or $frac{720}{143}$ minutes or $frac{43200}{143}$ seconds is $5$ minutes and $2tfrac{14}{143}$ seconds for the first swappable position after 12 o'clock, and then any integer multiple will do. Thirteen times this ($1$ hour, $5$ minutes and $27tfrac{3}{11}$ seconds) and you get the first matching position.
$endgroup$
– Henry
Aug 24 '11 at 10:30
$begingroup$
@Henry, you're right: at $x=360/143$, the minute hand is at 30.20979... degrees past 12 o'clock; divide by 6 to get the number of minutes past 12 o'clock. I will edit.
$endgroup$
– Gerry Myerson
Aug 25 '11 at 4:47
add a comment |
$begingroup$
A visual proof. Every intersection point of the black grid is a solution.
Explanation. On the $x$-axis the position of the hours clock hand, on the $y$ axis the position of the minutes clock hand. When the short hand goes between hour n to n+1, the long hand makes a complete turn (from 0 to 12). If you exchange the hands, you exchange $x$ and $y$ coordinates... so you look for intersection of the graph with its simmetry with respect to the diagonal of the square domain.
answered Jan 11 '17 at 5:45
Emanuele PaoliniEmanuele Paolini
17.9k22052
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$begingroup$
Beautiful! From this picture, you can easily see that there are 12*12 points of intersection because there are twelve lines crossing twelve lines. This means that there are 144-1 = 143 distinct times where swapping the hour and minute hand give a valid time (because we don't want to count midnight twice).
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– user326210
Jan 18 '18 at 5:47
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x$ be the position of the hour hand, as measured in degrees clockwise from 12 o'clock. So, for example, at 1 o'clock, $x=30$. Let $y$ be the position of the minute hand; then $yequiv12xpmod{360}$, because the minute hand spins 12 times as fast as the hour hand. In order for $(y,x)$ to be a valid pair of positions for (hour hand, minute hand), we must also have $xequiv12ypmod{360}$. Putting these together, we get $xequiv144xpmod{360}$, which is $143xequiv0pmod{360}$, which has the solutions $x=0,360/143,720/143,1080/143,dots$.
$x=360/143$ is $12times360/143=30.20979dots$ minutes past 12 o'clock; 30 minutes, 12 and four-sevenths seconds after 12 o'clock. And then any integer multiple of that will do.
EDIT: As Henry points out in a comment, the 2nd paragraph contains an error. $x=360/143$ is $12times360/143=30.20979dots$ degrees past 12 o'clock, but it is $2times360/143$ minutes past 12, which is (as Henry says) 5 minutes, $2{14over143}$ seconds after 12.
answered Aug 24 '11 at 3:08
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
$begingroup$
How many solution in this case?
$endgroup$
– KevinBui
Aug 24 '11 at 3:50
$begingroup$
@DKahnh - 143. 360*143/143 = 360 ≡ 0 (mod 360).
$endgroup$
– Rex Kerr
Aug 24 '11 at 4:14
7
$begingroup$
The situation can be viewed geometrically as follows. The handles' position can be specified giving the angles from 12 o' clock. Thus the clock space is the torus $T=S^1times S^1$ and the valid time positions describe the curve $phi(theta)=(theta, 12theta)$ in $T$. Swapping the handles gives an automorphism of the clock space $T$ which transforms the curve $phi(theta)$ into the curve $psi(theta)=(12theta,theta)$. The solutions of the problem are the intersections of the two curves.
$endgroup$
– Andrea Mori
Aug 24 '11 at 8:46
1
$begingroup$
I don't understand the second paragraph. $frac{12}{143}$ hours or $frac{720}{143}$ minutes or $frac{43200}{143}$ seconds is $5$ minutes and $2tfrac{14}{143}$ seconds for the first swappable position after 12 o'clock, and then any integer multiple will do. Thirteen times this ($1$ hour, $5$ minutes and $27tfrac{3}{11}$ seconds) and you get the first matching position.
$endgroup$
– Henry
Aug 24 '11 at 10:30
$begingroup$
@Henry, you're right: at $x=360/143$, the minute hand is at 30.20979... degrees past 12 o'clock; divide by 6 to get the number of minutes past 12 o'clock. I will edit.
$endgroup$
– Gerry Myerson
Aug 25 '11 at 4:47
add a comment |
$begingroup$
Let $x$ be the position of the hour hand, as measured in degrees clockwise from 12 o'clock. So, for example, at 1 o'clock, $x=30$. Let $y$ be the position of the minute hand; then $yequiv12xpmod{360}$, because the minute hand spins 12 times as fast as the hour hand. In order for $(y,x)$ to be a valid pair of positions for (hour hand, minute hand), we must also have $xequiv12ypmod{360}$. Putting these together, we get $xequiv144xpmod{360}$, which is $143xequiv0pmod{360}$, which has the solutions $x=0,360/143,720/143,1080/143,dots$.
$x=360/143$ is $12times360/143=30.20979dots$ minutes past 12 o'clock; 30 minutes, 12 and four-sevenths seconds after 12 o'clock. And then any integer multiple of that will do.
EDIT: As Henry points out in a comment, the 2nd paragraph contains an error. $x=360/143$ is $12times360/143=30.20979dots$ degrees past 12 o'clock, but it is $2times360/143$ minutes past 12, which is (as Henry says) 5 minutes, $2{14over143}$ seconds after 12.
answered Aug 24 '11 at 3:08
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
$begingroup$
How many solution in this case?
$endgroup$
– KevinBui
Aug 24 '11 at 3:50
$begingroup$
@DKahnh - 143. 360*143/143 = 360 ≡ 0 (mod 360).
$endgroup$
– Rex Kerr
Aug 24 '11 at 4:14
7
$begingroup$
The situation can be viewed geometrically as follows. The handles' position can be specified giving the angles from 12 o' clock. Thus the clock space is the torus $T=S^1times S^1$ and the valid time positions describe the curve $phi(theta)=(theta, 12theta)$ in $T$. Swapping the handles gives an automorphism of the clock space $T$ which transforms the curve $phi(theta)$ into the curve $psi(theta)=(12theta,theta)$. The solutions of the problem are the intersections of the two curves.
$endgroup$
– Andrea Mori
Aug 24 '11 at 8:46
1
$begingroup$
I don't understand the second paragraph. $frac{12}{143}$ hours or $frac{720}{143}$ minutes or $frac{43200}{143}$ seconds is $5$ minutes and $2tfrac{14}{143}$ seconds for the first swappable position after 12 o'clock, and then any integer multiple will do. Thirteen times this ($1$ hour, $5$ minutes and $27tfrac{3}{11}$ seconds) and you get the first matching position.
$endgroup$
– Henry
Aug 24 '11 at 10:30
$begingroup$
@Henry, you're right: at $x=360/143$, the minute hand is at 30.20979... degrees past 12 o'clock; divide by 6 to get the number of minutes past 12 o'clock. I will edit.
$endgroup$
– Gerry Myerson
Aug 25 '11 at 4:47
add a comment |
$begingroup$
Let $x$ be the position of the hour hand, as measured in degrees clockwise from 12 o'clock. So, for example, at 1 o'clock, $x=30$. Let $y$ be the position of the minute hand; then $yequiv12xpmod{360}$, because the minute hand spins 12 times as fast as the hour hand. In order for $(y,x)$ to be a valid pair of positions for (hour hand, minute hand), we must also have $xequiv12ypmod{360}$. Putting these together, we get $xequiv144xpmod{360}$, which is $143xequiv0pmod{360}$, which has the solutions $x=0,360/143,720/143,1080/143,dots$.
$x=360/143$ is $12times360/143=30.20979dots$ minutes past 12 o'clock; 30 minutes, 12 and four-sevenths seconds after 12 o'clock. And then any integer multiple of that will do.
EDIT: As Henry points out in a comment, the 2nd paragraph contains an error. $x=360/143$ is $12times360/143=30.20979dots$ degrees past 12 o'clock, but it is $2times360/143$ minutes past 12, which is (as Henry says) 5 minutes, $2{14over143}$ seconds after 12.
answered Aug 24 '11 at 3:08
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
Let $x$ be the position of the hour hand, as measured in degrees clockwise from 12 o'clock. So, for example, at 1 o'clock, $x=30$. Let $y$ be the position of the minute hand; then $yequiv12xpmod{360}$, because the minute hand spins 12 times as fast as the hour hand. In order for $(y,x)$ to be a valid pair of positions for (hour hand, minute hand), we must also have $xequiv12ypmod{360}$. Putting these together, we get $xequiv144xpmod{360}$, which is $143xequiv0pmod{360}$, which has the solutions $x=0,360/143,720/143,1080/143,dots$.
$x=360/143$ is $12times360/143=30.20979dots$ minutes past 12 o'clock; 30 minutes, 12 and four-sevenths seconds after 12 o'clock. And then any integer multiple of that will do.
EDIT: As Henry points out in a comment, the 2nd paragraph contains an error. $x=360/143$ is $12times360/143=30.20979dots$ degrees past 12 o'clock, but it is $2times360/143$ minutes past 12, which is (as Henry says) 5 minutes, $2{14over143}$ seconds after 12.
answered Aug 24 '11 at 3:08
Gerry MyersonGerry Myerson
147k8149302
edited Jul 18 '17 at 22:00
answered Aug 24 '11 at 3:08
Gerry MyersonGerry Myerson
147k8149302
answered Aug 24 '11 at 3:08
Gerry MyersonGerry Myerson
147k8149302
answered Aug 24 '11 at 3:08
Gerry MyersonGerry Myerson
147k8149302
147k8149302
$begingroup$
How many solution in this case?
$endgroup$
– KevinBui
Aug 24 '11 at 3:50
$begingroup$
@DKahnh - 143. 360*143/143 = 360 ≡ 0 (mod 360).
$endgroup$
– Rex Kerr
Aug 24 '11 at 4:14
7
$begingroup$
The situation can be viewed geometrically as follows. The handles' position can be specified giving the angles from 12 o' clock. Thus the clock space is the torus $T=S^1times S^1$ and the valid time positions describe the curve $phi(theta)=(theta, 12theta)$ in $T$. Swapping the handles gives an automorphism of the clock space $T$ which transforms the curve $phi(theta)$ into the curve $psi(theta)=(12theta,theta)$. The solutions of the problem are the intersections of the two curves.
$endgroup$
– Andrea Mori
Aug 24 '11 at 8:46
1
$begingroup$
I don't understand the second paragraph. $frac{12}{143}$ hours or $frac{720}{143}$ minutes or $frac{43200}{143}$ seconds is $5$ minutes and $2tfrac{14}{143}$ seconds for the first swappable position after 12 o'clock, and then any integer multiple will do. Thirteen times this ($1$ hour, $5$ minutes and $27tfrac{3}{11}$ seconds) and you get the first matching position.
$endgroup$
– Henry
Aug 24 '11 at 10:30
$begingroup$
@Henry, you're right: at $x=360/143$, the minute hand is at 30.20979... degrees past 12 o'clock; divide by 6 to get the number of minutes past 12 o'clock. I will edit.
$endgroup$
– Gerry Myerson
Aug 25 '11 at 4:47
add a comment |
$begingroup$
How many solution in this case?
$endgroup$
– KevinBui
Aug 24 '11 at 3:50
$begingroup$
@DKahnh - 143. 360*143/143 = 360 ≡ 0 (mod 360).
$endgroup$
– Rex Kerr
Aug 24 '11 at 4:14
7
$begingroup$
The situation can be viewed geometrically as follows. The handles' position can be specified giving the angles from 12 o' clock. Thus the clock space is the torus $T=S^1times S^1$ and the valid time positions describe the curve $phi(theta)=(theta, 12theta)$ in $T$. Swapping the handles gives an automorphism of the clock space $T$ which transforms the curve $phi(theta)$ into the curve $psi(theta)=(12theta,theta)$. The solutions of the problem are the intersections of the two curves.
$endgroup$
– Andrea Mori
Aug 24 '11 at 8:46
1
$begingroup$
I don't understand the second paragraph. $frac{12}{143}$ hours or $frac{720}{143}$ minutes or $frac{43200}{143}$ seconds is $5$ minutes and $2tfrac{14}{143}$ seconds for the first swappable position after 12 o'clock, and then any integer multiple will do. Thirteen times this ($1$ hour, $5$ minutes and $27tfrac{3}{11}$ seconds) and you get the first matching position.
$endgroup$
– Henry
Aug 24 '11 at 10:30
$begingroup$
@Henry, you're right: at $x=360/143$, the minute hand is at 30.20979... degrees past 12 o'clock; divide by 6 to get the number of minutes past 12 o'clock. I will edit.
$endgroup$
– Gerry Myerson
Aug 25 '11 at 4:47
$begingroup$
How many solution in this case?
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– KevinBui
Aug 24 '11 at 3:50
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How many solution in this case?
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– KevinBui
Aug 24 '11 at 3:50
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@DKahnh - 143. 360*143/143 = 360 ≡ 0 (mod 360).
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– Rex Kerr
Aug 24 '11 at 4:14
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@DKahnh - 143. 360*143/143 = 360 ≡ 0 (mod 360).
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– Rex Kerr
Aug 24 '11 at 4:14
7
7
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The situation can be viewed geometrically as follows. The handles' position can be specified giving the angles from 12 o' clock. Thus the clock space is the torus $T=S^1times S^1$ and the valid time positions describe the curve $phi(theta)=(theta, 12theta)$ in $T$. Swapping the handles gives an automorphism of the clock space $T$ which transforms the curve $phi(theta)$ into the curve $psi(theta)=(12theta,theta)$. The solutions of the problem are the intersections of the two curves.
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– Andrea Mori
Aug 24 '11 at 8:46
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The situation can be viewed geometrically as follows. The handles' position can be specified giving the angles from 12 o' clock. Thus the clock space is the torus $T=S^1times S^1$ and the valid time positions describe the curve $phi(theta)=(theta, 12theta)$ in $T$. Swapping the handles gives an automorphism of the clock space $T$ which transforms the curve $phi(theta)$ into the curve $psi(theta)=(12theta,theta)$. The solutions of the problem are the intersections of the two curves.
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– Andrea Mori
Aug 24 '11 at 8:46
1
1
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I don't understand the second paragraph. $frac{12}{143}$ hours or $frac{720}{143}$ minutes or $frac{43200}{143}$ seconds is $5$ minutes and $2tfrac{14}{143}$ seconds for the first swappable position after 12 o'clock, and then any integer multiple will do. Thirteen times this ($1$ hour, $5$ minutes and $27tfrac{3}{11}$ seconds) and you get the first matching position.
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– Henry
Aug 24 '11 at 10:30
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I don't understand the second paragraph. $frac{12}{143}$ hours or $frac{720}{143}$ minutes or $frac{43200}{143}$ seconds is $5$ minutes and $2tfrac{14}{143}$ seconds for the first swappable position after 12 o'clock, and then any integer multiple will do. Thirteen times this ($1$ hour, $5$ minutes and $27tfrac{3}{11}$ seconds) and you get the first matching position.
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– Henry
Aug 24 '11 at 10:30
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@Henry, you're right: at $x=360/143$, the minute hand is at 30.20979... degrees past 12 o'clock; divide by 6 to get the number of minutes past 12 o'clock. I will edit.
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– Gerry Myerson
Aug 25 '11 at 4:47
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@Henry, you're right: at $x=360/143$, the minute hand is at 30.20979... degrees past 12 o'clock; divide by 6 to get the number of minutes past 12 o'clock. I will edit.
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– Gerry Myerson
Aug 25 '11 at 4:47
add a comment |
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A visual proof. Every intersection point of the black grid is a solution.
Explanation. On the $x$-axis the position of the hours clock hand, on the $y$ axis the position of the minutes clock hand. When the short hand goes between hour n to n+1, the long hand makes a complete turn (from 0 to 12). If you exchange the hands, you exchange $x$ and $y$ coordinates... so you look for intersection of the graph with its simmetry with respect to the diagonal of the square domain.
answered Jan 11 '17 at 5:45
Emanuele PaoliniEmanuele Paolini
17.9k22052
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Beautiful! From this picture, you can easily see that there are 12*12 points of intersection because there are twelve lines crossing twelve lines. This means that there are 144-1 = 143 distinct times where swapping the hour and minute hand give a valid time (because we don't want to count midnight twice).
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– user326210
Jan 18 '18 at 5:47
add a comment |
$begingroup$
A visual proof. Every intersection point of the black grid is a solution.
Explanation. On the $x$-axis the position of the hours clock hand, on the $y$ axis the position of the minutes clock hand. When the short hand goes between hour n to n+1, the long hand makes a complete turn (from 0 to 12). If you exchange the hands, you exchange $x$ and $y$ coordinates... so you look for intersection of the graph with its simmetry with respect to the diagonal of the square domain.
answered Jan 11 '17 at 5:45
Emanuele PaoliniEmanuele Paolini
17.9k22052
$endgroup$
$begingroup$
Beautiful! From this picture, you can easily see that there are 12*12 points of intersection because there are twelve lines crossing twelve lines. This means that there are 144-1 = 143 distinct times where swapping the hour and minute hand give a valid time (because we don't want to count midnight twice).
$endgroup$
– user326210
Jan 18 '18 at 5:47
add a comment |
$begingroup$
A visual proof. Every intersection point of the black grid is a solution.
Explanation. On the $x$-axis the position of the hours clock hand, on the $y$ axis the position of the minutes clock hand. When the short hand goes between hour n to n+1, the long hand makes a complete turn (from 0 to 12). If you exchange the hands, you exchange $x$ and $y$ coordinates... so you look for intersection of the graph with its simmetry with respect to the diagonal of the square domain.
answered Jan 11 '17 at 5:45
Emanuele PaoliniEmanuele Paolini
17.9k22052
$endgroup$
A visual proof. Every intersection point of the black grid is a solution.
Explanation. On the $x$-axis the position of the hours clock hand, on the $y$ axis the position of the minutes clock hand. When the short hand goes between hour n to n+1, the long hand makes a complete turn (from 0 to 12). If you exchange the hands, you exchange $x$ and $y$ coordinates... so you look for intersection of the graph with its simmetry with respect to the diagonal of the square domain.
answered Jan 11 '17 at 5:45
Emanuele PaoliniEmanuele Paolini
17.9k22052
edited Jun 13 '17 at 15:24
answered Jan 11 '17 at 5:45
Emanuele PaoliniEmanuele Paolini
17.9k22052
answered Jan 11 '17 at 5:45
Emanuele PaoliniEmanuele Paolini
17.9k22052
answered Jan 11 '17 at 5:45
Emanuele PaoliniEmanuele Paolini
17.9k22052
17.9k22052
$begingroup$
Beautiful! From this picture, you can easily see that there are 12*12 points of intersection because there are twelve lines crossing twelve lines. This means that there are 144-1 = 143 distinct times where swapping the hour and minute hand give a valid time (because we don't want to count midnight twice).
$endgroup$
– user326210
Jan 18 '18 at 5:47
add a comment |
$begingroup$
Beautiful! From this picture, you can easily see that there are 12*12 points of intersection because there are twelve lines crossing twelve lines. This means that there are 144-1 = 143 distinct times where swapping the hour and minute hand give a valid time (because we don't want to count midnight twice).
$endgroup$
– user326210
Jan 18 '18 at 5:47
$begingroup$
Beautiful! From this picture, you can easily see that there are 12*12 points of intersection because there are twelve lines crossing twelve lines. This means that there are 144-1 = 143 distinct times where swapping the hour and minute hand give a valid time (because we don't want to count midnight twice).
$endgroup$
– user326210
Jan 18 '18 at 5:47
$begingroup$
Beautiful! From this picture, you can easily see that there are 12*12 points of intersection because there are twelve lines crossing twelve lines. This means that there are 144-1 = 143 distinct times where swapping the hour and minute hand give a valid time (because we don't want to count midnight twice).
$endgroup$
– user326210
Jan 18 '18 at 5:47
add a comment |
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1
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Enumeration works: 12:00, 1:05, 2:10, 3:15... the pattern should be clear. On the other hand, depending on the analog clock's mechanism, the latter times might no longer have the hands to be swappable.
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– J. M. is not a mathematician
Aug 24 '11 at 2:57
2
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At 1:05, the minute hand's on the 1, the hour hand's a little past it; if you swap, the hour hand's exactly on the 1, the minute hand's a little past it, and that's not a valid position.
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– Gerry Myerson
Aug 24 '11 at 3:10
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For the clock I have, that happens on "later times" (e.g. 9:45). I guess it does depend on the clock.
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– J. M. is not a mathematician
Aug 24 '11 at 3:20
1
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@Gerry Myerson - Hopefully J.M. meant the hands-pointing-in-the-same-direction positions and just rounded (i.e. every 12/11 hours). This is correct but incomplete, as your answer shows (every 13th solution of yours is a same-direction solution).
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– Rex Kerr
Aug 24 '11 at 3:22
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Yes, @Rex has it. I wasn't thinking of the "non-coincident hands" solutions, and I'm not in the mood for the needed arithmetic... hence I left it as a comment.
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– J. M. is not a mathematician
Aug 24 '11 at 3:38