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Inverting Clocks!

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How many times in a day can we invert the hour and minute hands, and have the clock still showing a "possible" time?

Note:

Invert = replace each one with the other

Possible time = time that can be seen on a clock

Ex: if it's 6 o'clock, the hour hand is on 6 and the minute hand is on 12. Now if we invert the hands, we get the hour hand on 12 and the minute hand on 6. This is not a "possible" time because at 12:30 the hour hand is midway between 12 and 1.

Again, there is a complicated way to solve it and another much simpler way!

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Using your example, I'd say never.

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How many times in a day can we invert the hour and minute hands, and have the clock still showing a "possible" time?

Note:

Invert = replace each one with the other

Possible time = time that can be seen on a clock

Ex: if it's 6 o'clock, the hour hand is on 6 and the minute hand is on 12. Now if we invert the hands, we get the hour hand on 12 and the minute hand on 6. This is not a "possible" time because at 12:30 the hour hand is midway between 12 and 1.

Again, there is a complicated way to solve it and another much simpler way!

Consider the hour between noon and one o'clock. The hour hand starts at the zero-minute position and moves as far as the five-minute position. Including twelve o'clock itself, when the hour and minute hand coincide, the minute hand will pass a "bit past the hour" point for one o'clock, two o'clock, and so on through eleven o'clock, for a total of twelve times in the first hour when the hour and minute hands can be swapped and still yield a possible time. For the second hour, when the hour hand is between five and ten minutes past the hour, same story -- twelve possible positions. The is true completely through the ten o'clock hour, so the first eleven hours each give twelve swappable positions.

For the eleven o'clock hour, this will still succeed while the minute hand passes between twelve and one, between one and two, and so on, all the way to when it passes between ten and eleven (that is, eleven times). But for the last five minute period, the minute hand will not reach the correct position that the hour hand would be at until the hour hand itself has reached the twelve o'clock position, which of course takes us back to our original, and already-counted, configuration. Thus, the eleven o'clock hour will only have eleven such positions.

Therefore, the total number of swappable positions is (11 * 12) + 11, or 143 different distinct configurations where the hour and minute hand can be swapped and still yield a valid time.

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cool

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286 times a day.

First, when the minute and hour hands coincide.

That happens 11 times between midnight and noon and another 11 times between noon and midnight.

These 22 daily coincidences are exactly 12/11 of an hour, or about 65.45 minutes, apart.

But those aren't the only times.

... at 1/5 the speed of a normal second hand. Let's call that imaginary hand the fifth hand.

Beginning at midnight, when the fifth hand will have made its first revolution, five minutes will have elapsed,

and the minute hand will have moved from 12 to 1. Thus the fifth hand and the minute hand behave

together just as the minute and hour hands do: At each moment in time, the fifth hand occupies the

position a minute hand would occupy if the hour hand were at the position of the minute hand.

That fact provides an elegant solution to the question.

The minute and hour hands can be interchanged at those times when the fifth hand and the hour hand coincide.

As the hour hand completes a full revolution from midnight to noon, the minute hand makes 12 complete

revolutions, and the fifth hand makes 144 complete revolutions -- 12 for each of the minute hand's revolutions.

Thus at 143 equally spaced intervals of 12/143 hours, or about 5.035 minutes, the fifth and hour hands coincide -

signaling that the hour and minute hand positions can legally be interchanged.

Between noon and midnight, another 143 coincidences occur, making a total of 286 times a day the hour and minute

hands can be legally interchanged.

Note: since 143 = 11 x 13, it follows that between any two of the 11 times that the minute and hour hands coincide,

there are 12 additional times that the fifth hand [moving at 12 times the speed of the minute hand] coincides with

the hour hand.

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My reasoning was sound, I think, but I didn't answer the question. Bonanova did.

The question was how many times during the day were the hands swappable, not how many different configurations were possible. Since a clock only shows twelve hours and the question applied to 24 hours, my answer was half the correct answer.

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How many times in a day can we invert the hour and minute hands, and have the clock still showing a "possible" time?

... there is a complicated way to solve it and another much simpler way!

Interested to know roolstar's approach.

Care to share, roolstar?

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As for spoxjox:

Consider the hour between noon and one o'clock. The hour hand starts at the zero-minute position and moves as far as the five-minute position. Including twelve o'clock itself, when the hour and minute hand coincide, the minute hand will pass a "bit past the hour" point for one o'clock, two o'clock, and so on through eleven o'clock, for a total of twelve times in the first hour when the hour and minute hands can be swapped and still yield a possible time. For the second hour, when the hour hand is between five and ten minutes past the hour, same story -- twelve possible positions. The is true completely through the ten o'clock hour, so the first eleven hours each give twelve swappable positions.

I'm not sure your reasonning is correct unless I'm missing something here. it looks to me like you said: "the minute hand and the hour hand will coincide TWELVE times EVERY hour! And that, of course is simply not true. In fact the minute and hour hands will only coincide ONCE every hour. In fact you yourself said "a bit past the hour", this means we are already in the 1 o'clock.

Now bonanova:

Imagine a clock hand that moves at 1/5 the speed of a normal second hand. Let's call that imaginary hand the fifth hand.

Beginning at midnight, when the fifth hand will have made its first revolution, five minutes will have elapsed,

and the minute hand will have moved from 12 to 1. Thus the fifth hand and the minute hand behave

together just as the minute and hour hands do: At each moment in time, the fifth hand occupies the

position a minute hand would occupy if the hour hand were at the position of the minute hand.

Great reasonning. We simply created a "fifth hand" that goes 12 times faster than the minute hand; exactly like the minute hand goes 12 times faster than the hour hand. That was my "much simpler approach" too. Simple doesn't always mean easy to grasp though.

Another approach however is Mechanical and based on Geometry not on logic alone. I prefer it because we can easilly calculate each possible position with a simple 2 equations system.

Can you figure out what I mean?

Well. I started by a simple definition of "possible" time on a clock. At any moment, the clock display follows a fairly simple rule:

Let M be the angle between the MINUTE hand and the 12 o'clock position.

Let H be the angle between the HOUR hand and the hour's number position (like if it's 2:30, the hour number position is the 2 position)

Well at any given time, (equation 1) M/360 = H/30 ==> H = M/12 (can you find your reasonning here bonanova?)

So if we need to calculate when we can invert them, we need another equation:

If we divide the clock into 12 sectors (the sector between 12 and 1 = 0; the sector between 1 and 2 = 1......)

Let A be the number of the sector where the hour hand is

Let B be the number of the sector where the minute hand is

A & B = {0,1,2,3,...,11} 12 possiblilities each

(equation 2) is the inverted (equation 1): (Ax30 + H)/360 = (M - Bx30)/30

(equation 2): (Ax30 + H)/12 = M - Bx30

But remember, A & B are defined. So if we what to know how many times we can invert them, we have to give A & B all possible solutions and resolve the system with the 2 equation (equation 1 & 2) and 2 variables (H & M).

Since there are 12 possible A and 12 possible B, there seems to be 144 possibilities.

But if you solve the A = 11 and B = 11 possibilities. the answer is M = 360 and H = 30 which is 12:00.

So in reality, there are 143 solutions every 12 hours.

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Another approach however is Mechanical and based on Geometry not on logic alone. I prefer it because we can easilly calculate each possible position with a simple 2 equations system.

Can you figure out what I mean?

... to write the equations for the angles from vertical of the minute hand and of the "fifth" hand and plot them.

Interchangeability boils down to the "fifth" aligning with the hour hand.

You can thus ignore the minute hand altogether.

The fifth hand travels at 144 = 12x12 times the speed of the hour hand,

meaning angle[5] = 144 x angle[H] where 5 and H stand for the two hands.

Practically, angle[5] can be kept in the range [0, 360) so angle[5] = (144x angle[H]) modulo 360 degrees

Again, you're just left with the fact that as H travels from 0 to 360 once, 5 does it 144 times.

The hands coincide at Hour-hand angular intervals of 360/143 = 2.5174825174825174825174825174825 degrees [approximately].

For purposes of answering the question, you just need the number 143 [for each 12-hour interval].

If you want the angular positions of the hands, they're this:

angle[H] = [360/143]n where n = 0, 1, 2, 3, ... 142 degrees [for each 12-hour interval]

angle[M] = (12 x angle[H]) modulo 360 degrees.

If you want the precise times, they're this:

time = 12 x angle[H]/360 hours.

Then convert fractions of hours to minutes and seconds [exercise left to the reader].

You could write equations for the minute hand angle too,

but I didn't think it was as direct to find the moments of interchangeability that way.

Nice puzzle!

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As for spoxjox:

Consider the hour between noon and one o'clock. The hour hand starts at the zero-minute position and moves as far as the five-minute position. Including twelve o'clock itself, when the hour and minute hand coincide, the minute hand will pass a "bit past the hour" point for one o'clock, two o'clock, and so on through eleven o'clock, for a total of twelve times in the first hour when the hour and minute hands can be swapped and still yield a possible time. For the second hour, when the hour hand is between five and ten minutes past the hour, same story -- twelve possible positions. The is true completely through the ten o'clock hour, so the first eleven hours each give twelve swappable positions.

I'm not sure your reasonning is correct unless I'm missing something here. it looks to me like you said: "the minute hand and the hour hand will coincide TWELVE times EVERY hour! And that, of course is simply not true. In fact the minute and hour hands will only coincide ONCE every hour. In fact you yourself said "a bit past the hour", this means we are already in the 1 o'clock.

Yes, you are correct; you misunderstood me. As the parts in blue indicate, I am referring to the idea that at about 12:05:30, 12:11:00, 12:16:30, etc. the minute hand will reach a point where, if it were swapped with the hour hand, the clock would read a possible time. Thus, there are twelve such positions for each of eleven hours, and eleven such positions for the twelfth hour, giving 143 total for each twelve-hour period.

Merry Christmas!

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Good reasoning guys!

That was my first post on this forum and I enjoyed these conversation.

If you liked this puzzle, just look for my next post soon my friend.

Merry Christmas and a happy new year

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