riddle - clock hands

shahar

shahar

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How many times the minutes hand is on the hours hand in 24 hours in analog clock?
 
shahar said:
How many times the minutes hand is on the hours hand in 24 hours in analog clock?
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Is this a 12-hour clock, or a 24-hour clock? When are you starting the 24-hour period? And are you including both the start and end of the 24 hours, if the hands are together then?

And are you presenting this to us as a riddle, or asking for the answer? In either case, please tell us your own thoughts about it.
 
In a regular clock - analog clock not digital.
I only know 12-hours clock that from 1 to 12 I don't know analog clock of 24 hours only digital clock. Is there a along clock of 24-hours?!
 
The question is about 12-hours-number clock. I look in Google. The clock of 24 hours is very rare or even is not exist in Israel at all.
 
shahar said:
In a regular clock - analog clock not digital.
I only know 12-hours clock that from 1 to 12 I don't know analog clock of 24 hours only digital clock. Is there a along clock of 24-hours?!
Click to expand...

I was, of course, really asking you to show some sort of work, which starts with thinking about what a clock is.

Have you thought about the problem at all?
 
I think in this way:
there is route of 12x for the hour to every second x.
How can I continue...
I try:
x = 12x
But the solution is:
x + 1= 12x
Why?
Why the solution need to add 1 to the route of the hour that is x.
Here the solution:
בוא נניח שהמחוג הקטן עובר מרחק x, בזמן הזה הגדול עובר מרחק 12x וכדי שיהיו באותו מקום הגדול צריך לבצע הקפה יותר מהקטן, אז המשוואה שלנו תהיה x+1=12x⟹x=1/11 x+1=12⟹x=1/11 (זה בערך 5 דקות ו-27 שניות) אנחנו מתחילים כששניהם על 12 ואז בכל שעה אנחנו מזהים מפגש אחד (למשל בשעה 01:05:27). ככה אפשר לגלות שיש לנו 22 מפגשים במשך יממה
 
shahar said:
Here the solution:
בוא נניח שהמחוג הקטן עובר מרחק x, בזמן הזה הגדול עובר מרחק 12x וכדי שיהיו באותו מקום הגדול צריך לבצע הקפה יותר מהקטן, אז המשוואה שלנו תהיה x+1=12x⟹x=1/11 x+1=12⟹x=1/11 (זה בערך 5 דקות ו-27 שניות) אנחנו מתחילים כששניהם על 12 ואז בכל שעה אנחנו מזהים מפגש אחד (למשל בשעה 01:05:27). ככה אפשר לגלות שיש לנו 22 מפגשים במשך יממה
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It's all Greek to me :)
 
shahar said:
I think in this way:
there is route of 12x for the hour to every second x.
How can I continue...
I try:
x = 12x
But the solution is:
x + 1= 12x
Why?
Why the solution need to add 1 to the route of the hour that is x.
Here the solution:
בוא נניח שהמחוג הקטן עובר מרחק x, בזמן הזה הגדול עובר מרחק 12x וכדי שיהיו באותו מקום הגדול צריך לבצע הקפה יותר מהקטן, אז המשוואה שלנו תהיה x+1=12x⟹x=1/11 x+1=12⟹x=1/11 (זה בערך 5 דקות ו-27 שניות) אנחנו מתחילים כששניהם על 12 ואז בכל שעה אנחנו מזהים מפגש אחד (למשל בשעה 01:05:27). ככה אפשר לגלות שיש לנו 22 מפגשים במשך יממה
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The translation, via Google, is

Let's assume that the small hand moves a distance of x, during this time the big hand moves a distance of 12x and in order for them to be in the same place the big one needs to make more laps than the small one, so our equation will be x+1=12x⟹x=1/11 x+1=12⟹x=1/ 11 (that's about 5 minutes and 27 seconds) We start when both are at 12 and then every hour we detect one encounter (for example at 01:05:27). This way you can find out that we have 22 meetings during a day​

What do you not understand about this?

Now, they failed to define x fully, which could be part of your difficulty; it appears to be the number of full rotations made by the minute hand, from one meeting of the hands to the next; they are calling those "laps" (as translated). Do you see that the hour hand has to make a full rotation plus the distance the minute hand has moved, in order to meet again?

Did you notice that if you solve x = 12x, the only solution is x = 0? So your idea is clearly wrong.
 
Yes. I understand. I need to refer to the hands like two runners. So, the addition of 1 to the x is because he did one cycle as they encountered. Do I right?
 
shahar said:
Yes. I understand. I need to refer to the hands like two runners. So, the addition of 1 to the x is because he did one cycle as they encountered. Do I right?
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Yes, this is like asking how many times two runners will meet if one makes 24 laps around a track, while the other goes in the same direction at 1/12 the speed. They meet each time the second "laps" the other, which means that he has gone 1 lap more than the other.
 
Does this really need any algebra or analogies about one runner lapping another? I accept these are mathematical "approaches" to the problem but surely it's such a simple question that even a ten year-old (like my granddaughter) could tell you the answer is 24 (or 25 if you include both the starting time and the finishing time)?
 
The Highlander said:
Does this really need any algebra or analogies about one runner lapping another? I accept these are mathematical "approaches" to the problem but surely it's such a simple question that even a ten year-old (like my granddaughter) could tell you the answer is 24 (or 25 if you include both the starting time and the finishing time)?
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Such a simple problem that you got it wrong? Certainly!

That's part of the value of math: It can correct underthinking (sometimes at the risk of overthinking).

The correct answer is 22 (or 23 if you count both midnights). That's the provided answer, which is correct. Here are those 23 times (every 12/11 of an hour):

0:00:00​
1:05:27​
2:10:55​
3:16:22​
4:21:49​
5:27:16​
6:32:44​
7:38:11​
8:43:38​
9:49:05​
10:54:33​
12:00:00​
13:05:27​
14:10:55​
15:16:22​
16:21:49​
17:27:16​
18:32:44​
19:38:11​
20:43:38​
21:49:05​
22:54:33​
0:00:00​

So, what is your explanation for your answer? Presumably you have some justification for it.
 
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Dr.Peterson said:
So, what is your explanation for your answer? Presumably you have some justification for it.
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Absolutely no justification whatsoever! I just jumped in without giving proper, full consideration to the problem, thinking that there would be a crossover of the hands for each 'hour' on the clock without, thereby, realizing that when it came to the eleventh hour the hands had moved so far forward that they crossed at the 12! So there was no crossover 'attributable' to 11 am or pm.

Thus fully validating your comment:-
Dr.Peterson said:
That's part of the value of math: It can correct underthinking (sometimes at the risk of overthinking).
Click to expand...
Mea Culpa! ?
Maybe my granddaughter would have made a better job of it. Doh! ?
 
The Highlander said:
without giving proper, full consideration to the problem, thinking that there would be a crossover of the hands for each 'hour' on the clock without, thereby, realizing that when it came to the eleventh hour the hands had moved so far forward that they crossed at the 12! So there was no crossover 'attributable' to 11 am or pm.
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Yes, that is the non-mathematical, more qualitative approach I have taken to this problem in the past, which is sufficient for some people to understand, but leaves others unsure. Some people are unconvinced by the math and need something like this to increase their confidence -- so often both approaches are needed together to give a good answer.

So this is a valuable contribution to the discussion!

(To tell the truth, I was suspecting that your comment was tongue-in-cheek, intentionally illustrating how "thinking like a ten-year-old" can lead to a confident but wrong answer.)
 
Dr.Peterson said:
(To tell the truth, I was suspecting that your comment was tongue-in-cheek, intentionally illustrating how "thinking like a ten-year-old" can lead to a confident but wrong answer.)
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Nope! I'm just like an (over)confident ten year-old most times. ?
 
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I checked it on her little (teaching) clock; the hands move (manually) in the correct relationship. ?
So I was then able to see my error (alluring though it was).
clock.jpg​
 
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