Four Clocks Problem: Find Time When All 4 Clocks are Same

[dipinsingh]

Four 12 hour clocks are lined up. The first reads 10:41 and loses 2hours and 24 minutes a day. The second reads 4:50 and loses 45 minutes a day. The third reads 3:45 and gains 1hour and 20 minutes a day. The fourth reads 1:05 and keeps perfect time. How long will it be before all four clocks reads exactly the same time?
.

Ap has given the answer as 709 days. Can anybody improve the answer and find less no of days when four clocks read exactly same time?

 

[davee123]

Can anybody improve the answer and find less no of days when four clocks read exactly same time?

You mean -12 days?

It seems like if you're going by the accuracy of the minute hands, and assuming that each clock starts off at 0 seconds, they actually reach the same time at 12:59 towards the end of the 708th day, are fade in and out of synch until 1:05 on the 709th day (as close as they'll ever get to being in synch), and fade back out of synch by 1:11 on the 709th day.

DaveE

 

[dipinsingh]

You mean -12 days?

It seems like if you're going by the accuracy of the minute hands, and assuming that each clock starts off at 0 seconds, they actually reach the same time at 12:59 towards the end of the 708th day, are fade in and out of synch until 1:05 on the 709th day (as close as they'll ever get to being in synch), and fade back out of synch by 1:11 on the 709th day.

DaveE

Sir,
pls check.
The first reads 10:41 and loses 2hours and 24 minutes a day

I do not think 12 days is correct.
First watch will loose 24 hrs 288 mins in 12 days.
We can leave 24 hrs.
It will show time less by 288 mins from 10:41.or 4 hrs 48 min less than 10:41.i.e. 5:53 where as perfect one will show correct time after exactly 12 days i.e. 1:05.

 

[davee123]

I do not think 12 days is correct.

Note, "negative twelve", not "twelve". You suggested that there was a number of days *less* than 709 when all the clocks would read the same time. However, I don't believe that to be feasible, assuming that time is in the future. Well, ok, it IS feasible, but it's kind of silly. And that's at 708 days, where I noted above.

As a test, I wrote a program that increments the clocks by 1 second at a time (sometimes 2, and sometimes 0, depending on the clock), and compares all their readouts (12-hour, not 24-hour), and verified that indeed they reach the same time at 709 days, and are in synch for several small increments of time before and after the 709 day mark, meaning that yes indeed, they'll match on the 708th day. And assuming that their accuracy hasn't changed, they were also in synch 12 days before the start time.

DaveE

 

[dipinsingh]

Note, "negative twelve", not "twelve". You suggested that there was a number of days *less* than 709 when all the clocks would read the same time. However, I don't believe that to be feasible, assuming that time is in the future. Well, ok, it IS feasible, but it's kind of silly. And that's at 708 days, where I noted above.

As a test, I wrote a program that increments the clocks by 1 second at a time (sometimes 2, and sometimes 0, depending on the clock), and compares all their readouts (12-hour, not 24-hour), and verified that indeed they reach the same time at 709 days, and are in synch for several small increments of time before and after the 709 day mark, meaning that yes indeed, they'll match on the 708th day. And assuming that their accuracy hasn't changed, they were also in synch 12 days before the start time.

DaveE

Question here is for future time only.

 

[dipinsingh]

Please note that I have not said it is feasible. There is one brainteaser asked by someone. Ap has replied for that as 709 days. I have asked the best persons in this forum to check whether there is any scope for improvement as I could not find any better answer.

 

[dipinsingh]

Note, "negative twelve", not "twelve". You suggested that there was a number of days *less* than 709 when all the clocks would read the same time. However, I don't believe that to be feasible, assuming that time is in the future. Well, ok, it IS feasible, but it's kind of silly. And that's at 708 days, where I noted above.

As a test, I wrote a program that increments the clocks by 1 second at a time (sometimes 2, and sometimes 0, depending on the clock), and compares all their readouts (12-hour, not 24-hour), and verified that indeed they reach the same time at 709 days, and are in synch for several small increments of time before and after the 709 day mark, meaning that yes indeed, they'll match on the 708th day. And assuming that their accuracy hasn't changed, they were also in synch 12 days before the start time.

DaveE

I think -12 is also not correct.
Pls check second watch only for checking.
The second reads 4:50 and loses 45 minutes a day.

12 days before, it will be 9 hrs plus from 4:50 . Then time will be 13:50 0r 1:50.
But the correct clock will show time of 1:05.
So -12 is also not correct.
Pls check for future if you can!

 

[dingpud]

Not sure if this matters, but do all of the clocks start in the A.M. or in P.M.?

I wasn't sure what was meant by 12 hour clocks...I am now guessing that there is no a.m. or p.m. If there was, 709 days would not be correct because the first clock would read 1:05 p.m. 2nd reading 1:05 a.m., 3rd reading 1:05 p.m., and fourth reading 1:05 a.m.

 

[dipinsingh]

Not sure if this matters, but do all of the clocks start in the A.M. or in P.M.?

I wasn't sure what was meant by 12 hour clocks...I am now guessing that there is no a.m. or p.m. If there was, 709 days would not be correct because the first clock would read 1:05 p.m. 2nd reading 1:05 a.m., 3rd reading 1:05 p.m., and fourth reading 1:05 a.m.

12 hour clocks means , these are analog , clocks where u can not differentiate between A.M. and P.M.
709 is correct in this case.
But my request here is to check whether these clocks will show same time before 709 days( taking future case only) ?

 

[dingpud]

well, it seems like in this case that it won't happen. you have confirmation from 3 people now.

There are days, prior to day #709 when 3 of the 4 clocks read the same time, but not all 4.

Definitley a neat teaser...post more if you have them...

 

[dipinsingh]

well, it seems like in this case that it won't happen. you have confirmation from 3 people now.

There are days, prior to day #709 when 3 of the 4 clocks read the same time, but not all 4.

Definitley a neat teaser...post more if you have them...

Brainteaser for you ( Posted on forum also)

Weighing...
On a Christmas tree there were two blue, two red and two white balls. All seemed the same, however in each colour pair one ball was heavier. All three lighter balls were the same weight, just like all three heavier balls.
Using a pair of scales twice, identify the lighter balls.

 

[dingpud]

I am guessing that I can't assume that just picking the balls up will give away which ones are the light ones and which ones are the heavy ones?

That would be too easy, right?

 

[dipinsingh]

I am guessing that I can't assume that just picking the balls up will give away which ones are the light ones and which ones are the heavy ones?

That would be too easy, right?

I appreciate your ability to find the heavy balls by just guessing, but can u help others to find it by using a balance??

 

[dingpud]

I am going to take a crack at this:
Assign arbitrary subscripts to each ball as the following:
W1 W2
R1 R2
B1 B2

Step 1:
Place B1+B2+R1 on the first scale.
Place W1+W2+R2 on the second scales.
Whichever scales reads the higher value contains the heavy red ball. Record the high weight measurement and record as m_alpha
Label the heavy ball as R_heavy
Label the light ball as R_light
(Here is where a possible technicallity may occur)

Step 2:
Use only 1 scale for this measurement:
Place on the scale B1+W1+R_heavy
If the reading on the scale is greater than m_alpha then B1 and W1 are both heavy while B2 and W2 are both light.
If the reading on the scale is less than m_alpha then B1 and W1 are both light while B2 and W2 are both heavy.
If the reading on the scales is equal to m_alpha then use the fourth scale with the same set-up as the third weigh in, however, swap out B1 with B2.
Apply the same logic rules as above, and you should be able to differentiate between the heavy and light balls.

Let me know what you think...

I wanted to give the answer in pure mathematical / algebraic form, but don't know when I'll get time to attempt that.

 

[doodle]

Four 12 hour clocks are lined up. The first reads 10:41 and loses 2hours and 24 minutes a day. The second reads 4:50 and loses 45 minutes a day. The third reads 3:45 and gains 1hour and 20 minutes a day. The fourth reads 1:05 and keeps perfect time. How long will it be before all four clocks reads exactly the same time?

Ap has given the answer as 709 days. Can anybody improve the answer and find less no of days when four clocks read exactly same time?

Spoiler

In exactly 389 days later, all the 4 clocks will read 1:05.

 

[dipinsingh]

Spoiler

In exactly 389 days later, all the 4 clocks will read 1:05.

I think it is not correct.
Pls check the statement.
The third reads 3:45 and gains 1hour and 20 minutes a day.

I think u have solved with condition of loosing time 1 hr 20 min.
Pls check.
Excuse me if i m wrong.

 

[dingpud]

Has anyone been able to verify my take at weighing the Christmas balls?

 

[dipinsingh]

I am going to take a crack at this:
Assign arbitrary subscripts to each ball as the following:
W1 W2
R1 R2
B1 B2

Step 1:
Place B1+B2+R1 on the first scale.
Place W1+W2+R2 on the second scales.
Whichever scales reads the higher value contains the heavy red ball. Record the high weight measurement and record as m_alpha
Label the heavy ball as R_heavy
Label the light ball as R_light
(Here is where a possible technicallity may occur)

Step 2:
Use only 1 scale for this measurement:
Place on the scale B1+W1+R_heavy
If the reading on the scale is greater than m_alpha then B1 and W1 are both heavy while B2 and W2 are both light.
If the reading on the scale is less than m_alpha then B1 and W1 are both light while B2 and W2 are both heavy.
If the reading on the scales is equal to m_alpha then use the fourth scale with the same set-up as the third weigh in, however, swap out B1 with B2.
Apply the same logic rules as above, and you should be able to differentiate between the heavy and light balls.

Let me know what you think...

I wanted to give the answer in pure mathematical / algebraic form, but don't know when I'll get time to attempt that.

Please check . I am pasting ur solution.
Step 2:
Use only 1 scale for this measurement:
Place on the scale B1+W1+Rheavy
If the reading on the scale is greater than malpha then B1 and W1 are both heavy while B2 and W2 are both light.

There are no ways to measure absolute value of malpa as suggested by you. Only six balls and a balance with two pans. No weights are available.

 

[dipinsingh]

Has anyone been able to verify my take at weighing the Christmas balls?

Please check . I am pasting ur solution.
Step 2:
Use only 1 scale for this measurement:
Place on the scale B1+W1+Rheavy
If the reading on the scale is greater than malpha then B1 and W1 are both heavy while B2 and W2 are both light.
....
Now
There are no ways to measure absolute value of malpa as suggested by you. Only six balls and a balance with two pans. No weights are available.

 

[dingpud]

OK, so it is a balance scale, and not necessarily a weighing scale.

Alright, I'll have to look at it again then...

 

[dipinsingh]

OK, so it is a balance scale, and not necessarily a weighing scale.

Alright, I'll have to look at it again then...

Yes, Sir.
There is only one balance scale with two pans. No weights available.
You can use scale twice only.
Please excuse me if language of my question created any confusion.

 

[dingpud]

Language, not a problem...

Mathematics is the universal language, right?

 

[doodle]

I think it is not correct.
Pls check the statement.
The third reads 3:45 and gains 1hour and 20 minutes a day.

I think u have solved with condition of loosing time 1 hr 20 min.
Pls check.
Excuse me if i m wrong.

You are absolutely correct. :)

Btw, here's my take on the christmas tree problem...
Let the blue, red and white balls be labelled A, B, C, respectively. Weigh A1+B1 against A2+C1. If A1+B1 == A2+C1, weigh A1 versus A2. If A1 > A2, then the heavier balls are A1, B2 and C1; otherwise, it is A2, B1 and C2. If A1+B1 > A2+C1, weigh A1+A2 against B1+C1 next. If A1+A2 > B1+C1, then the heavier balls are A1, B2, C2; if A1+A2 < B1+C1, the heavier balls are A1, B1, C1; otherwise, if A1+A2 == B1+C1, the heavier balls are A1, B1, C2. If, in the first step, A1+B1 < A2+C1, weigh A1+A2 against B1+C1 next. If A1+A2 > B1+C1, then the heavier balls are A2, B2, C2; if A1+A2 < B1+C1, the heavier balls are A2, B1, C1; otherwise, if A1+A2 == B1+C1, the heavier balls are A2, B2, C1.