What time is it?

[needimprovement]

If the hour and minute hands are equidistant from the 6 hour, what is the exact time?

[DJBruce]

At 12:00 the minute hand and the hour hand occupy the same spot, and there for are equidistant from the 6 on the clock.

[Mr Skeptic]

 

It is one of ...

12:00, 12 and 12/13ths of an hour

1 and 1/11th of an hour, 1 and 11/13ths of an hour

2 and 2/11th of an hour, 1 and 10/13ths of an hour

3 and 3/11th of an hour, 1 and 9/13ths of an hour

4 and 4/11th of an hour, 1 and 8/13ths of an hour

5 and 5/11th of an hour, 1 and 7/13ths of an hour

6 and 6/11th of an hour, 1 and 6/13ths of an hour

7 and 7/11th of an hour, 1 and 5/13ths of an hour

8 and 8/11th of an hour, 1 and 4/13ths of an hour

9 and 9/11th of an hour, 1 and 3/13ths of an hour

10 and 10/11th of an hour, 1 and 2/13ths of an hour

11 and 1/13ths of an hour

 

[needimprovement]

 

At 12:00 the minute hand and the hour hand occupy the same spot, and there for are equidistant from the 6 on the clock.

 

Sorry, Not the right answer.

[DJBruce]

Sorry, Not the right answer.

 

I am fairly certain that it is one of the solutions to question.

[Moontanman]

Does anyone really know what time it is.... does anybody really care....about time.....

[needimprovement]

I am fairly certain that it is one of the solutions to question.

Not really: The minute hand is taller than the hour hand (this is a trick question after all).

 

Disregarding the above and just concentrating on angles, I would say: at any time the minute hand and the hour hand overlap, which would be 22 times in 24 hours.

[Mr Skeptic]

Not really: The minute hand is taller than the hour hand (this is a trick question after all).

 

Nope, that would make it an ill-defined unsolvable problem. If you want that, you need to mention how much longer the minute hand is.

 

Disregarding the above and just concentrating on angles, I would say: at any time the minute hand and the hour hand overlap, which would be 22 times in 24 hours.

 

Not good enough, that misses over half of the solutions.

[rigney]

If the hour and minute hands are equidistant from the 6 hour, what is the exact time?

 

Ok! beat me over the head with it, but could it be five forty five, plus or minus a tick???

Edited August 18, 2010 by rigney

[Klaynos]

Not really: The minute hand is taller than the hour hand (this is a trick question after all).

 

Not on my clock.

[DJBruce]

Here is another clock one that has an answer:

 

Assuming a regular analog clock, what is the angle between the minute and the hour hand at 3:15?

Edited August 20, 2010 by DJBruce

[needimprovement]

Here is another clock one that has an answer:

 

Assuming a regular analogy clock, what is the angle between the minute and the hour hand at 3:15?

Assuming that "analogy" means analog, 7.5°.

[DJBruce]

Assuming that "analogy" means analog, 7.5°.

 

Opps, sorry about that typo I did indeed mean analog. Yep, nice job NI.

[needimprovement]

Any takers?

[Sisyphus]

Well, Mr Skeptic already gave the complete set of correct answers in post 3 for a fully analog clock with smooth motion. If the minute hand ticks but the hour hand is a smooth motion, then DJBruce's answer is the only correct one, assuming you don't count mid-tick, in which case the question is unsolvable without more information. If both hands "tick," it is simpler:

 

 

12:00

1:05

1:55

2:10

2:50

3:15

3:45

4:20

4:40

5:25

5:35

6:00

7:25

7:35

8:20

8:40

9:15

9:45

10:10

10:50

11:05

11:55

 

 

However, I don't know that I've ever seen a clock like that.

 

And if you're going to ask when the ends of the hands are equidistant when the hands are different lengths, then that's unanswerable unless you know the ratio between the minute hand, hour hand, and clock face radius.

[John Cuthber]

At least one part of both hands (the pivot points) are coincident and so always at the same distance from anything (in terms of the x and y dimensions) including the 6.

The condition is, therefore, met at all times.

 

(Incidentally, my clock is a projection clock so the hour and minute hands actually do occupy the same plane in case anyone was wondering.)

 

Since the condition specified is always met the question simplifies to "what's the exact time?"

Well, the forum software timestamps posts reasonably accurately. I think it's about 19:22 (BST).

[needimprovement]

Well, Mr Skeptic already gave the complete set of correct answers in post 3 for a fully analog clock with smooth motion. If the minute hand ticks but the hour hand is a smooth motion, then DJBruce's answer is the only correct one, assuming you don't count mid-tick, in which case the question is unsolvable without more information. If both hands "tick," it is simpler:

 

 

12:00

1:05

1:55

2:10

2:50

3:15

3:45

4:20

4:40

5:25

5:35

6:00

7:25

7:35

8:20

8:40

9:15

9:45

10:10

10:50

11:05

11:55

 

 

However, I don't know that I've ever seen a clock like that.

 

And if you're going to ask when the ends of the hands are equidistant when the hands are different lengths, then that's unanswerable unless you know the ratio between the minute hand, hour hand, and clock face radius.

Sorry, the wrong answer. Try harder.

[John Cuthber]

The question is not amenable to a correct answer because, as has been shown, there are many possible answers depending on how the question is interpreted.

 

I think it falls to you to "try harder" to ask a clearer question rather than expecting us to try harder to answer the one you have set.

[needimprovement]

The question is not amenable to a correct answer because, as has been shown, there are many possible answers depending on how the question is interpreted.

 

I think it falls to you to "try harder" to ask a clearer question rather than expecting us to try harder to answer the one you have set.

Sorry to disappoint you John Cuthber, the details provided are sufficient to solve this problem. Only it is a hard nut to crack.

 

This is similar to "prove 2x2=5"

 

I know there are lot of math wizards on the SFN, who can easily solve this problem, hence I am holding back the solution, which is quite simple to crack this hard nut.

Edited August 30, 2010 by needimprovement

[DJBruce]

Sorry to disappoint you John Cuthber, the details provided are sufficient to solve this problem. Only it is a hard nut to crack.

 

This is similar to "prove 2x2=5"

 

I know there are lot of math wizards on the SFN, who can easily solve this problem, hence I am holding back the solution, which is quite simple to crack this hard nut.

 

Wait you are saying that this problem is completely fallacious and does not have an actual solution just like the proof of 2x2=5?

[needimprovement]

Wait you are saying that this problem is completely fallacious and does not have an actual solution just like the proof of 2x2=5?

No. The problem have actual solution. When I say similar to prove 2x2=5, I mean "disguised with simplicity".

[DJBruce]

No. The problem have actual solution. When I say similar to prove 2x2=5, I mean "disguised with simplicity".

 

Without you accurately describing the problem there are numerous correct solutions through out this thread. Therefore, could you please restate the problem so that we figure out which of the right answers you want?

 

The error in that proof is not that disguised it is fairly obvious and easy to spot.

[John Cuthber]

Great, now, not only do I know that the original problem does not have a unique solution (several perfectly valid solutions have been supplied), but I don't understand why Needsimprovement thinks that asking me to prove that 2X2=5 is in any way helpful.

Does he mean just plain wrong.

Does he mean that while everybody else uses 4 for the answer to that question, he has decided to use the symbol "5" for the number of dots here ....

Did he mean to ask me to prove that 2X2 =4 and cocked it up?

 

The trouble is that 'When I say similar to prove 2x2=5, I mean "disguised with simplicity". ' doesn't really mean anything.

[needimprovement]

Great, now, not only do I know that the original problem does not have a unique solution (several perfectly valid solutions have been supplied), but I don't understand why Needsimprovement thinks that asking me to prove that 2X2=5 is in any way helpful.

Does he mean just plain wrong.

Does he mean that while everybody else uses 4 for the answer to that question, he has decided to use the symbol "5" for the number of dots here ....

Did he mean to ask me to prove that 2X2 =4 and cocked it up?

 

The trouble is that 'When I say similar to prove 2x2=5, I mean "disguised with simplicity". ' doesn't really mean anything.

I am providing the answer. The third super power of 9, is a 'GIGANTIC' number which has remained un-calculated, despite the use of super computers!! (to the best of my knowledge)

 

2 is equal to 3 is also solved in the same way as 2x2 = 5

 

since 4 -10 = 9 - 15

there fore 2^2 - 2x5 = 3^2 - 3x5

there fore 2^2 - 2x2x5/2 + (5/2)^2 = 3^2 - 2x3x5/2 + (5/2)^2

there fore (2-5/2)^2 = (3-5/2)^2 take sq.root on both sides

there fore 2 -5/2 = 3 -5/2

there fore 2 = 3 QED

 

If you give a close look, you'll understand the 'Modus-Operandi' behind this calculation. This method can be used to prove a number equal to its next number viz: 2=3, (2x2) 4=5, (3x3) 9=10 and so on.

 

Let me explain the modus operandi: say we want to prove 9=10 (three threes are ten)

 

1. Multiply both the numbers to obtain the negative difference of the equation so 9x10=90 therefore -90 is the difference

 

2. Square the numbers on both sides and add (-90)

there fore 81-171 = 100-190

there fore 9^2 -9x19 = 10^2 -10x19 (now multiply and divide the neg.no.by 2)

there fore 9^2 -2x9x19/2 = 10^2 -2x10x19/2

 

3. Now add the positive difference(+90) plus 1/4 on both sides i.e. +90.1/4=361/4=(19/2)^2

there fore 9^2 -2x9x19/2 + (19/2)^2 = 10^2 -2x10x19/2 + (19/2)^2

there fore (9-19/2)^2 = (10-19/)^2

there fore 3x3 -19/2 = 10 -19/2

there fore 3x3 = 10 QED

 

In the nutshell, the whole creation is to nullify everything, leaving (-1/2)^2 = (+1/2)^2

 

------------------------------------------

 

The answer is 2, easy for those who remember the strings!!

 

Like pythegorian number 3^2+4^2 = 5^2, there is another beautiful string :

 

10^2+11^2+12^2 = 13^2+14^2 The sum equals to 365 and sqr of 365 is 133225 another memorable number, multiplied by 2 gives 266450.

[DJBruce]

I am completely lost:

 

- Does the answer have to do with tetration?

- Does the answer somehow have to do with the two fallacious "proofs" you showed?

- The answer has to do with a Pythagorean Triple?