The answer is 264.

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Ah,I totally forgot to exclude the cases when the two would hands would overlap.A partial solution to this problem can be found here

Rahul Saha
- 7 years, 4 months ago

Honestly,i have been gaining anything from this site and it's simply because the explanation are not clear enough

ogedengbe abdullah
- 7 years, 2 months ago

It's really difficult. I thought one can't tell the correct time when the one hand is over another hand.

Partho Kunda
- 7 years, 4 months ago

Yep, that's what I did

Sam Thompson
- 7 years, 4 months ago

It is easy

The only way you dont get tricked by the clock is when the hand is overlap with the other hand

So there are 12 possible overlap in one loop of clocks, and the total possible position is 12 x 12

In one loop you will get tricked [(12x12) - 12 = 132] times

so in one day you get 132 x 2 that count
**
264
**

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Different values of n and n' are- 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11

All units are-- Displacements x & y in radians, Velocity in rad/s

It is obvious that when displacement of hr hand from "n" is x rad, displacement of min hand from "0" is 12x rad.

**
How do I know the correct time ?
**

Assume A hand as hr hand.Find its displacement from closest "n". According to this, displacement of min hand(B) from "0" should be 12x.

Look at the clock, if B is pointing at displacement of 12x from "0" then our assumption is correct and we know the time as we know the hr and min hands.If B is pointing somewhere else then it simply means that our assumption was wrong and thus not A but B is the hr hand and hence we know the time.

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Where will problem arise ?
**

I do above steps and my assumption is correct,i.e, For A pointing towards P, B points towards Q; So A is hr hand and B is min hand. -----(1)

But I just want to check what happens if I consider B as hr hand.

So I repeat above procedure with B as hr hand and find for B pointing towards Q, A points towards P. This implies B is hr hand and A is min hand. -----(2)

But (1) and (2) contradict and I won't be able to determine the time. (That is configuration of hands remains same irrespective of which is selected as hr hand.)

So the question asks how many times in a day(period of 24 hrs) will I come across such situation ?

Let us formulate this mathematically.

For any time in day--

So y=12x-n' $\frac{\pi}{6}$ -----(3)

For ambiguous situation, we know that configuration of hands remains same irrespective of which is selected as hr hand. We consider both as hr hands one by one.We have--

So ambiguity arises when 12y=n $\frac{\pi}{6}$ +x -----(4)

Combining (3) and (4) we get all values of x which would result in ambiguous time.

So x= $\frac{2\pi}{143}$ ( $\frac{n}{12}$ +n')

But there is an exception, following configurations obtained from above formula won't result in ambiguous time.

There will be one such configuration every hour. -----(5)

How many distinct values of x do we have minus the exception...that would be our answer.

*
* For a period of 12 hours *
*

n stands for hr hand and can have values 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11; i.e 12 values

n' stands for min hand, so for each value of n, n' will have values 0 to 11 except the value where n'=n ( accounting for the exception mentioned in
**
(5)
**
which occurs once every our).So we have 11 values of n' for every value of n.

No. of distinct values of x = 11x12=132

So 132 ambiguous times in 12 hrs.

This implies 264 ambiguous times in 24 hrs.
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ANS
**

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Thank you for writing such a clear solution! Upvoted.

Daniel Liu
- 7 years, 3 months ago

(5), There are 11 such configurations in a 12 hour period

Ajinkya Shivashankar
- 4 years, 7 months ago

**
do not
**
overlap each other.There are total
$143$
places where there is a real time even after exchange and among them
$11$
overlap each other.Also we can distinguish between AM and PM, so the answer is
$2(143-11)=\boxed{264}$

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how can we find 143 places?

Partho Kunda
- 7 years, 4 months ago

Let the hour hand be A, the minute hand be B, then hand B moves 12 times faster than hand A. When A and B are interchangeable, then we won't be able to tell the time. Let's say we fix a certain time in the day. If hand A reaches hand B, B should also move 12 times as fast as hand A, and if hand B meets with the original hand A, then the 2 hands can switch positions. Thus, we introduce another hand C, such that this hand is 12 times as fast as hand B, and if hand C coincides with hand A, we won't be able to tell the time!

Thus, number of times hand A coincide with hand C in a day: 2*144-2= 286.

However, when all 3 hands coincide with each other, we will still be able to tell the time because whether A and B switch positions it will still be the same time.

Number of times B coincide with A and C: 2*12-2=22.

Thus, the answer is 286-22=264.

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Let the actual time on such a case be H hours M minutes where H = {0, 1, 2, ..., 11} and 0 <= M < 60. Let's signify a hand at 12 to be at a displacement of 0 degrees.

Angular Velocity of the Hour hand = 0.5 degrees/minute Angular velocity of the Minute hand = 6 degrees/minute which is 12 times faster than the Hour hand.

So, at the given time the Hour hand would be at an angular displacement of = (30H + 0.5M) degrees. And, the Minute hand would be at an angular displacement of = 6M degrees.

Such a time would only be confusing if there's a correct time when the Minute hand is at (30H + 0.5M) degrees when the Hour hand is at 6M degrees.

Thus, we have to solve for the following equation. 6M - (30H + 0.5M)/12 = 30k where k = {0, 1, 2, ..., 11} ==> M = (60/143) * (H + 12k)

Now, (H + 12k) can have 12 * 12 = 144 values as (H,k) = {(0,0), (0,1), ..., (0,11), (1,0), ..., (11,10), (11,11)} First, we have to exclude the largest value of (H,k) = (11,11) which makes M = 60 which is NOT permitted. Also, there would be 11 cases in the 12-hour period when the Hour hand and the Minute hand would overlap and hence, no confusion in the time. That makes the total of (144 - 1 - 11) = 132 confusing time in a 12-hour period making it to 132 * 2 = 264 in the entire day.