Try solving these easy problems by deductive thinking

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Offline Alan McDougall

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Hello,

A little logic problem I hope it is posted in the correct place, If not please move it

x: I forgot how old your three kids are.
y: The product of their ages is 36.
x: I still don't know their ages.
y: The sum of their ages is the same as your house number.
x: I still don't know their ages.
y: The oldest one has red hair.
x: Now I know their ages!

How old are they
« Last Edit: 01/07/2008 13:08:35 by Alan McDougall »
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blakestyger

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Re: Try solving these easy problems by deductive thinking
« Reply #1 on: 30/06/2008 20:12:18 »
There are two twins aged 2 and the eldest is 9.

Bit of a chestnut, eh?
« Last Edit: 30/06/2008 20:36:44 by blakestyger »

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Offline Alan McDougall

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Re: Try solving these easy problems by deductive thinking
« Reply #2 on: 01/07/2008 09:12:42 »
Blake,

From the statement that the product of their ages is 36 the possibilities of the three individual ages are:

1,1,36
1,2,18
1,3,12
1,4,9
1,6,6
2,2,9
2,3,6
3,3,4

From the statement that the sum equals the house number it is possible to eliminate all but two possibilities. The sums of the rest are unique and would allow for an immediate answer. For example if the house number were 16 the ages must be 1, 3, and 12. The two remaining possibilities are 2, 2, and 9; or 1, 6, and 6.
After the clue that the oldest has red hair you can eliminate 1, 6, and 6 because the oldest two have the same age thus there is no oldest son. The only remaining posibility is 2, 2, and 9.

Answer
Their ages are 2, 2, and 9

You were correct maybe next time something much more difficult.

Do you know the 12 ball oddball problem?


Alan
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Offline rosalind dna

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Re: Try solving these easy problems by deductive thinking
« Reply #3 on: 01/07/2008 11:18:20 »
Alan what do you do if the family concerned might live in their home, that only has a name on it's address not a number in it's roadway??
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Offline Alan McDougall

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Re: Try solving these easy problems by deductive thinking
« Reply #4 on: 01/07/2008 12:06:29 »
Rosalin,

Quote
Alan what do you do if the family concerned might live in their home, that only has a name on it's address not a number in it's roadway??

You will not be able to solve the problem in the way I suggested, let me think about this and come back to you
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Offline lightarrow

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Re: Try solving these easy problems by deductive thinking
« Reply #5 on: 01/07/2008 12:15:01 »
Do you know the 12 ball oddball problem?
That is?

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Offline Alan McDougall

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« Reply #6 on: 01/07/2008 13:11:48 »
OK

You notice I have changed thread title

Here are three more easy problems before we try something really difficult


try these  easy tests of logical thinking.

Their are twelve identical featureless balls of equal volume and size. One differs minutely in mass or weight if you like.

The task is to establish in three weighing steps using a very sensitive balance scale, like the scale of Libra. Which ball is different and if it is heaver or lighter. You can use any combination , three and three, four against four, anything you like but you must solve the problem in three weighs. You cant use a bathroom scale this would be useless.

O O O O O O O O O O O O

 Problem two

A person is rowing his bout upstream in a river flowing at three miles an hour at seven miles per hour relative, to the bank of the river. His hat falls off and only after 45 minutes does he notice this. He immediately turns around and rows at the same speed to get his beloved hat. (disregard the time taken for turning around for the purpose of this test).

How long does it take for him to catch up to his hat and retrieve it??


Last problem

A man sentenced  to death is given a choice. He is put in a room with two PC computers, one is programed to only lie, and the other programed to only tell the truth. There are two exit doors , one leading to the death chamber and the other to freedom. He is only allowed to key in "one question" to only "one of the computers" and by this one question, he must establish the door to freedom or face execution.

Give it a go

alan
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Offline lightarrow

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« Reply #7 on: 01/07/2008 14:59:03 »
OK

You notice I have changed thread title

Here are three more easy problems before we try something really difficult


try these  easy tests of logical thinking.

Their are twelve identical featureless balls of equal volume and size. One differs minutely in mass or weight if you like.

The task is to establish in three weighing steps using a very sensitive balance scale, like the scale of Libra. Which ball is

different and if it is heaver or lighter. You can use any combination , three and three, four against four, anything you like

but you must solve the problem in three weighs. You cant use a bathroom scale this would be useless.

O O O O O O O O O O O O
3 groups of 4 balls; one group in a plate of the scale, another in the second plate; this way we identify the group with the lighter ball (if the 2 groups in the scale have the same weight, then the third group is the one we are looking for). We divide this group in two sub-groups of 2 balls and we put them on the scale, so we identify the subgroup with the lighter ball. With the last measure (one ball in each plate) we identify the single ball.
Quote
Problem two

A person is rowing his bout upstream in a river flowing at three miles an hour at seven miles per hour relative, to the bank

of the river. His hat falls off and only after 45 minutes does he notice this. He immediately turns around and rows at the

same speed to get his beloved hat. (disregard the time taken for turning around for the purpose of this test).

How long does it take for him to catch up to his hat and retrieve it??
If I understood the english correctly, he goes at 3 miles/h with respect to the river and the river flows at 4 miles/h. In a ref. frame still with respect to the river, he looses the hat, goes ahead for 45 minutes and then turns back. Of course he'll take the same 45 minutes to return to the point he lost the hat regardless of his speed and on the river's speed.
Quote
Last problem

A man sentenced  to death is given a choice. He is put in a room with two PC computers, one is programed to only lie, and the

other programed to only tell the truth. There are two exit doors , one leading to the death chamber and the other to freedom.

He is only allowed to key in "one question" to only "one of the computers" and by this one question, he must establish the

door to freedom or face execution.
"If I asked the other PC which door leads to the death chamber, what did it answer?" The answer is the door to freedom.

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Offline BenV

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« Reply #8 on: 01/07/2008 16:04:01 »
Quote
3 groups of 4 balls; one group in a plate of the scale, another in the second plate; this way we identify the group with the lighter ball (if the 2 groups in the scale have the same weight, then the third group is the one we are looking for). We divide this group in two sub-groups of 2 balls and we put them on the scale, so we identify the subgroup with the lighter ball. With the last measure (one ball in each plate) we identify the single ball.

How does this work if we don't know whether the odd ball is lighter or heavier?  Surely, we would need to add an extra weighing stage to work this out - ie group A is lighter than group B - this is either because there's a light ball in A or a heavy ball in B.  We then need to test one against group C - If it's equal to B, the 'odd' ball is light, if it's equal to A the 'odd' ball is heavy.  But that's used up one of our tests, and now we only have one left to determine which one of the 4 it could be.

If we get lucky and A & B are identical - we know that the odd ball is in group C.  So we split C into C1 and C2, and C1 comes out lighter - we split that and find they're equal - we then know that the odd ball is one of C2, and that it's heavy, but we've run out of tests.

This can't just rely on getting lucky, can it?

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Offline lightarrow

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« Reply #9 on: 01/07/2008 17:45:12 »
Quote
3 groups of 4 balls; one group in a plate of the scale, another in the second plate; this way we identify the group with the lighter ball (if the 2 groups in the scale have the same weight, then the third group is the one we are looking for). We divide this group in two sub-groups of 2 balls and we put them on the scale, so we identify the subgroup with the lighter ball. With the last measure (one ball in each plate) we identify the single ball.

How does this work if we don't know whether the odd ball is lighter or heavier?
Right, I lost the part where he says we don't know if it's lighter or heavier... [:I]
Quote
  Surely, we would need to add an extra weighing stage to work this out - ie group A is lighter than group B - this is either because there's a light ball in A or a heavy ball in B.  We then need to test one against group C - If it's equal to B, the 'odd' ball is light, if it's equal to A the 'odd' ball is heavy.  But that's used up one of our tests, and now we only have one left to determine which one of the 4 it could be.
Maybe we could do this way, but don't know if it's considered one more measure (for a total of 3) or two more measures: when we have identified the odd group of 4, let's say we have found it's lighter, we put 1 ball in each plate; if we are lucky and one is lighter, it's finished; if they weight the same, we add another ball to each plate so we discover which of the last two is lighter. (I know, it seems cheating...  [:)])

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Offline BenV

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« Reply #10 on: 01/07/2008 17:57:00 »
I have seen this before without the heavier/lighter thing, so I think we can let you off for missing that bit!

It does seem like cheating - and this sort of thing shouldn't rely on luck.

Come on then Alan - how does this one work?

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Offline Alan McDougall

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« Reply #11 on: 02/07/2008 01:35:35 »
Lightarrow,

Respectfully you are incorrect you assumed the odd ball was lighter, it could be lighter or heavier, the problem is more complex than your solution.

Ben

Luck must play no part in the solution it must be fool proof (no pun intended)

Regards

Alan
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Offline lightarrow

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« Reply #12 on: 03/07/2008 19:30:22 »
Lightarrow,

Respectfully you are incorrect you assumed the odd ball was lighter, it could be lighter or heavier, the problem is more complex than your solution.

Ben

Luck must play no part in the solution it must be fool proof (no pun intended)

Regards

Alan
Thank you for the "Respectfully". I wasn't able to find the solution. Can you give us the answer?

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Offline LeeE

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« Reply #13 on: 04/07/2008 00:00:28 »
Me too - I can solve it in two weighings if I cheat and use either gravitational attraction or momentum, but not purely via weighing.
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Offline Alan McDougall

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« Reply #14 on: 04/07/2008 04:54:23 »
Good morning people,

Here is the solution(s) to the 12 odd ball problem!

You could use chalk to identify ball potially lighter or heaver or both

Before I explain some conventions in setting out the solution to the Twelve Balls problem, here again is a statement of the task.

There are twelve balls, all the same size, shape and color. All weigh the same, except that one ball is minutely different in weight, but not noticeably so in the hand. Moreover, the odd ball might be lighter or heavier than the others.

 Your challenge was to discover the odd ball and whether it is lighter or heavier.

You must use a beam balance only, and you are restricted to three weighing operations.

Note:  by starting with 6 /6 or, 5/ 5 or 3/ 3. or 2/2 or 1/1  it is impossible to solve the problem

The only way to solve this rather complex problems is by starting by weighing 4 balls against 4 as you will see by the solution below

 Conventions:

 At every weighing one of three things theoretically can happen: the pans can balance, the left pan can go down or the left pan can go up.

 It will be necessary to refer to a given ball as definitely normal (N), potentially heavier (H) or potentially lighter (L). Often our identification of a ball in this way will be as part of a group (= This group contains a heavier/lighter ball), and will depend on what we learn from a previous weighing. At the start, all balls have a status of unknown (U).

To show at each weighing what is being placed in each pan, represent the situation as per the following examples:
.
First Weighing   UUUU UUUU

Pans balance   All these Us are now known to be Ns; the odd ball is one of the remaining unweighed four (call them UUUU from now on).

Proceed to Second Weighing: Case 1

Left pan down   

One of the four balls in the left pan  might be heavier (call them HHHH from now on) or one of the four balls in the right pan might be lighter (call them LLLL from now on).

Proceed to Second Weighing Case 2

Left pan up   

One of the four balls in the left pan might be lighter (call them LLLL from now on) or one of the four balls in the right pan is heavier (call them HHHH from now on).

Proceed to Second Weighing Case 2

Second Weighing   

Case 1   UUU NNN
Pans balance   All these Us are now known to be Ns; the odd ball is the remaining unweighed U, but we dont yet know if its heavier or lighter than normal.

Proceed to Third Weighing Case 1

Left pan down   

One of these Us is heavier than normal, but we dont yet know which one (call them HHH from now on).


Proceed to Third Weighing Case 2

Left pan up   

One of these Us is lighter than normal, but we dont yet know which one (call them LLL from now on).

Proceed to Third Weighing Case 3

Case 2   HHL HLN
Pans balance   

All these Hs and Ls are now known to be Ns; the odd ball is one of the remaining unweighed H or two Ls.

Proceed to Third Weighing Case 4

Left pan down   

The odd ball is one of the left two Hs or the right L.

Proceed to Third Weighing Case 5

Left pan up   

The odd ball is either the right H or the left L.

Proceed to Third Weighing Case 6

Third Weighing   
Case 1   U N
Pans balance   Not possible

Left pan down   The odd ball is this U, and its heavier

Left pan up   The odd ball is this U, and its lighter

Case 2   H H
Pans balance   The odd ball is the remaining unweighed H (heavier)

Left pan down   The odd ball is the left H (heavier)

Left pan up   The odd ball is the right H (heavier)

Case 3   L L

Pans balance   The odd ball is the remaining unweighed L (lighter)

Left pan down   The odd ball is the right L (lighter)

Left pan up   The odd ball is the left L (lighter)

Case 4   L L

Pans balance   The odd ball is the remaining unweighed H (heavier)

Left pan down   The odd ball is the right L (lighter)

Left pan up   The odd ball is the left L (lighter)
Case 5   H H

Pans balance   The odd ball is the remaining unweighed L (lighter)


Left pan down   The odd ball is the left H (heavier)

Left pan up   The odd ball is the right H (heavier)

Case 6   H N
Pans balance   The odd ball is the remaining unweighed L (lighter)

Left pan down   The odd ball is this H (heavier)

Left pan up   Not possible

 


The Truth remains the Truth regardless of our beliefs or opinions the Truth is always the Truth even if we know it or do not know it (The Truth remains the Truth)

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Offline lightarrow

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« Reply #15 on: 04/07/2008 13:21:26 »
First Weighing   UUUU UUUU
...
Left pan down   
One of the four balls in the left pan  might be heavier (call them HHHH from now on) or one of the four balls in the right pan might be lighter (call them LLLL from now on).
Proceed to Second Weighing Case 2
...
Second Weighing   
...
Case 2   HHL HLN
Where did you take that "N" ball?

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Offline Alan McDougall

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« Reply #16 on: 04/07/2008 15:47:23 »
Quote
Where did you take that "N" ball?

I assume from where did Get the N ball, I got it from the unweighed group of 4 balls that I knew were normal n balls. After using it I put it aside.

Cut out 12 peaces of paper and imagine you are doing the problem on an imaginary scale. You will see the solution I gave is the correct one. 
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Offline lightarrow

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« Reply #17 on: 05/07/2008 08:03:20 »
Quote
Where did you take that "N" ball?
I assume from where did Get the N ball, I got it from the unweighed group of 4 balls that I knew were normal n balls. After using it I put it aside.
Right, thank you.
Very interesting problem! Thank you for posting it. I'm waiting for the next ones!

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Offline Alan McDougall

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« Reply #18 on: 06/07/2008 08:32:14 »
Oh by the way people I said the problems were easy and this is not true about the odd ball problem.

I was given that problem over 30 years ago my a work colleague (and solved it myself).

I have posed the problem to numerous People since then and only one person in all that time was able to give the correct answer.

If you did not solve it believe me you are in good company.

Regards

Alan

I will think of another and come back
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Offline Bored chemist

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« Reply #19 on: 06/07/2008 16:13:51 »
Just to confuse people further. Did you know it's possible to answer the question about the balls without any weighings i.e. without the scales or even in zero gravity?
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Offline lightarrow

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« Reply #20 on: 06/07/2008 19:18:20 »
Just to confuse people further. Did you know it's possible to answer the question about the balls without any weighings i.e. without the scales or even in zero gravity?
Making a ball at rest collide with another ball and see if this last one stops or goes ahead or goes back after collision?

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Offline Bored chemist

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« Reply #21 on: 07/07/2008 06:56:30 »
Yes, basically play snooker with them and the odd one should show up.
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Offline lightarrow

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« Reply #22 on: 07/07/2008 10:50:28 »
Yes, basically play snooker with them and the odd one should show up.
Ok, then, for the others readers I explain:
If you collide a ball A at rest with another ball B of the same dimension and mass (the two balls' centers of mass aligned with the trajectory) then after collision B ball stops completely; if instead the B ball were lighter, after collision it bounce back (slower); if it were heavier, it keeps going in the same sense (slower).

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Offline Alan McDougall

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« Reply #23 on: 07/07/2008 15:07:01 »
LA,

Correct but it might be difficult with balls that differ minutely.
And it would not solve the 12 ball problem .
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Offline Bored chemist

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« Reply #24 on: 07/07/2008 19:02:40 »
"Correct but it might be difficult with balls that differ minutely."
Same as a set of scales then.

"And it would not solve the 12 ball problem ."
Yes it does.
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Offline Alan McDougall

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« Reply #25 on: 07/07/2008 19:43:19 »
"No it does not" If luck is not on your side your solution might require as much as 36 bouncing combinations before finding the odd ball.

I am aware the less massive ball will bounce back. In my youth I also played snooker.

Alan
« Last Edit: 07/07/2008 19:45:03 by Alan McDougall »
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Offline lightarrow

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« Reply #26 on: 07/07/2008 19:54:54 »
"No it does not" If luck is not on your side your solution might require as much as 36 bouncing combinations before finding the odd ball.
Actually, with only one shot, if you put all the balls in a row, quite distant one another, and you shoot the first (or the last). [:)]

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Offline Alan McDougall

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« Reply #27 on: 07/07/2008 22:46:15 »
Sorry it makes no sense!
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Offline lightarrow

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« Reply #28 on: 08/07/2008 13:59:41 »

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Offline Alan McDougall

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« Reply #29 on: 08/07/2008 16:03:38 »
It still makes no sense you are only allowed three weights/bounced/shots

Shots

If I understand you correctly, you want to bounce the balls against each other,hopefully by careful observation you would see the ball with less mass  bounce futher back than the one with the greater mass.

To do this you can choose and ball and bounce it against any other ball. If luck is not on your side it would need 11 bounces to slove the problem. Not three weighs on a pivot scale as in my 12 odd ball solution
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Offline Bored chemist

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« Reply #30 on: 08/07/2008 19:49:39 »
The whole point is that this way I need precisely zero weighings, I might need a dozen shots or so (I'm alousy player) but nobody said anything about them in the original question. In principle I could, as Lightarrow says, do it with one shot.I could do this in zero gravity with no weighings so that's less than three or do you somehow think 3 is not more than zero?

Let's make this as clear as I can.
I can solve the problem of finding the odd ball and whether it's light or heavy with zero weighings.
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Offline Alan McDougall

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« Reply #31 on: 09/07/2008 08:11:04 »
OH!!Man come on!! be real, your experiment solution would require a trip the the space atation. It is not practicle and completely irrelevant to my 12 all problem.
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Offline lightarrow

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« Reply #32 on: 09/07/2008 23:28:39 »
Alan, can you post the other problems you said to have? I find them interesting.

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Offline Alan McDougall

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« Reply #33 on: 10/07/2008 09:30:31 »
lightarrow,

Ok here is three more, I hope they are of interest!

Try these three problems and if they are found of interest I will pose more

1) Two brothers share a flock of x sheep. They take the sheep to the market and sell each sheep for $x. At the end of the day they put the money from the sales on the table to divide it equally. All money is in $10 bills, except for less than ten excess $1 bills. One at a time they take out $10 bills. The brother who draws first also draws last. The second brother complains about getting one less $10 bill so the first brother offers him all the $1 bills. The second brother still received a total less than the first brother so he asks the first brother to write him a check to balance the things out. How much was the check


2) Four dogs occupy the four corners of a square with side of length a. At the same time each dog starts walking at the same speed directly toward the dog on his left. Eventually all four dogs will converge at the center of the square. What path does each dog follow and what is the distance each dog walks until he reaches the center?

   
3) Five pumpkins are weighed two at a time in all ten sets of two. The weights are recorded as 16, 18,19, 20, 21, 22, 23, 24, 26, and 27 pounds. All individual weights are also integers. How much does each pumpkin weigh?

Some of the problems are from internet, most are mine collected over the years.  I love problems of deductive logical thinking.

We can go the route of the really difficult but that would be no fun I think

Regards

Alan
The Truth remains the Truth regardless of our beliefs or opinions the Truth is always the Truth even if we know it or do not know it (The Truth remains the Truth)

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Offline Bored chemist

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« Reply #34 on: 10/07/2008 18:47:48 »
"OH!!Man come on!! be real, your experiment solution would require a trip the the space atation. It is not practicle and completely irrelevant to my 12 all problem."
Nonsense.
At most, it would require a trip to the local pool hall. In fact anywhere with a reasonably flat table or floor would do.
The fact is that, if you happened to be on a space station, my method would still work, but your's wouldn't.
It's perfectly practical; what's impractical about playing snooker or pool?

Your original question was just a puzzle- any solution that meets the stated objective is valid. In terms of minimising the number of weighings, my solution beats yours.
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Offline Alan McDougall

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« Reply #35 on: 11/07/2008 00:02:39 »
Yes I wanted to say it could be done "your way" on earth. But your suggestion has nothing to do with the puzzle. To say anything that meets the objective is silly and outside the confines of the odd ball problem.

Regards

Alan
The Truth remains the Truth regardless of our beliefs or opinions the Truth is always the Truth even if we know it or do not know it (The Truth remains the Truth)

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Offline lightarrow

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« Reply #36 on: 13/07/2008 12:55:24 »
2) Four dogs occupy the four corners of a square with side of length a. At the same time each dog starts walking at the same speed directly toward the dog on his left. Eventually all four dogs will converge at the center of the square. What path does each dog follow and what is the distance each dog walks until he reaches the center?
The distance walked is a. The path is an exponential spiral with respect to the square's center (a/2;a/2):

r(t) = -a/2 exp[-θ(t)]*{cos[θ(t)] + sin[θ(t)];cos[θ(t)] - sin[θ(t)]} + (a/2;a/2)

|r(t) - (a/2;a/2)| = a/sqrt(2) exp[-θ(t)]

About the distance walked my reasoning is probably not simple:
after a time dt the dog (I considered the one at the left down corner in the origin of coordinates) walks for vdt up and vdt at right so he is now at a distance sqrt[(a-vdt)^2 + (vdt)^2] ~ a - vdt so it's now in a corner of a new square whith side of lenght: a - vdt; so the differential variation of the square's side lenght is da = (a - vdt) - a = -vdt and this equation is valid in every moment. Integrating from 0 to t:

∫da = ∫-vdt --> a(t) - a = -vt --> a(t) = a - vt

When they meet, at the instant T, they are in a square of side lenght 0, so:

0 = a - vT --> T = a/v

Since the speed v is constant, the total lenght L is simply v*T:

L = v*a/v = a.

About the path followed, my reasoning was even less simple, because I solved it analitically (complicated computations), so I assume there must be a simpler way.

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Offline Bored chemist

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« Reply #37 on: 13/07/2008 17:34:43 »
"To say anything that meets the objective is silly and outside the confines of the odd ball problem."
Yet it still solves the problem of spotting the oddball, and it does so with fewer weighings.
« Last Edit: 13/07/2008 17:43:40 by Bored chemist »
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Offline Alan McDougall

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« Reply #38 on: 14/07/2008 09:34:51 »
Lightarrow,

Lightarrow,

I like you approach to this not so simple little problem






Hard Solution
It is intuitive that at all times the dogs will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0).

Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5).

Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1.

Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.

Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).

So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e-x.

At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2-1/2. So r(x)=2-1/2*e-x.

The arc length is given by the formula:

Integral from 0 to infinity of ((f(x))2+(f'(x))2)1/2 =

Integral from 0 to infinity of ((2-1/2*e-x)2 + (-2-1/2*e-x)2)1/2 =

Integral from 0 to infinity of (2-1*e-2x + 2-1*e-2x)1/2 =

Integral from 0 to infinity of e-x =
-e-x from 0 to infinity =
=0 - (-e0) =
0 -(-1) = 1 The path the dogs take is called a logarithmic spiral.

It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e-2*pi/21/2 = 0.03055663 times it's size before the revolution.
 
Easy Solution
David Wilson suggested the more simple solution which follows...
Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.

Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs.

He also provided the following graphics.
For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)).

So a pentagon will have angles of 108. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines.

In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180)=-0.309 units towards dog A.

The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472.

What do you think?

Alan

The Truth remains the Truth regardless of our beliefs or opinions the Truth is always the Truth even if we know it or do not know it (The Truth remains the Truth)

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Offline Alan McDougall

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« Reply #39 on: 15/07/2008 06:01:00 »
OK,

How about this little puzzle.

A horse is tied to a stake with a rope 5 metres in length, 10 metres in front of him is a bunch of nice hay. The hay is firmly tied down and can not be moved.

The horse somehow gets to the hay and begins to eat. How does he do it?

Regards
Alan
The Truth remains the Truth regardless of our beliefs or opinions the Truth is always the Truth even if we know it or do not know it (The Truth remains the Truth)

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Offline Alan McDougall

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« Reply #40 on: 15/07/2008 06:15:28 »
Now try this

Please don't muse over the request act immediately you read what you must do

Before anything get a pencil and paper ready. please don't scroll down yet

OK scroll down now:.................? far down please!!
































































































Try to think of two objects or shapes "one inside the other" under the circumstances I suggested and "IMMEDIATELY DRAW THEM"
as soon as the image pops into your mind.

Note don't reveal what you drew I will attempt to predict that later, after a few replies, such as OK Alan I have my drawing!!
The Truth remains the Truth regardless of our beliefs or opinions the Truth is always the Truth even if we know it or do not know it (The Truth remains the Truth)

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Offline BenV

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« Reply #41 on: 15/07/2008 09:50:38 »
OK,

How about this little puzzle.

A horse is tied to a stake with a rope 5 metres in length, 10 metres in front of him is a bunch of nice hay. The hay is firmly tied down and can not be moved.

The horse somehow gets to the hay and begins to eat. How does he do it?

Regards
Alan

Forgive me if I'm being too logical about this one, but horses are pretty good at chewing through rope.  If was a chain, I'd have to think about it again...

Oh, and I've thought of a shape within a shape.

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lyner

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« Reply #42 on: 15/07/2008 13:23:06 »
Perhaps someone takes pity on him?

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Offline Alan McDougall

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« Reply #43 on: 15/07/2008 13:39:58 »
No no guys with repect there is a more logical answer. The horse must get to the hay all on his own, no outside help permitted

Regards

Alan
The Truth remains the Truth regardless of our beliefs or opinions the Truth is always the Truth even if we know it or do not know it (The Truth remains the Truth)

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Offline BenV

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« Reply #44 on: 15/07/2008 15:00:48 »
So "he chews through the rope" doesn't score any points?

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Offline Bored chemist

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« Reply #45 on: 15/07/2008 20:18:56 »

I don't know how to hide an answer, but I doubt this is the answer being looked for.
The puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought was that he chews through, or breaks the rope.)
« Last Edit: 15/07/2008 20:24:06 by Bored chemist »
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Offline LeeE

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« Reply #46 on: 15/07/2008 20:28:50 »
OK,

How about this little puzzle.

A horse is tied to a stake with a rope 5 metres in length, 10 metres in front of him is a bunch of nice hay. The hay is firmly tied down and can not be moved.

The horse somehow gets to the hay and begins to eat. How does he do it?

Regards
Alan

The horse is standing at the extreme length of the rope, five meters from the stake.  The horse is facing the stake, and five meters further away in the same direction is the hay.  The horse walks the ten meters across the 'circle' and eats the hay.
...And its claws are as big as cups, and for some reason it's got a tremendous fear of stamps! And Mrs Doyle was telling me it's got magnets on its tail, so if you're made out of metal it can attach itself to you! And instead of a mouth it's got four arses!

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Offline Alan McDougall

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« Reply #47 on: 16/07/2008 00:23:25 »
BoardChemist,

Quote
I don't know how to hide an answer, but I doubt this is the answer being looked for
.

"" to the mind guess"" of prediction test in  "blue letters", This is post separate from the horse hay one. 

Quote
The puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought
was that he chews through, or breaks the rope.)

Yessssssss!! The above is the correct answer a little silly I know but many people most often just presume the skate is firmly hamered into the ground

Regards

Alan
The Truth remains the Truth regardless of our beliefs or opinions the Truth is always the Truth even if we know it or do not know it (The Truth remains the Truth)

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Offline BenV

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« Reply #48 on: 16/07/2008 17:57:06 »
I'm fairly sure I know the answer to the pumpkins one, just haven't yet given myself the time to write it out! Back with you soon on that...  and knowing horses, a horse would have chewed through the rope anyway, even if the stake wasn't in the ground.  They just like to be a pain.

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Offline Pumblechook

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« Reply #49 on: 16/07/2008 21:26:24 »
Lost track of all that..

1)Weigh 4 against 4..   Either balance or not.

2)Swap 1 from one pan to the other...  Replace 3 balls with 3 un-weighed balls.

Lock at change of state between first and second weighing.  In all cases you can find the odd one..  Might have to indentify one as a control. 

It come down to one of two balls, lighter or heavier... Weigh one against a control.

OR one of the three where you now know you are looking for a lighter or heavier.  Weigh any two together.