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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 »


 

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 »
 

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
 

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??
 

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
 

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?
 

Offline Alan McDougall

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Try solving these easy problems by deductive thinking
« 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
 

Offline lightarrow

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Try solving these easy problems by deductive thinking
« 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.
 

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?
 

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...  :))
 

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?
 

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
 

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?
 

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.
 

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

 


 

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?
 

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. 
 

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!
 

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
 

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?
 

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?
 

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.
 

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).
 

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 .
 

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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« Reply #24 on: 07/07/2008 19:02:40 »

 

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