Puzzles

[firewalker99]

I agree. It's like saying prove that 1 = 0.9999999999.....~

NB: 1 = 0.9999999...~

Precisely, exactly. No argument.

That on the other hand is an easy one
suppose x = 0.99...

10x = 9.99...
-x = -0.99...
------------------
9x = 9
==> x=1 ==> 0.99... = 1

 

[CYOF]

Yes I thought some time more over that riddle, but "my son" still seems to fit the question...?
Obviously not MY son, but the son of the person speaking that riddle.

Ah, yes , it is His Son.

But another good lesson, we must be very clear, the same as with trading, confusion will cause errors.

My Son and His Son are not the same, even though in your mind they were, but in many others they will be mean exactly what they say.

 

[frugi]

Here's one I dug up from ages ago...

Four people have hats atop their heads: two white; two black.
They all know that there are two white and two black hats in total.
They are all looking to the left and cannot see those to their right.
They cannot see their own hats.
The unfortunate that lurks behind the brick wall is invisible to all, as he is to them, but everyone knows he is there.
Nobody will be deliberately silent.

They can't speak to each other

Eventually one of them pipes up and says with full certainty what colour his/her hat is. Who and why?

Here is a piccie:

http://www.trade2win.com/boards/attachment.php?attachmentid=8632

 

[new_trader]

Here's one I dug up from ages ago...

Four people have hats atop their heads: two white; two black.
They all know that there are two white and two black hats in total.
They are all looking to the left and cannot see those to their right.
They cannot see their own hats.
The unfortunate that lurks behind the brick wall is invisible to all, as he is to them, but everyone knows he is there.
Nobody will be deliberately silent.

They can't speak to each other

Eventually one of them pipes up and says with full certainty what colour his/her hat is. Who and why?

Here is a piccie:

http://www.trade2win.com/boards/attachment.php?attachmentid=8632

I can't see the piccie, so I am going to assume I understand the puzzle and have the answer. The 2nd man pipes up and says he is wearing the opposite colour of the 1st man.

The reason: If the 3rd man could see 2 white hats or 2 black hats he would know without doubt that he must have the opposite colour and would call out regardless of the fact that he can't see the man behind the wall. The fact that he sees 1 black and 1 white is obvious by his silence. The 1st man can't see any hats, so he can't call out and neither can the man behind the wall.

 

[frugi]

You've got it, with no picture too. Nice one! Although (I think) your numbering convention is different from mine.

So just to tidy up loose ends for t'others, here's the piccie attached again. Let's number the stick men 1 - 4 from the left.

I'm assuming from the above that in this case you'd say #3 will pipe up.

 

[blackcab]

here's the piccie attached again

Good one frugi.

Here's a classic: a rope passes over a pulley. On one end of the rope is a 140lb weight and on the other is a man weighing exactly 140lb. The man starts to climb the rope. What happens to the weight?

 

[frugi]

Blackcab, that reminds me of the one where there is an aeroplane on a conveyor belt. As the pilot fires up the engines the plane's wheels turn (as the plane is being pushed forward by the engines) but the conveyor belt goes backwards, always matching the wheels' speed. Can the plane take off?

In your scenario when the two bodies are at rest it does not matter how high off the ground the two masses are in relation to each other (ignoring the negiible difference in gravity?) so once he has finished climbing the weight will be where it was when he started. But during the climb when he pulls down with his arms he will exert an extra force (in addition to his weight), thus every time he climbs up a bit the weight will rise in sympathy (by the length of rope he pulls through his hands) until he relaxes his arms whereupon it will drop back down to where it was. I bet I'm horribly wrong as physics was definitely not a forté. :

One more: In a coin throwing game, you win a pound for every head thrown before you throw the first tail. How much do you expect to win?

 

[firewalker99]

Good one frugi.

Here's a classic: a rope passes over a pulley. On one end of the rope is a 140lb weight and on the other is a man weighing exactly 140lb. The man starts to climb the rope. What happens to the weight?

Good for some thought blackcab, nice one!

As the monkey is pulling on the rope, hence exercising more downwards force (his own weight + the pulling strength); the weight will go up, not?

 

[blackcab]

One more: In a coin throwing game, you win a pound for every head thrown before you throw the first tail. How much do you expect to win?

It looks like the sum to infinity of 1/2^k, so it would be one pound - is that right?

 

[blackcab]

Good for some thought blackcab, nice one!

As the monkey is pulling on the rope, hence exercising more downwards force (his own weight + the pulling strength); the weight will go up, not?

Actually I posted that because I remembered it from a book of my father's I used to enjoy as a child - Pastimes and Diversions. But after I posted I looked it up on Google and it turns out to be a horrible question full of unrealistic assumptions like frictionless pulleys and massless ropes, and generally all too much like Applied Maths A level for me to enjoy :cheesy:

 

[frugi]

All I can remember from Applied Maths* is why the cat slid off the roof: it didn't have a big enough mu. :cheesy:

Sorry I don't actually know the answer to the coin one!

*not something I'm proud of - indeed I despair of people who actually boast about being poor at maths.

 

[firewalker99]

The Book Worm

A book worm wants to eat its way through an encyclopedia consisting of 25 volumes. Each volume has a carton cover of 1cm thick. The books are placed in a book case in the normal way (standing vertical, so you can read the title on the book spine). The first 4 parts are 6 centimeters thick each (without the cover), the following 12 are 9 centimeters each, and the other books are 4,5cm. The worm starts to eat it's way from the front of volume 1. How many centimeters of paper and carton has he eaten when he reaches the back of volume 25?

This one looks easy, but there's one thing most people overlook...

 

[trendie]

A book worm wants to eat its way through an encyclopedia consisting of 25 volumes. Each volume has a carton cover of 1cm thick. The books are placed in a book case in the normal way (standing vertical, so you can read the title on the book spine). The first 4 parts are 6 centimeters thick each (without the cover), the following 12 are 9 centimeters each, and the other books are 4,5cm. The worm starts to eat it's way from the front of volume 1. How many centimeters of paper and carton has he eaten when he reaches the back of volume 25?

This one looks easy, but there's one thing most people overlook...

this is a good one, FW!!
There is a variation whereby the distance from vol-1 to vol-25 is used as a fuse-burn timer to trigger a bomb, and the question is how long before the bomb goes off.

 

[stoic]

Had a go at this, solved it on my second attempt, couldn't be bothered to read your solution firewalker99 since it seemed all over the place. This problem is an exercise in maximising as much information at each step as possible:

Solution to the 12 ball problem:

weight 4 balls against 4 balls: two things happen:

1) they balance, we are now only interested in the 4 remaining balls, goto 2A

2) they do not balance, goto 2B

2A: name remaining balls 1,2,3,4. weight 1 and 2, against 3 and a dummy ball (taken from the 8 that balanced. again 2 things happen. If this balances, we know that ball 4 is the culprit and we weight that against a dummy ball to ascertain whether it is lighter or heavier. If it does not balance goto 3A

2B: number the balls on the right scale 1,2,3,4 and on the left 5,6,7,8. Note which way the scale tilted, without loss of generality assume the right side is heavier (we can reverse the numbering otherwise). We now weight 1,5,6 against 7,2,8. Again 2 things happen. If they don't balance goto 3B, But if they balance we know that balls 3 and 4 are the culprits, so all we do is to balance ball 4 against a dummy ball (ball 1 say) and if it balances we know that ball 4 is lightest, if not then ball 3 will be the lightest (we can also weight 3 against 4)

3A: we not the which side is heavier, assume it is the right side (same treatment for left side). We not balance balls 2 and 3 against 2 dummy balls. 3 things happens. If they balance we know that ball 1 is the lightest ball. If the right side is still heavy we know that ball 2 is the lightest. If the left side is heavy we know that ball 3 is the heaviest.

3B: if the right side is still heavier then we know that balls 1,7,8 are the culprits, if the left side is heavier we know that balls 5,6,2 are the culprits. We now need to deal with 3 balls, so assume that 1,7,8 are the culprits (treaments of 5,6,2 is similar), Weight balls 1,7 against 2 dummy balls. 3 things happen. If they balance we know 8 is the heaviest (since the weighing before the right side was heavier). If the right side is still heavier we know that ball 1 is the lightest. If the left side is heavier we know that ball 7 is the heaviest

 

[blackcab]

I just caught sight of stoic saying he's done the 12 balls one but am not reading it - I'm stuck on it. Is there definitely a solution?

 

[firewalker99]

I just caught sight of stoic saying he's done the 12 balls one but am not reading it - I'm stuck on it. Is there definitely a solution?

yes stoic's response is excellent... there is a solution and it's a nice one imo
I finally found it too, but was too lame to write it out completely here...
credits to stoic!

 

[new_trader]

Had a go at this, solved it on my second attempt, couldn't be bothered to read your solution firewalker99 since it seemed all over the place. This problem is an exercise in maximising as much information at each step as possible:

Solution to the 12 ball problem:

weight 4 balls against 4 balls: two things happen:

1) they balance, we are now only interested in the 4 remaining balls, goto 2A

2) they do not balance, goto 2B

2A: name remaining balls 1,2,3,4. weight 1 and 2, against 3 and a dummy ball (taken from the 8 that balanced. again 2 things happen. If this balances, we know that ball 4 is the culprit and we weight that against a dummy ball to ascertain whether it is lighter or heavier. If it does not balance goto 3A

2B: number the balls on the right scale 1,2,3,4 and on the left 5,6,7,8. Note which way the scale tilted, without loss of generality assume the right side is heavier (we can reverse the numbering otherwise). We now weight 1,5,6 against 7,2,8. Again 2 things happen. If they don't balance goto 3B, But if they balance we know that balls 3 and 4 are the culprits, so all we do is to balance ball 4 against a dummy ball (ball 1 say) and if it balances we know that ball 4 is lightest, if not then ball 3 will be the lightest (we can also weight 3 against 4)

3A: we not the which side is heavier, assume it is the right side (same treatment for left side). We not balance balls 2 and 3 against 2 dummy balls. 3 things happens. If they balance we know that ball 1 is the lightest ball. If the right side is still heavy we know that ball 2 is the lightest. If the left side is heavy we know that ball 3 is the heaviest.

3B: if the right side is still heavier then we know that balls 1,7,8 are the culprits, if the left side is heavier we know that balls 5,6,2 are the culprits. We now need to deal with 3 balls, so assume that 1,7,8 are the culprits (treaments of 5,6,2 is similar), Weight balls 1,7 against 2 dummy balls. 3 things happen. If they balance we know 8 is the heaviest (since the weighing before the right side was heavier). If the right side is still heavier we know that ball 1 is the lightest. If the left side is heavier we know that ball 7 is the heaviest

Mostly correct, but you made a typo in 2B

But if they balance we know that balls 3 and 4 are the culprits, so all we do is to balance ball 4 against a dummy ball (ball 1 say) and if it balances we know that ball 4 is lightest, if not then ball 3 will be the lightest (we can also weight 3 against 4)

If ball 4 balances with a known good ball then it is ball 3 which is odd light.

 

[stoic]

Mostly correct, but you made a typo in 2B



If ball 4 balances with a known good ball then it is ball 3 which is odd light.

actually 2B should read: number the balls on the leftscale 1,2,3,4 and on the right 5,6,7,8, but then again I was making a "stream of conciousness" proof to see how well my thoughts translated to the keyboard. I left the mistake there on purpose waiting to see if anyone spots it.

This problem is interesting in that most people will try to do a weighing should be giving more information away. Take for example step 2A, you could just weight 1 and 2 against 3 and 4 but that has terrible repercussions. It's only when I realised that then I solved the problem. This is an exercise in being seeking out an optimal path, instead of being lazy

 

[blackcab]

I've just done the 12 balls puzzle. What a *******! But a good *******.

I haven't read any of the posts about it, so here's my own home brew solution.

Key:
"K" = a ball that you know for certain is not the odd ball
"U" = a ball that may be the odd ball that is light (goes Up)
"D" = a ball that may be the odd ball that is heavy (goes Down)

(Weigh 1) Weigh 4 balls on each side

If they balance then the odd ball is in the other 4, so do this:

Put aside one of those other 4

(Weigh 2) Weigh the remaining 3 against any 3 of the balls in Weigh 1

If they balance then the odd ball is the one you put aside. To find out if it's heavier or lighter, weigh it against any of the other balls (Weigh 3) to get the answer

Otherwise (they don't balance in Weigh 2) the odd ball is one of the 3, and because you weighed them against 3 'known' balls, you now know if it's heavier or lighter. To find which one it is, weigh one of them against another (Weigh 3). If they don't balance, that tells you which ball it is, and if they do balance, you know it's the other one of the 3​

Otherwise (they don't balance in Weigh 1), do this:

Put aside one ball from the side that went up in Weigh 1 and two balls from the side that went down, remembering which ball came from which side (call them UDD). That leaves you with UUU | DD on the scales

Swap around one U and one D so that you have DUU | DU on the scales

Add one of the balls from the 4 that you have not weighed yet to the side that has only 2 balls, leaving 3 balls on each side: DUU | DUK

(Weigh 2) Weigh them

If they balance then the odd ball is one of the UDD trio that you just put aside, and to find which one, weigh the two Ds against each other (Weigh 3). If they don't balance, then the odd ball is the one that goes down and is heavier. If they do balance then the odd ball is the U from that trio and is lighter

Otherwise, if the sides go up/down in Weigh 2 in the same way that they did in Weigh 1, then it's one of the 3 balls that you did not swap around in Weigh 2 (i.e. it's one of the trio UUD). To find which one, weigh the two Us against each other (Weigh 3). If they don't balance, then the odd ball is the one that goes up and is lighter. If they do balance then the odd ball is the D from that trio and is heavier

Otherwise (the sides go up/down in Weigh 2 in the opposite way that they did in Weigh 1), the odd ball is one of the pair of balls that you swapped around in Weigh 2 (UD). To find which one, weigh the U against any of the other balls (except the D of that pair) (Weigh 3). If they do not balance then the odd ball is the U and is lighter, otherwise the odd ball is the D of the pair and is heavier​

 

[blackcab]

You have two identical barrels, one half full of water and the other half full of whisky. You take one teaspoon of water from the water barrel, put it into the whisky barrel and give it a stir. Then you take one teaspoon of this mixture from the whisky barrel and put it into the water barrel. Which of the barrels contains more of its original contents? Ignore physical properties of liquids like density.

This always used to stump me.