Pint glass challenge/problem

[abskebabs]

Hi everybody, I found this in a book full of mathematical puzzles and I thought it would be enjoyable for people to have a go at it. Ok now imagine you have a 10 pint glass, a 7 pint glass a tap and nothing else which you could use as a container. Do not worry how much fluid the tap can give you, for the purposes of this question I will say it is water and there is a potentially infinite amount of it, but you can deem it to be whatever you're imagine sees fit;) .

 

Can you obtain exactly 9 pints of water using only the 3 things available to you?

It is possible to empty the glasses, fill them using the tap, or by pouring the contents of one into the other. However, the puzzle cannot be solved by pouring only partial amounts of water into the 10 pint glass and trying to measure 9 pints, as you are required to get EXACTLY 9 pints, and this would be an unreliable form of measurement as these glasses have no scales on them.

 

If you can do that then try to get 8 pints. If you can do that, see if you get either of the 2 results by employing different methods. Can you try getting other amounts with different glasses, like say 1 pint from an 8 pint glass, 5 pint glass and a tap? Are you beginning to see a pattern emerging?

 

Now once you have done all that you can get to the most interesting part of this thread. It is an investigation into the general problem. I would advise tackling the above part of this post before before attempting this.

 

So imagine now that you have a glass of A pints and another glass of B pints( so both of these are unknowns).Now imagine that the only thing you know is that the A pint glass is larger than the B pint glass. This is a condition I am setting. Now investigate what happens if you try to fill one with the other, and continue to do this. What do you observe? Can you comment on this? Do you have to set certain conditions to decide on what outcomes are possible? When you do this please relay the results of your investigation. In the mean time I will be attempting to tackle this general problem myself.

 

Have fun!

[YT2095]

pour the 10 into the 7, 3 times, you`ll get 3 pints come out each time making 9 pints.

10-7=3

3x3=9

 

for 8, you`de pour the 7 into the 10, 4 times and get 2 lots of 4 making 8.

7+7=14

14-10=4

4x2 (coz you do it twice) = 8

 

EDIT: I`m of the opinion that 1,2 and 5 are not possible without a 3`rd vessel, as 8 and 9 outlined above went on the floor as the correct amounts.

[JohnF]

Fill the 10

Use the full 10 to fill the 7 - leaving 3 in the 10

Empty the 7

Pour the 3 from the 10 into 7 - the 7 now holds 3 and the 10 is empty

Fill the 10

Use the 10 to fill the 7 - It will take 4 this time leaving 6 in the 10

Empty the 7 and pour the 6 from the 10 in to the 7

Fill the 10

Fill the 7 from the 10 - It takes 1 this time leaving 9 in the 10

[John Cuthber]

Since I can deem the tap to dispense whatever I want I will have it dispense 10%v/v alcohol. Not only does that mean that to get 9 pints of water all I have to do is fill the 10 pint jug (OK there's something else there too, but it's 9 pints of water.), but I can get pleasntly pickled later on and start thinking about numerology.

[JohnF]

Continuing from post #3

 

Now there is 9 in the 10 and the 7 is full

Empty the 7

Fill the 7 from the 10 - the 10 had 9 in it so now there is 2 in the 10

Empty the 7

Pour the 2 from the 10 in to the seven

Fill the 10

Fill the 7 from the 10 - The 7 had 2 in it so it takes 5

There is now 5 in the 10

Empty the 7

Pour the five from the 10 in to the 7

Fill the 10

Fill the 7 from the 10 - it takes 2 leaving 8 in the 10

[YT2095]

so we have 10, 9 and 8 established.

fill the 7 don`t empty it.

 

now we have 10,9,8 AND 7 established

[grifter]

pour the 10 pint glass (filled) in to the 7 pint glass, twice, the excess gets you 6...

 

2(10-7) = 6

 

general rule would equal

 

X(A-B)=3X

 

re arranges to give

 

X/3(A-B)=X

 

where A is 10 and B is 7...

 

so if you wanted to find how many times to fill the 7p glass from the 10p glass to get X number of pints, substitute the value of pints for X

 

e.g. 15 pints would equal

15/3(10-7)

5(10-7)

 

and yes, no suprises it equals 15

 

hope this isn't to obvious but just incase anyone hadn't allready worked it out

[JohnF]

pour the 10 pint glass (filled) in to the 7 pint glass, twice, the excess gets you 6...

 

You can't do this. How do you hold the 2 lots of 3 pints?

 

Pour from the 10 in to the 7 leaving 3 in the 10 pint glass. You can now either fill the 10 pint glass or empty it and you will then have either...

 

1. An empty 10 pint glass and a full 7 pint glass.

or

2. 3 pints in the 10 pint glass and a full 7 pint glass.

[YT2095]

can I Edit this thread and change the water into Beer, that way I`m SURE you will all be much more careful with the Maths (for a while).

[abskebabs]

I must say I am pleasantly surprised at the reception this thread is receiving. If only the other thread I placed in the maths part of this forum got this much attention:rolleyes: .

 

Anyway, John F seems to have done this puzzle the same way I did, and kudos to him for working it out. YT I think I see what you and the Grifter are getting at in terms of multiplying the original difference in volume of the 10 pints by the No of times the 7 pint glass is filled using the 10 pint glass, though I have had difficulties relating the differences observed in the quantities as you go through dividing 10 pints between the 2 glasses to what you have stated.

 

Nevertheless the rule Grifter has stated is of limited scope, and is not general. In the case mentioned above it only works if you refill the 7 pint glass 3 times, i.e. when the amount in the is 9. If I was to rely on your formula, then the 4th time I fill the 10 pint glass and tip it over to the 7 pint I would get 12 pints, which is clearly nonsense as I can't even get this amount into the 10 pint glass.

 

Also the rules states that this is true for any values A and B. This is not true, and it can be seen that the number 3 written is just the difference between 10 and 7 in the specific case where A=10 and B=7.

Therefore your statement is really just:

[math]X(A-B)=X(A-B)=R[/math]

Where I have put R to represent the remaining amount in the 10 pint glass after every "step".

 

To illustrate, I will rewrite your formula:

[math]3X=X(A-B)[/math]

You said this was valid regardless of the values of A and B, but it cannot be general, as it doesn't work with any other values of A-B!

 

In case I may have jumped to conclusions I will provide further commentary in a little while.

[Pre4edgc]

Well, from what I see with both of your examples (10 and 7, 8 and 5) They are both seperated by three, which would mean that in order to get a certain amount, they would end up with a multiple of three in both containers. ie Eight goes into five. Three in eight, five in five. Empty five. Pour three into five. Fill eight. Pour two into five. Six in eight, five in five. Both end up with multiples of the difference. So would that mean the same if you had ten and six instead of ten and seven? I think that the amounts you can get are directly restricted to the multiples of the difference of the two amounts... In my opinion at least.

 

Okay. TIme for an update. Yeah, maybe it's been just ten minutes, but I may have found a formula that tells the number of times you can pour before you get the amount maxed by B in A. I repeated it, and it worked just fine. I had it where (A-B)^2=B. I tested it with a slightly different equation rather than the 10 and 7. I turned it to 10 and 6. And it still worked! I worked it out, and after a repeated 16 steps for that last one, it turned out that you'd get 6 pints in the 10 pint container, thus causing a repeat in the pattern. It works for the 10 and 7 one as well. I mainly had it to where the amount in the 10 container changed each time to an amount less than ten after it filled the other container. Like when ten filled 7, leaving 3 in it. That I said was step 1. Then when you emptied 7, filled it with the three, refilled the ten, and dumped four into the seven, and left 6 in the ten, it became step 2. I haven't worked a way to get an exact answer of steps to get a particular number, seeing as the ten and six one only has four different amounts...

[grifter]

where A is 10 and B is 7

 

i did actually write the parameters of A and B and yes i realize it is of limited scope and somewhat hopeless, thanks for pointing that out I think perhaps my lapse in intelligence was from experimenting with the formula using beer

[abskebabs]

As I have pretty much wasted today doing absolutely nothing useful apart from briefly browsing the section on vector calculus and electrostatics in my electrodynamics book today, I thought I'd shed some light on what ocurred during my own investigation into the problem I had earlier mentioned here. Who knows I may even be able to focus my mind in the process and then get some revision done:-p !(I even make myself laugh sometimes)

 

Ok, 1st of all I will go through my solution of the 8 pint 5 pint problem for you.

My aim will not be to show you how to get 1 pint though, but to show you what happens if you keep cycling through. Hopefully if you're reading this you've already made a good attempt at solving the problem yourself so I'm not giving away any "spoilers".

 

I will use the LHS bracket of the expression I write down to represent the amount in a 5 pint glass, and the RHS to represent the amount in the 3 pint glass(In hindsight I should have originally given simpler examples with a difference in volume different to 3 pints, but hell with it). By convention(unless mentioned otherwise) from here on you will see the glass that is recognised as holding a larger amount to be the one on the left. You should be able to figure out from the steps when I empty the contents of one glass into another and when I just empty a glass on to the hypothetical floor. I will begin with both glasses empty and will fill the 8 and use that to try and fill the 5. I will proceed from here henceforth:

[math](0,0)\rightarrow(5,0)\rightarrow(2,3)\rightarrow(2,0)\rightarrow(0,2)

\rightarrow(5,2)\rightarrow(4,3)\rightarrow(4,0)\rightarrow(1,3)\rightarrow(1,0)[/math]

[math](1,0)\rightarrow(0,1)\rightarrow(5,1)\rightarrow(3,3)\rightarrow(3,0)\rightarrow(0,3)

\rightarrow(5,3)[/math]

 

Notice we reach the end of our cycle(or just before the beginning depending o n your perspective), because if we carry out the same procedure again, we loop back to the conditions we had at the beginning. The No of stages it takes to get to this point is important(also notice, just before completion the amount in the large glass is always the same as the max amount in the small glass, tho I know thats kinda obvious! ) and something to keep in mind when attempting to see what happens generally. I think also if you keep in mind the answer you have for the following question, it may help you when investigating the general problem: Would the No of stages it takes to solve the above problem differ if we had a 6 pint glass and a 2 pint glass, and carried out a similiar procedure? For that matter would the No of stages differ if we had any multiple of the above ratio between the sizes of the glasses?

 

When tackling the general problem, which I urge you to try, I would recommend laying out your work in the form shown above or in some similiar clear way. You may notice you have to make "choices" along the way concerning possible relative magnitudes between quantities involved(I am referring to more than the obvious initial choice of whether unknown quantity A is larger than B).

 

I may be back to ask some slightly more advanced questions on this problem a little later when I have further advanced my own investigation and noticed patterns within the problem.

 

Happy hunting!

[abskebabs]

i did actually write the parameters of A and B and yes i realize it is of limited scope and somewhat hopeless, thanks for pointing that out I think perhaps my lapse in intelligence was from experimenting with the formula using beer

lol. I think I should try that. I once tried doing my physics homework at the pub, but unfortunately it didn't turn out to be as productive as initially intended!

[Pre4edgc]

Well, about my original hypothesis... It isn't right... I tested it with some other amounts, and it didn't seem to work. I can't get a formula to determine the number of times it can take to get back to square one with any given amount, but like abskebabs, I realize the number of times it takes may be related to the multiple of the differences between A and B. I don't think a formula for this is possible, but perhaps a shortcut through that process. I'll keep some updates while I continue looking through this.

[Comandante]

Hey, something just occured to me.. what book is this?

[abskebabs]

The book is called "Mathematical Puzzling" by A. Gardiner