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Alex's 55-pint Birthday Bonanza


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#1 bonanova

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Posted 30 October 2007 - 09:46 PM

Happy Birthday to me ... HBTM ... etc sang Alex as he strolled into Morty's
last night and then announced, Boys, I've got a real treat for ya tonight!

First, take a look at this. And he hung from the ceiling a complicated,
interconnected set of five scales, whose 10 balance pans he'd labeled
A, B, C, ... H, I, J.

Posted Image

Now the barkeep gave me one request, bein' it's my birthday on Saturday, and all.

He's gonna draw out fifty-five cold pints. Then, if we can place a different number
of pints
into each of these balance pans, we'll have a real party, cuz the drinks
will be on the house!

Every pan gets a pint to start with, and all five scales have to balance.

Alright boys, let's get at it ... we have until midnight to get this done.
Otherwise we pay for the drinks.

Feel free to help Alex, Jamie, Davey and the boys celebrate!

Edit for clarity
The marks on the bars are equally spaced.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#2 Writersblock

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Posted 30 October 2007 - 10:07 PM

ok.
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#3 Writersblock

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Posted 31 October 2007 - 04:22 AM

I think there might be an error here. A lever exerts the force of pressure times length. So J will weigh 8J. H-I will weigh 2(H+I) and E+F+G must = A+B+C+D. Also, A+B+C+D+E+F+G must = 1/2 of 8J+2(H+I). IF this is true, then A+B+C+D and E+F+G must both be fractions.
8J+(2(H+I)). Must be even if J, H, and I are whole numbers. Therefore A+B+C+D and E+F+G must both be odd. Odd+odd =even. Therefore if A+B+C+D and E+F+G are equal and 1/2 of an odd number, they must be a fraction. Right
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#4 bonanova

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Posted 31 October 2007 - 06:02 AM

A+B+C+D+E+F+G must = 1/2 of 8J+2(H+I). = 4J + (H+I)
IF this is true, then A+B+C+D and E+F+G must both be fractions Right? ;)

Spoiler for solution

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#5 Writersblock

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Posted 31 October 2007 - 07:00 AM

Yeah, I get you. I was really close but messed up by leaving 1/2 of the left hand weight off when I tested H and I for what they need to be. Let me give it another shot.
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#6 Writersblock

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Posted 31 October 2007 - 07:06 AM

Yeah, EZ when I don't muff my math. :blush:

Spoiler for solution


Hmmm. 11:00 PM - LET'S PARTY! :lol:
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#7 Writersblock

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Posted 31 October 2007 - 07:11 AM

AS a P.S. - I'd love to see this beer mobile hanging from a bar ceiling!
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#8 Writersblock

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Posted 01 November 2007 - 06:57 PM

So a pint weighs about a pound and a tenth. If you assume a force of 3 pints will upset the balance enough to spill all the drinks, how do you get the pints down to drink them?
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#9 bonanova

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Posted 01 November 2007 - 10:46 PM

So a pint weighs about a pound and a tenth. If you assume a force of 3 pints will upset the balance enough to spill all the drinks, how do you get the pints down to drink them?

Get Alex, Jamie and Davie to remove 2 or 3 glasses at a time,
keeping things balanced at each step.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell




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