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Old July 8 2009, 06:33 PM   #16
ShamelessMcBundy
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Re: Tuesday Night Quiz!

NickInABox wrote: View Post
What is this spin-off bullshit!
It can be the next big thing. Could even become a franchise.
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Old July 8 2009, 06:57 PM   #17
Zion Ravescene
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Re: Tuesday Night Quiz!

Jadzia wrote: View Post
NickInABox wrote: View Post
What is this spin-off bullshit!
I thought it would make a nice change to counting the vowels in a set of place names

Is nobody but Zion and RJ going to make a guess??
I think your overtly complex maths puzzle scared everyone away.
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Old July 8 2009, 07:40 PM   #18
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Re: Tuesday Night Quiz!

Jadzia wrote: View Post
Brian the engineer disassembles an old Warcom press brake, and begins playing with the gears he finds inside it.

He has 3 cogs: A, B, and C. All of which have at least 8 teeth, and no cog has more than 40 teeth.

Cog B is the biggest and has the same number of teeth as cogs A and C added together.

Brian arranges the cogs as shown below. A red dot is drawn at the top of each of the cogs, and cog A has a crank handle attached to it, also shown in the diagram.




As the crank handle is turned, Brian notices the dots rotate out of their positions at different speeds.

He continues to turn the crank handle in the same direction until all three of the dots simultaneously appear back in the positions they started in, and then stops turning.

Brian reports that during the turning, cog C made 45 revolutions more than cog A.

How many teeth do each of the cogs have?

Answers via PM.


This explains why the Warcom was inoperative this morning...

And this gear ratio stuff is too much like work I tell you!
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Old July 8 2009, 07:42 PM   #19
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Re: Tuesday Night Quiz!

Plecostomus wrote: View Post
Jadzia wrote: View Post
Brian the engineer disassembles an old Warcom press brake, and begins playing with the gears he finds inside it.

He has 3 cogs: A, B, and C. All of which have at least 8 teeth, and no cog has more than 40 teeth.

Cog B is the biggest and has the same number of teeth as cogs A and C added together.

Brian arranges the cogs as shown below. A red dot is drawn at the top of each of the cogs, and cog A has a crank handle attached to it, also shown in the diagram.




As the crank handle is turned, Brian notices the dots rotate out of their positions at different speeds.

He continues to turn the crank handle in the same direction until all three of the dots simultaneously appear back in the positions they started in, and then stops turning.

Brian reports that during the turning, cog C made 45 revolutions more than cog A.

How many teeth do each of the cogs have?

Answers via PM.


This explains why the Warcom was inoperative this morning...

And this gear ratio stuff is too much like work I tell you!
I recommending solving them with a few Gyros. Of both sorts.
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Old July 8 2009, 08:25 PM   #20
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Re: Tuesday Night Quiz!

*nails sign to press-brake*

ACHTUNG!
ALLES TURISTEN UND NONTEKNISCHEN LOOKENPEEPERS!
DAS BENDINGSCHNITCLE IST NICHT FÜR DER GEFINGERPOKEN UND MITTENGRABEN! ODERWISE IST EASY TO SCHNAPPEN DER SPRINGENWERK, BLOWENFUSEN UND POPPENCORKEN MIT SPITZENSPARKSEN.

IST NICHT FÜR GEWERKEN BEI DUMMKOPFEN!

DER RUBBERNECKEN SIGHTSEEREN KEEPEN DAS COTTONPICKEN HÄNDER IN DAS POCKETS MUSS.

ZO RELAXEN UND WATSCHEN DER BLINKENLICHTEN.
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Old July 8 2009, 09:30 PM   #21
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Re: Tuesday Night Quiz!

ShamelessMcBundy wrote: View Post
Math is evil and needs to be nuked from orbit. It's the only way to be sure.
Some math is nice - like dividing up slices of cake.
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Old July 9 2009, 12:14 AM   #22
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Re: Tuesday Night Quiz!

Zion Ravescene wrote: View Post
I recommending solving them with a few Gyros. Of both sorts.
Gyros make my gears loose.
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Old July 9 2009, 12:30 AM   #23
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Re: Tuesday Night Quiz!

To anyone who is struggling, here's how to get started...

Suppose the teeth counts are a, b and c. We know that b=a+c.

So that's tooth ratios of a : (a+c) : c.

Those are the only two variables you need; you don't need to introduce any more.

Once you get the logic set up, it unfolds beautifully to give the solution, and it's all elementary algebra.

If cog A does 1 revolution... then what?
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Old July 9 2009, 12:33 AM   #24
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Re: Tuesday Night Quiz!

That's an oxymoron.
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Old July 9 2009, 12:33 AM   #25
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Re: Tuesday Night Quiz!

Jadzia wrote: View Post
To anyone who is struggling, here's how to get started...

Suppose the teeth counts are a, b and c. We know that b=a+c.

So that's tooth ratios of a : (a+c) : c.

Those are the only two variables you need; you don't need to introduce any more.

Once you get the logic set up, it unfolds beautifully to give the solution, and it's all elementary algebra.
You're right. I definitely did it the hard way.
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Old July 9 2009, 12:37 AM   #26
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Re: Tuesday Night Quiz!

I came out with the same answers as before.

I just can't do Math.
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Old July 9 2009, 11:22 PM   #27
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Re: Tuesday Night Quiz!

WINNERS:
Zion was the only one with the correct answer.


ANSWERS:
Cog A has 28 teeth; Cog B has 36 teeth; Cog C has 8 teeth.


SOLUTION:
As promised, an elegant solution to the puzzle.

1. Suppose the teeth counts are a, b and c. We know that b=a+c.

So the tooth ratios are a : (a+c) : c

2. Now work out the turning ratios:

If cog A does 1 revolution, cog B does a/b revolutions = a/(a+c) revolutions. Meanwhile cog C does a/c revolutions.

So the turning ratios are 1 : a/(a+c) : a/c

Now let us suppose that a,b,c are triply coprime. If we multiply the number of teeth on each cog by some positive integer k, that will not affect the mechanics since the turning ratios will remain the same. (eg, a/c = ka/kc)

What this condition ensures is that there is no smaller set of cogs which solves the problem.

I will continue to count the teeth as a,b,c, although a,b,c are now triply coprime. I allow myself to scale a solution up by some positive integer k, without loss of generality.


3. Scale up the turning ratios by (a+c).c to eliminate the denominators in the turning ratios.

c.(a+c) : a.c : a.(a+c)

Of course, the cogs can do any multiple of these turnings and end up back in their original configuration. But it is implied in the question, that these are the least such numbers, so the numbers of turnings must have no common factor (otherwise we could divide by that factor and find a solution with less turnings).

So let m be the highest common factor of these three numbers:
m=hcf{ a.(a+c), a.c, c.(a+c) }.

Since we assumed that a,(a+c),c are triply coprime, it clearly follows that m must be equal to unity.

Now let's move from ratios to actual numbers of turnings.

So,
the number of turns made by cog A is c.(a+c)
the number of turns made by cog B is a.c
the number of turns made by cog C is a.(a+c)

We have now ensured that this will create a solution with a minimal number of turnings as long as a,(a+c),c are triply coprime.


4. We are told that cog C rotates 45 times more than cog A.

So c.(a+c) + 45 = a.(a+c)

Rearranging gives:

(a+c).(a-c) = 45

If we know what (a+c) and (a-c) are, then we can work out what a and c are. Basic simultaneous equations there. a = [(a+c) + (a-c)] / 2 and c = [(a+c) - (a-c)] / 2


5. The prime factorisation of 45 is 3*3*5.

Consider how these factors are shared between (a-c) and (a+c), and what that means for a and c.

There are only three ways of doing this:

(a+c)*(a-c) ... a , c ... b=a+c
45 * 1 ... a=23, c=22 ... b=45
15 * 3 ... a=9, c=6 ... b=15
9 * 5 ... a=7, c=2, ... b=9

45*1 -- The first of these gives too big result for b. Since b can be no bigger than 40. We can only scale up by k, not down, so this is no good. Were it not for that limiting condition this would have been a valid solution.

15*3 -- In the second case, a,b,c are not coprime, but have a common factor of 3. We are relying on them being coprime. So this is no good. [This means that as C makes 45 turns more than cog A, the starting configuration will be met thrice, not once.]

9*5 -- In the third case, the figures are coprime, however, the figures are too small for the bounds. We can scale up as provision was given to do this. Cog C is the smallest and needs to be at least 4 times bigger to give the minimum 8 teeth.

Scaling up by 4 gives a=28, b=36, c=8. This is now a valid solution.

Are there other solutions? Scaling up by 5 or more would give b at least 45. This is too big, so there are no other solutions.

Q.E.D.
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Old July 9 2009, 11:25 PM   #28
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Re: Tuesday Night Quiz!

Can I have my gears back now? I need to get the press running again.
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Old July 9 2009, 11:34 PM   #29
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Re: Tuesday Night Quiz!

That was an infintely simpler solution than the one I came up with, with the same result. Mine relied more on luck, though.
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Old July 10 2009, 12:42 AM   #30
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Re: Tuesday Night Quiz!

Zion Ravescene wrote: View Post
That was an infintely simpler solution than the one I came up with



Well, anyway, I was close. My answer had three numbers. But now my brain won't stop hurting.
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