Barcallica
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For clarity I probably should have written that using a different variable since I was re-scaling the r terms.
If you're given a set of terms {r1, r2, r3} which are rational, then you can write them in terms of integer numerators and denominators:
r1 = r1n / r1d, r2 = r2n / r2d, r3 = r3n / r3d
You can then multiply each of the terms {r1, r2, r3} by a scaling factor r1d*r2d*r3d to generate a new set of scaled terms {s1, s2, s3} where
s1 = r1 * (r1d*r2d*r3d), s2 = r2 * (r1d*r2d*r3d), s3 = r3 * (r1d*r2d*r3d)
Then s1 = (r1n/r1d) * (r1d*r2d*r3d) = r1n*r2d*r3d, which is the product of three integers, so s1 is an integer, and likewise for the other s terms. Then you can easily solve the problem with the integers {s1, s2, s3}, and since they only differ from the r terms by a scaling factor, the same solution will work for the r terms.
Ah! I knew you wouldn't do that. I should have understood the solution before finding error. Nicely done
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Getting things started by solving for cases where {r1, r2, r3} are all rational numbers, but not solved for cases where they're irrational.
Notice that if you have any set of numbers {r1, r2, r3} with which you can find a solution, that solution will also work if the same numbers are multiplied by any scaling factor {k*r1, k*r1, k*r3}.
First I'll consider the case where {r1, r2, r3} are all rational. In that case, you can write them as numerator / denominator of integer terms: r1 = r1n/r1d; r2 = r2n/r2d; r3 = r3n/r3d. Multiply those numbers by r1d*r2d*r3d, so now r1 = r1n*r2d*r3d; r2 = r2n*r1d*r3d; r3 = r3n*r1d*r2d and all of those terms are integers. Now you can find a solution with those integers, and since it only differs from the original {r1, r2, r3} by a scaling factor, the solution with the integers will also work for the original rational numbers.
And it's easy to find a solution if you're dealing with integers. Just take r1 and r2, and let r1 be the smaller of the two. Keep subtracting r1 from r2, and r2 will eventually reach zero (in which case you're done) or will reach some number x smaller than r1, in which case you can repeatedly subtract x from r1 until r1 reaches zero or some number smaller than x, and keep repeating until one of them reaches zero.
I don't have a solution if there are irrational numbers, but this might be a start. Notice that since {r1, r2, r3} are all positive and a1*r1 + a2*r2 + a3*r3 = 0, that means that one or two of the {a1, a2, a3} terms are negative. If two of those terms are negative then you can multiply by the scaling factor -1 to make it so one of the terms is negative. I will use that fact, and without loss of generality say that a3 is the negative term, so I'll rewrite the equation as
a1*r1 + a2*r2 = a3*r3
where now we know that the {a1, a2, a3} terms are all non-negative integers. The r terms might be negative after multiplying by -1.
Now I'll multiply all of the r terms by the scaling factor a3 to generate the set of scaled terms {s1, s2, s3} so that
s1 = a3*r1
s2 = a3*r2
s3 = a3*r3 = a1*r1 + a2*r2.
This is of the from s1 = nx, s2 = ny, s3 = bx + cy, where {n, b, c} are integers. This seems like it's working its way toward a solution, but I don't see how to finish it off from this point.
How multiplying r1 = r1n/r1d by r1d*r2d*r3d is r1 = r1n*r2d*r3d. You tried to prove r1, r2, r3 are integers, which is impossible.
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Call the ten original balls a, b, c, ..., h, i, and j.
The difference balls have individual weights of a-b, b-c, ...,h-i, i-j, j-a and a total weight of zero.
The difference balls thus have individual weights of zero. (The original balls have the same weight.)
Any two groups of zero-weight balls weigh the same.
difference balls all have weight more than zero. If a>b<c, then first 2 difference balls weigh a-b and c-b etc...
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For area, had to check the web on how to calculate the radius of incircle
So, the first circle has radius 10/3
So the first circle covers 2/3 of the altitude length
Without going into details, I would say by symmetry that each circle covers the next 2/3rd
So the second circle would be of radius = 1/3(10 - 20/3) = 10/9
ratio of each consecutive circle radius = 1/3
So sum of areas = pi(r1² + r2² + ...)
= pi.(10/3)² / (1 - 1/9) = (25/4)pi
The first circle has radius 10/3 but doesn't cover 2/3 of the altitude length
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What is the sum of circumferences of the infinite stack of circles in the pitcure?
What is the sum of areas?
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Since the prisoners doesn't appear to be around,
1. What is the beginning states of the lamps.
2. Can I use off/on then on/off as a signal?
3. How many times can they act a day? (it says they are shuffled each day)
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If Bi > Bi+1, put the green ball in pile 1
If Bi+1 > Bi, put the green ball in pile 2
At the end, both piles will weigh the same
Effectively the weight of green ball = absolute (Bi - Bi+1)
By sorting the piles as above, we take care of the absolute value of this subtraction.
Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero.
I don't understand this solution. What is Bi?
Bi is a blue ball at position i (i is from 1 to 10). For the tenth ball bi is blue ball 10 and Bi+1 is blue ball 1
Good catch. I had done it very clumsy way.
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If Bi > Bi+1, put the green ball in pile 1
If Bi+1 > Bi, put the green ball in pile 2
At the end, both piles will weigh the same
Effectively the weight of green ball = absolute (Bi - Bi+1)
By sorting the piles as above, we take care of the absolute value of this subtraction.
Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero.
I don't understand this solution. What is Bi?
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If Bi > Bi+1, put the green ball in pile 1
If Bi+1 > Bi, put the green ball in pile 2
At the end, both piles will weigh the same
Effectively the weight of green ball = absolute (Bi - Bi+1)
By sorting the piles as above, we take care of the absolute value of this subtraction.
Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero.
But weight difference is always positive.
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There are 10 blue balls on a circle, all have different mass. I put green balls between every 2 neighboring blue balls, which weighs the difference of those 2 blue balls. Prove that I can put those green balls in two groups that weigh same.
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Thanks
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1. bonanova
2. dee_tot
4. araver - voting for Marksmanjay
5. Marksmanjay
7. onetruth
9. Auramyna -TRAPPED in kitchen cupboard for N3, D3
10. Barcallica - voting for onetruth
6. phil1882 -DEAD- executed as Hallucinator
8. Boquise -DEAD- executed as Homicidal Maniac
3. plasmid -DEAD- poisoned by Killers
11. mtamburini -RELEASED as ParanoiacVoting for onetruth, simply because she is not indy. We don't wanna lynch the pretender, do we sanes?
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Dr. Fate: Y-san
Roster:
1) Phil1882
2) Nana77
3)
4)
5)
6)
7)
8)
9)
10)
11)
12) Kikacat123
13) Araver27182818284
14)
15) Barca
Back-ups:
1)
2)
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bonanova,
I see no sense in you disallowing 12 or 123, because each of those numbers have strict increases
in their digits.
He didn't disallow 12 or 24, he correctly disallowed 112, 224 etc. Without strcict increase, the asnwer would be infinite.
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I couldn't find a reason why the killers would refrain from killing two nights in a row. So I assume game is continueing without killers???
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I know who the pretender is, if I diclose that info, would the sane release me? I wonder.
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How many natural numbers are there, whose digits increase left to right? (this is from a book, so some of you might already know this one)
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Bonanova is replacing Freeown.
We are looking for a replacement for Flame.
Night prolonged?
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How about this one.
First prisoners will convey even/odd info (f=odd t=even).
Phase 1.
a -> a -> a -> a ->
<- b -> b -> b ->
<- c <- c - > c ->
Now they all know each others even/odd. And should know their one of TFTFF, FFFTT, FTTFF ...
and can uniquely name them LF, MF, RF, LT, RT (L=left R=right)
Phase 2.
T - F=2
T - T=4
F - F=1
F - T= 3 or 5
first LT sends out f or t, which will determine 2 or 4.
second LF will send out f or t.
third MF will send out f or t.
Phase 3.
Now left of the two who got 3 or 5 will send f if 3, T if 5.
Total 26 signals. And this is time independent as all actions will be determined by order of receiving and sending signals.
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With bainanova's {1, 2, 3, 4, 5} <=> {F, T, FF, TF, FT} my solution yeilds best scenario=12 worst=21 signals (T+F).
And it isn't C, for example, have to make b right after hearing b. Because all actions have order => A will make a only after and right after hearing 3 signal from B, and B knows he will hear a and do nothing until then etc .
Actually there will be 12 signals overall and it's matter of distributing it evenly without suspicion and I think it is possible given enough time (but this is subjective).
Anyway I get your idea that we should look for solution that requires minimal action.
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Host: Asylum Warden: Kikacat123
Co-host: Deputy Warden: nana77
1. FreeOwn
2. Flamebirde - voting for Phil
4. araver - voting for mtamburini
5. Marksmanjay
6. phil1882 - voting for onetruth
7. onetruth - voting for Phil
9. Auramyna -trapped in vent for N2, D2-
10. Barcallica - voting for mtamburini
11. mtamburini - voting for onetruth
DEAD:
3. plasmid - DEAD - poisoned by Killers
8. Boquise - DEAD - executed as Homicidal Maniac
making it 3 way tie once again.
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I invite all of you to join the bw.
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Roster:
Host: Asylum Warden: Kikacat123
Co-host: Deputy Warden: nana77
1. FreeOwn
2. Flamebirde
3. plasmid -DEAD- poisoned by Killers
4. araver
5. Marksmanjay
6. phil1882
7. onetruth
8. Boquise -DEAD- executed as Homicidal Maniac
9. Auramyna -trapped in vent for N2, D2-
10. Barcallica - voting for onetruth
11. mtamburini
Another Prisoner Problem 2...
in New Logic/Math Puzzles
Posted
(with regard to 2nd answer) Then can't it be solved with similar fashion. Switching once is odd, switching twice is even etc...
Anyway, this is what I was asking with second question.