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On November 26 2004 06:10 BigBalls wrote:
Phil was short in money. So he sent his sister amy a message as follows:
$ WI.RE
MO.RE
________
MON.EY
how much should amy send phil?
about this problem.....i have found 3 solutions.....however, my uncle has a jewish friend who is absolutely brilliant and he says 2 of the solutions cancel each other out. I kind of see what he means but i dont agree with it. Someone either refute or support his claim.
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On November 25 2004 21:53 baal wrote:
yeah im answering what the question was suppoused to be, if it wasnt obvious enough, someone reposted it in the first pages of this huge thread.
and i insulted cayou.... Muuuffffaaaassaaaa!!!!! :p
but this whole time when u were calling us idiots, we specifically pointed out that we were answering what the question was in fact asking and not what it should have been asking.
That's not very nice. or fair. And makes you look more the idiot than us.
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On November 26 2004 06:10 BigBalls wrote:
Phil was short in money. So he sent his sister amy a message as follows:
$ WI.RE
MO.RE
________
MON.EY
how much should amy send phil?
This might be a solution.
E=2, Y=4, R=6, I=7, O=0, N=8, W=9, M=1
So he asked for 108.62 units.
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***********************************Possible spoiler*************************
On November 26 2004 06:10 BigBalls wrote:
Phil was short in money. So he sent his sister amy a message as follows:
$ WI.RE
MO.RE
________
MON.EY
how much should amy send phil?
Thanks for your posts Bigballs,
I'm not sure but assuming it substituted letters for numbers, and same letters represent certain numbers it could be
$WI.RE
_MO.RE
=======
MON.EY
$95.74
+10.74
=====
106.48
?
i started off the left side knowing M can only be 1 cos u cant get anything over $200 from two two-digit dollars and "O" would have to be zero and "W" 9 to carry on with 9+1 etc
there could be other combinations for rest of the numbers i guess..
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well, those are both solutions i got, i also have a 3rd.
why cant it be two of them?
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I think each letter stands for a certain number, like in the one dollar column, you see a "I" plus the letter "O" which stand for a number and zero and add up to a different number "N".
If "O" was zero you can't get "I" and "N" to be different(EDIT: mathematically i mean), the difference has to be made by the cents carrying a 1 off to the dollars column...i think your uncle's friend may be talking about this..may be only one solution takes this into account..not sure
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By the way have everyone agreed on 2/3 chance of the other child being a boy?
Like Baal said, a good argument against 1/2 is that it would mean 75% of earth's population would be female, and last time I went went out for a dring and a good time, it seemed more like 60% were male 
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Solutions ive come up with:
95.74
10.74
106.48
92.74
10.74
103.48
97.62
10.62
108.24
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luhh, no, its agreed that 1/2 is the answer
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4th solution -
92.87
10.87
103.74
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5th solution -
95.87
10.87
106.74
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Sorry guys, problem with a question from my girlfriend's combinatorics assignment. Came up with a solution that I feel ok about (it's sort of ugly) but turns out to be not the correct answer according to the prof. Would really appreciate some help with a final answer and also a rough solution. I think there was a question like this or perhaps the same posted in this thread already but I can't seem to find it (or too lazy to wade thru all that).
How many triangles are there with integral sides with the following two properties:
1. Perimeter is 300
2. All sides must have different lengths.
Now the way I went about it (straight up counting way), I got 541. The prof sort of wanted to have the students use the inclusion-exclusion principle. Which should work fine but I was having trouble counting some stuff. I can give more details if someone wants.
Any help would be appreciated. Especially a final answer.
Edit: Never mind, just found my mistake and it is so so trivial.
Edit2: Correct answer should be 1801.
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He, I have a problem that's worded slightly different than the door-answering-girl one. It's part of a T.V. quiz in the Netherlands called "National Science Quiz" and it goes like this:
"You have a bag with one white ball in it. You randomly add a white or a red ball to it. Then you take one ball out of the bag, which turns out to be white. What is the chance that the remaining ball also is white?"
I don't think it's analogous to the girl-boy problem, because there is no possibility for a bag with two red balls (while there is a possibility for a boy-boy family).
I can see two approaches to the problem. The first is that if you repeat this experiment many times I have 50% bags with two white balls and 50% bags with one white and one red ball. If you take a white ball from the bag with two white balls, the chances of the other ball being white are 100%, and if you take the white ball from the bag with the white and red ball the chances for another white ball are zero percent. so 0,5*1+0,5*0 = 50% chance for a second white ball.
The second approach is reasoning that there is a 75% chance of getting a white ball on the first try and an 25% chance of getting a red one. Since you know that your first draw is a white ball, there's a 2/3 chance that it came from a white-white bag and a 1/3 chance it came from a white-red bag. So the chances for the second ball being a white one are 2/3.
But which one is the right solution and more importantly why. I can't see anything wrong with either reasoning, but one has to be wrong.
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The flaw in the first one is that you will get a white ball from a WW bag twice as often as you will get a white ball from a WR bag.
That means when you do the experiment 100 times, you will get:
50 white balls with WW
25 white balls with WR
25 red balls with WR (these don't count but still exist)
Out of the 75 whites you got, 2/3 come from a WW bag.
Your formula should be changed to:
2/3*1 + 1/3*0
instead of .5*1 + .5*0
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On November 28 2004 12:16 LaptopLegacy wrote:
He, I have a problem that's worded slightly different than the door-answering-girl one. It's part of a T.V. quiz in the Netherlands called "National Science Quiz" and it goes like this:
"You have a bag with one white ball in it. You randomly add a white or a red ball to it. Then you take one ball out of the bag, which turns out to be white. What is the chance that the remaining ball also is white?"
I don't think it's analogous to the girl-boy problem, because there is no possibility for a bag with two red balls (while there is a possibility for a boy-boy family).
I can see two approaches to the problem. The first is that if you repeat this experiment many times I have 50% bags with two white balls and 50% bags with one white and one red ball. If you take a white ball from the bag with two white balls, the chances of the other ball being white are 100%, and if you take the white ball from the bag with the white and red ball the chances for another white ball are zero percent. so 0,5*1+0,5*0 = 50% chance for a second white ball.
The second approach is reasoning that there is a 75% chance of getting a white ball on the first try and an 25% chance of getting a red one. Since you know that your first draw is a white ball, there's a 2/3 chance that it came from a white-white bag and a 1/3 chance it came from a white-red bag. So the chances for the second ball being a white one are 2/3.
But which one is the right solution and more importantly why. I can't see anything wrong with either reasoning, but one has to be wrong.
50%
The second reasoning is wrong, since it considers all 3 balls being potentially in the bag. However 1 white ball = 100%, the other is 50/50.
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Catyoul
France2377 Posts
As Karlsberg explained, it is 2/3 white for the second one.
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EPT
