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Old 10-05-2012, 05:57 PM   #9511
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Originally Posted by norway1456 View Post
If they are at $12M, Einstein should perhaps be around 10M, and Hendrix and Larson about $5M? Theodor Geisel (who was he, by the way) somewhere between
Very good! Einstein is indeed at $10 million. You're close on the others, but not within one million (go a little higher).

Theodor Geisel was Dr Seuss.
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Old 10-05-2012, 05:58 PM   #9512
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Neck and neck here!

1. Michael Jackson Bilbo1967 $170 million Fbone
2. Elvis Presley Daithi $55 million Fbone
3. Marilyn Monroe Hamlet53 $27 million Fbone
4. Charles Schultz Fbone $25 million Fbone
5= John Lennon Daithi $12 million norway1456
5= Elizabeth Taylor norway1456 $12 million norway1456
7. Albert Einstein pdurrant $10 million norway1456
8. Theodor Geisel Issybird
9= Jimi Hendrix norway1456
9= Stieg Larsson Daithi

Points:
Fbone 5 points
Norway1456 5 points
Daithi 3 points
Bilbo1967, Hamlet 53, issybird, pdurrant 1 point each
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Old 10-05-2012, 06:06 PM   #9513
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Since Hendrix and Larson was not at 5M, let's say 7 mill then (between 6 and 8 million).

I looked up Dr Seuss on wikipedia*, and found that he wrote the books about Horton. I think these books was the material for the Horton films (is this right?), quite popular films. He therefore should be at about 8 million (between 7 and 9 million)?

*Is this cheating? I did not look up the list itself, but I had to look up some facts about the guy to make a guess. It this is considered cheating, please ignore my guess

Last edited by Iznogood; 10-05-2012 at 06:07 PM. Reason: Typo
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Old 10-05-2012, 06:24 PM   #9514
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Quote:
Originally Posted by norway1456 View Post
Since Hendrix and Larson was not at 5M, let's say 7 mill then (between 6 and 8 million).

I looked up Dr Seuss on wikipedia*, and found that he wrote the books about Horton. I think these books was the material for the Horton films (is this right?), quite popular films. He therefore should be at about 8 million (between 7 and 9 million)?

*Is this cheating? I did not look up the list itself, but I had to look up some facts about the guy to make a guess. It this is considered cheating, please ignore my guess
Correct on all accounts! And you had an unbeatable lead even without Dr Seuss, so I'm happy to give you a pass on that.

1. Michael Jackson Bilbo1967 $170 million Fbone
2. Elvis Presley Daithi $55 million Fbone
3. Marilyn Monroe Hamlet53 $27 million Fbone
4. Charles Schultz Fbone $25 million Fbone
5= John Lennon Daithi $12 million norway1456
5= Elizabeth Taylor norway1456 $12 million norway1456
7. Albert Einstein pdurrant $10 million norway1456
8. Theodor Geisel Issybird $8 million norway1456
9= Jimi Hendrix norway1456 $7 million norway1456
9= Stieg Larsson Daithi $7 million norway1456

Points:
Norway1456 8 points
Fbone 5 points
Daithi 3 points
Bilbo1967, Hamlet 53, issybird, pdurrant 1 point each

Congratulations to Norway1456! Please go ahead and set the next quiz.
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Old 10-06-2012, 02:02 PM   #9515
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Imagine yourself as the detective in a forgery case. Laying in a row before you are 11 coins. You know that three of them are false and 8 are genuine. You also know that the three forged coins are laying together, side by side, somewhere in the row. The false coins are heavier than the genuine one. How many coins (at a minimum) do you have to weigh before you can point out the three false coins.
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Old 10-06-2012, 03:06 PM   #9516
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Two? That's a minimum, assuming you pick lucky.
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Old 10-06-2012, 03:14 PM   #9517
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Quote:
Originally Posted by norway1456 View Post
Imagine yourself as the detective in a forgery case. Laying in a row before you are 11 coins. You know that three of them are false and 8 are genuine. You also know that the three forged coins are laying together, side by side, somewhere in the row. The false coins are heavier than the genuine one. How many coins (at a minimum) do you have to weigh before you can point out the three false coins.
Four coins in just two weighings.
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Old 10-06-2012, 03:20 PM   #9518
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Four coins in just two weighings.
Here's how.

number the coins 1..11.
Weigh coin 3 against coin 9.
If they are the same weight, we know that they are both good coins (because they are more than three coins apart, they can't both be bad coins). So coins 1,2,10,11 must also be good coins (because the three bad coins are together). This leaves us with the five coins 4 ..8 to examine.
If they are not the same weight, we know which one is the bad coin, because it is heavier. So we now know that the bad coins are in the five coins 1..5 or 7..11.

So after one weighing of two coins we've identified five coins that must contain the three bad coins.

Weigh the two coins at the ends of the five coins (i.e. 1 and 5 or 4 and 8 or 7 and 11).

If the coins are the same weight, the bad coins are in the middle of the five. If one of them is bad, the bad coins are at that end of the row of five.

How did I do?
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Old 10-06-2012, 07:08 PM   #9519
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Brilliantly!
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Old 10-06-2012, 08:03 PM   #9520
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You are quite right. When I made the question, I had in mind a weight that measured the wieght of just one coin, you imagined to weigh them against each other, but that doesn't matter. The Grand Mouse has the floor
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Old 10-07-2012, 02:45 PM   #9521
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I thought of a better solution, if you know beforehand the weight of a good coin.

Weigh the middle one.
If it is a bad coin, the three bad coins must be in the five coins in the middle. We can weigh just two more coins and find all the bad ones.
If it is a good coin, the bad coins must be in the first five or the last five. And we know how to do that with weighing just two coins.

Bad coins found after weighing just three coins!

But this solution does require that we know how much a good coin weighs.
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Old 10-07-2012, 02:45 PM   #9522
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And now, a puzzle from me. Hmmm... I'll have a think.
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Old 10-08-2012, 11:58 PM   #9523
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While pdurrant is coming up with something here's a quick one to keep the thread moving:

How many squares are there on a standard checkerboard?
(The answer is not 64).
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Old 10-09-2012, 12:25 AM   #9524
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Do you have to count the squares made up of all the other squares?
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Old 10-09-2012, 12:53 AM   #9525
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Do you have to count the squares made up of all the other squares?
Yes, that is why I hinted the answer is not 64.
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