No. The master defines “on time” by forcibly removing people after the “right bell”. The master already knows the scheme that he is using. The members have to figure out which he is using.
I have told you several times with examples;
The fact that your color is a member of an infinite set is completely irrelevant and adds nothing to the syllogism other than distraction. Every color is a member of an infinite set at all times. So you cannot say that merely because that color is a member of an infinite set, it cannot be knowingly chosen.
Yes, James, I agree. That’s why I keep asking you to actually address the syllogisms I’m providing, which helpfully you now have, thanks.
Yes, the “right bell”. Not “any bell the Master arbitrarily decides should be the bell on which the logicians should leave”, but the bell that they are logically supposed to leave on. That statement by the Master only says that we don’t have to worry about illogical logicians, and it is common knowledge that we don’t have to worry about illogical logicians. It is similar to the statement in the Blue Eye problem that all the islanders are perfect logicians. Rather than simply decree it here, it is enforced by the Master, who I presume to be a perfect logician. It does not introduce caprice on the part of the Master, and a syllogism that would work with the assumption that all the logicians are perfect logicians, i.e. that they will immediately deduce anything deducible, will also work here.
Direct, and accurate. As I understand your criticism, the color of a given logician’s headband belongs to both infinite and finite sets (indeed, an infinite number of each), and so it’s not enough to say that it belongs to the infinite set of colors that can’t be seen. I think that’s right, though I think the syllogism can be salvaged. All it will take is a little more rigor on my part
You have to assume that there is only one possible algorithm and that the Master is using that one in order to know which logician should leave after which bell. Thus you must prove that there is only one possible algorithm (process for the elimination of logicians).
Write numbers from 1 to 6 into the cells of the diagram of size 6 x 6, so that each number occurs exactly once in each row and in each column. A brick must contain an odd and an even number. Two half-bricks in a row at the left and right edge of the diagram form one whole brick.
An uncle and his niece meet on a treat. The uncle says:"If one multiplies each age of three people, one obtains 2450. If one adds each age of the three people, one gets twice your age. “Well,” says the niece, “but that’s not enough to elicit the age of three people”. The uncle agrees and says: “This year one of those three people celebrated a very special birthday. I celebrated this very special birthday five years ago.”
Concerning the Uncle and Niece:
[tab]When I read “special birthday” I just dropped it. I dislike trying to figure out what someone might think of as a “special birthday”
But later, just making guesses, I came up with a “special birthday” being 50, in which case the people could be:
7
7
50
While the Uncle is 55 and the niece 32.
I have no idea of that is the kind of thing that you were expecting.[/tab]
And then the answer to the Sudoku puzzle (in case you suspect that it can’t be done):
[tab]3 9 4 | 5 2 8 | 1 6 7
2 5 6 | 1 7 4 | 8 9 3
7 8 1 | 9 6 3 | 4 5 2
[tab]For the age of the three persons (x, y, z):
x • y • z = 2450 = 2 • 5 • 5 • 7 • 7.
The “special birthday” is the 50th (= 2 • 5 • 5). The three persons are 50, 7, and 7 years old; the uncle is 55 years old (= 50 + 5); the niece is 32 years old (= (50 + 7 + 7) / 2).[/tab]
Your watch has stopped. So it does not work anymore. The little hand of the watch indicates approximately ten o’clock, and the big hand of the watch indicates approximately two o’clock. Both hands of the watch form an identical angle. When did your watch stop precisely?