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).
I gave you what you need in this post, this post and this post.
In the end, you must:
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.
Good speed!
A weighing problem.
On an ordinary beam balance the following shapes are in equilibrium:
• Circle and triangle with the square
• Circle with triangle and pentagon
• Two squares with three pentagons
How many triangles weighs a circle?
The age of an uncle and his niece.
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.”
How old are uncle and niece?
Sudoku champion:
[tab]4 3 6 5 1 2
2 1 3 6 4 5
5 2 4 3 6 1
3 4 2 1 5 6
1 6 5 2 3 4
6 5 1 4 2 3[/tab]
How many triangles weighs a circle?[tab]s = c+t
c = t+p
2s = 3p
s = 3/2 p
p = 2/3 s
c = t + 2/3 s = 2/3 c + 2/3 t
1/3 c = 2/3 t
c = 2t
1 circle = two triangles[/tab]
And speaking of Sudoku:
In case you do not know the rules:
Numbers 1 through 9 on each row, column, and square region of 9.
That is false, James, because each brick must contain an odd and an even number!
Oops … forgot that rule 
That is false too, James.
I had to interpret your English intention. Could you reword it, seeing how I am solving for it? Your “with” and “and” are ambiguous.
And the correction for the first one:[tab]2 3 6 5 1 4
4 2 3 6 5 1
6 1 4 3 2 5
3 5 2 1 4 6
1 4 5 2 6 3
5 6 1 4 3 2[/tab]
Englsih intention? Reword it?
Your equations show me that you have understood my text correctly.
Howsoever. Maybe the following three pictures will help:
Ahhh … I see that I left out a “t” 
[tab]s = c+t [that is the 1st pic]
c = t+p [that is the 2nd pic]
2s = 3p [that is the 3rd pic]
from 3rd:
s = 3/2 p
p = 2/3 s
from 2nd:
c = t + 2/3 s
from 1st:
c = t + 2/3 c + 2/3 t
c - 2/3 c = t + 2/3 t
1/3 c = t + 2/3 t
c = 5t[/tab]
And in case you missed it from above:
Well done, James. =D>
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]
9 6 7 | 4 3 5 | 2 8 1
5 1 3 | 2 8 6 | 7 4 9
8 4 2 | 7 1 9 | 6 3 5[/tab]
Well done. =D>
[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?
I hate always being the only kid in the class with his hand up.
Really?
Well I got deadlines and shit. Haven’t even been able to read this thread lately.
Who is the question directed to, the wearer of the watch or somebody else who sees the watch? If the latter, it depends on how fast he is moving when he observes the watch.
Right James?
JAMES! Wake up!