Pure maths, physics, chemistry, etc.: brain-training tasks that have nothing to do with trade [Part 2] - page 14
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No solution...Number the fields of the chessboard with numbers from 1 to 8 from left to right, in each row. After cutting out the corner square, the sum of all the numbers on the board is not divisible by 3. Whereas, the sum of the numbers covered by the 1x3 cardboard is divisible by 3.
And before we cut out?
And before we cut?
There is no solution...Number the fields of the chessboard with numbers from 1 to 8 from left to right, in each row. After cutting out a corner square, the sum of all the numbers on the board is not divisible by 3. Whereas, the sum of the numbers covered by the 1x3 cardboard is divisible by 3 .
63 is not divisible by 3 ??? why ???
ZS: Got it, stupid! )
Let me also post a problem from a famous forum.
The weight of the problem is 4.
The invaders, in a way known only to them, choose two different real numbers and write them on two pieces of paper. Then they invite Megamind to choose any piece of paper, look at the number written there and guess whether the number on the other piece of paper is higher or lower. Prove that Megamind has a strategy that will allow him to guess with more than 50% probability.
A guessing strategy with a probability of an exact answer greater than 50% exists (according to the moderators). I can't decide it myself.
Yes, I posted the exact same thing - it's already counted. Only I should add, that the sum of uncovered cells of the full board before the cardboard coverage is also divided by 3 (equals 288).
Sanek: it is not something like in the problem about the artilleryman, or something again confuse
There is the Monty-Python (-Hall) paradox - or the paradox of the two envelopes. But I frankly don't like the fact that all real numbers are considered there - instead of some segment.
Actually, there is a solution to the chessboard :-) I proved to my 5th grade maths teacher with a protractor in hand that the sum of the sides of a triangle is NOT equal to 180 degrees...
and from the same area you can also solve with a chessboard....
Let me also post a problem from a famous forum.
The weight of the problem is 4.
The invaders, in a way known only to them, choose two different real numbers and write them on two pieces of paper. Then they invite Megamind to choose any piece of paper, look at the number written there and guess whether the number on the other piece of paper is higher or lower. Prove that Megamind has a strategy that will allow him to guess with more than 50% probability.
A guessing strategy with a probability of an exact answer greater than 50% exists (according to the moderators). I can't solve it myself.
The point here is that the conditional probability that the second number is greater than the known number cannot be equal to the conditional probability that the second number is less than the known number. This implies that the probabilities of the occupants writing any number from + infinity to - infinity are constant, which means that the sum of the probabilities will be infinity. Hence the conditional probabilities are not equal to each other (0.5), which means that theoretically there is a way to guess more than 50% of the time.
The problem is actually "the paradox of the two envelopes".
P.S. While writing, Mathemat has already answered))
The task is in fact the "two envelope paradox"
People love paradoxes, regardless of education. They remind them of a happy childhood with Father Christmas and mannish bedtime stories.
I do not see this paradox, because the correct average when working with ratios is the geometric mean, not the arithmetic mean.