[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 226
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Индукция позволяет легко построить правильный алгоритм, сведя его к базе (2 стакана). Но доказывает ли она невозможность порчи? Я подумаю.
If the right algorithm can be corrupted, it's the wrong one.
:)
Now I finally understand why forex is doomed to wobble forever. It is because the number of pairs between currencies cannot be a degree of two...
!;)
Brilliant, MetaDriver (apart from the case of only two currencies)!
The answer to the problem is: the kid can't do it, if all but one glass is poured 100g, and the last one is poured 200g. Who can prove that the little guy can't do it?
It follows directly that even if there are only arbitrageurs left in fore, they will always have work ! :)
And since arbitrage is a completely risk-free activity (:as legend has it:), everybody will of course win! ;)
Mathemat писал(а) >>
The answer to the problem is: the little man will fail if all but one glass is poured 100g and the last glass is poured 200g. Who can prove that the kid can't do it?
Easy. 3100/30 = 310/3 = 103 + 1/3, which is not representable as a fractional finite binary number.
Actually the counterexample is pulled by the ears, the proof along with it -- the problem as a whole is more interesting.
OK, what if the last one has 130 grams (3030/30 = 101 exactly)?
Ага. ОК, а если в последнем будет 130 граммов (3030/30 = 101 ровно)?
You're mean!
;)
Well, let's reason. At least one example (29 glasses with a grams and one glass with b grams) let's try to solve in general case.
Let b = a + epsilon for certainty, and epsilon > 0 (though it probably doesn't matter). Then after the positive solution of the problem there should be exactly a + epsilon/30 in each glass.
On the other hand, how much milk can be in the glass after a finite number of steps?