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Is it possible under such conditions of the problem, to answer the proposed question?

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Имеем бесконечное количество яблок. Распространение их на бесконечное множество ofidentical число groups.In каждой группы 5 apples.Now от eachgroup вывезти 2 яблока. Осталось 3.

 

Theremainder множества разбит на бесконечное множество groups.In eachgroup и 7 яблок. Теперь возьмите за 2 яблока. Остается 5apples.

 

Theremainder множества разбит на бесконечное множество groups.In eachgroup, и 11 яблок. Теперь возьмите за 2 яблока. Остается 9apples.

 

 

 

 

Andso на неопределенный срок. Захват 2/n. Сохранить N-2/N) . N-простых чисел в order.Question: как manyapples останется? Конечного или бесконечного?

Edited by devami

Do you perhaps have an English version of the question, or what language is this so we can translate it?

As I read it (google translator iyf): we start with an infinite set of apples - partitioned into subsets of 5. From each of those, two are taken and the set is repartitioned into sets of 7 and two apples are again removed from each subset. This is repeated indefinitely, with partitions into subsets of prime sizes.

 

As I understood that: We start with apples {1,2,3,4,5,6,...} and in the first iteration we remove {5,6} U {10,11} U {15,16}... in the second iteration we remove {7,8} U {14,15} U {21,22}... in the third iteration we remove {11,12} U {22,23} U {33,44}...

 

The question is simple enough to understand, after infinite iterations, what has been allowed through this sieve?

 

The answer, I don't know - but some experimenting seems to show that a large amount of numbers do get through - so I'd be tempted to say that it's infinite.

 

I'm sure this can be solved though - so the answer to the question poised in the topic - is yes.

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[Цитата имя = 'Дата DJBruce' = '22 августа 2010 - 3:49 вечера 'метка = '1282484953' пост = '561153 ']

У вас может быть, на английском языке на вопрос, или на каком языке это так мы можем перевести это?

[/] Цитатой

 

Please!Original in Russian!Thank you in advance.

Zagadka.pdf

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