Statement
The trick as David reveals in the comments. Suppose we want to choose 11 cards with the property contrary to that, ie it is not possible to find two i, j, which satisfy the inequality i When we have these 11 letters, think of the least of them. Its value is at least 1. Therefore, the second must be greater than 2, because otherwise the property would not be met. Then it must be at least 3. Therefore, the third must be greater than or equal to 7 (= 2 3 - 1), for the same reason, the fourth largest or equal to 15 (= 2 4 - 1). Since the following letter to one that is greater than or equal to 2 n - 1 must be more than twice, which is 2 n + 1
- 2, its value must be greater than or equal to 2 n + 1 - 1, we see that value which occupies the position 11 must be at least 2 11 - 1, which is 2047. is obviously impossible to select a card higher than 2011, according to the statement, which demonstrated the problem.
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