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[ Information | Background | The CyberTeaser | Quantum News | Miscellaneous Stuff ] Solution to CyberTeaser B254The minimum value of n is 15. First, we prove that for 15 cards, the desired pair can be found. Suppose the contrary. Then the cards numbered 1 and 15 must be in different stacks, as must cards 1 and 3. Thus cards 3 and 15 are in the same stack. Therefore, cards 6 = 9 – 3 and 10 = 25 – 15 are in the other stack, which contradicts the assumption, since 6 + 10 = 16. Now we show that 14 cards can be distributed between the two stacks such that the sum of the numbers of any two cards of the same stack is not an exact square. Here is an example: 1, 2, 4, 6, 9, 11, 13 (the first stack) and 3, 5, 7, 8, 10, 12, 14 (the second stack). For any number of cards less than 14, the cards can be distributed between the two stacks in a similar way (with the desired condition holding true). CyberTeaser Archive
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