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[ Information | Background | The CyberTeaser | Quantum News | Miscellaneous Stuff ] Solution to CyberTeaser B187Yes, the required partition is possible. For instance, consider the 18 pairs of weights “equidistant from the ends”: 1 + 101, 2 + 100, … 18 + 84; and 32 similar pairs for the remaining 64 weights: 20 + 83, 21 + 82, 22 + 81, …, 51 + 52. If we take any 9 pairs from the first set and 16 pairs from the second set, we obtain the required partition. CyberTeaser Archive
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