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Assume N = 100, mine requires 9 times. The first egg is tried on 36, 64, 85, 100. Depending on the result of first egg drop, the second egg is tried on (8, 15, 21, 26, 30, 33, 35), (43, 49, 54, 58, 61, 63), (70, 75, 79, 82, 84), (90, 94, 97, 99). The third egg is just for linear search. The worse case is 9 drops, though it might be possible to reduce to 8 times.
- chenatuc October 27, 2008The idea follows from the two-egg dropping problem and this problem is reduced to solving the following inequality: sum_{k=1}^{n} k + sum_{k=1}^{n-1} k + ... + sum_{k=1}^{1} k > 100.