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Message: <blockquote data-quote="imhotep" data-source="post: 30934627" data-attributes="member: 562115">Noted this during the weekend but was busy away from home and also people should think and reply.Not difficult if you think. eg the frog problem.Lets say we had F frogs in the pond. The crane visits every Sunday and ate a fixed number of frogs, say x.The natural growth is a factor of 3.... and we assume that there are no natural deaths.Given constraint 100 ≤ x ≤ 150 First week - Start F frogs, Crane eats x, remainder F-xSecond week - Frogs multiply by a factor of 3, so Start count is 3(F-x). remainder 3(F-x) - xThird week Start count is 3 (3(F-x) -x) with a remainder 3 (3(F-x) -x) - xFourth week Start count is 3 (3 (3(F-x) -x) - x) with a remainder of 3 (3 (3(F-x) -x) - x) - xSince there's none left 3 (3 (3(F-x) -x) - x) - x = 0which simplifies to F = 40 x / 27with the constraint the ossible answer is F = 120 with x = 81.</blockquote>

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