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THUDandBLUNDER
Threefold Repetition

 Posted: Sun Sep 12, 2010 12:58 am    Post subject: 1 Quite simply, a monkey's mother is twice as old as the monkey will be when the monkey's father is twice as old as the monkey will be when the monkey's mother is less by the difference in ages between the monkey's mother and the monkey's father than three times as old as the monkey will be when the monkey's father is one year less than twelve times as old as the monkey is when the monkey's mother is eight times the age of the monkey, notwithstanding the fact that when the monkey is as old as the monkey's mother will be when the difference in ages between the monkey and the monkey's father is less than the age of the monkey's mother by twice the difference in ages between the monkey's mother and the monkey's father, the monkey's mother will be five times as old as the monkey will be when the monkey's father is one year more than ten times as old as the monkey is when the monkey is less by four years than one seventh of the combined ages of the monkey's mother and the monkey's father. If in a number of years equal to the number of times a monkey's mother is as old as the monkey, the monkey's father will be as many times as old as the monkey as the monkey is now, and assuming no a priori knowledge of the monkeys' longevity, find their respective ages. . .
THUDandBLUNDER
Threefold Repetition

 Posted: Sun Sep 26, 2010 10:42 am    Post subject: 2 I haven't tried to solve this one yet but the general method is straightforward enough. We have 3 equations in 3 unknowns. What could be simpler? (OK, 2 equations in 2 unknowns, duh). Let age of monkey = x Let age of mother = m Let age of father = f Considering the last paragraph, I get f + (m/x) = x[x + (m/x)] which simplifies to x^3 + (m - f)x - m = 0..........................(1) :
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