``` wu :: forums (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi) riddles >> medium >> Monkey Maths (Message started by: ThudanBlunder on Oct 30th, 2010, 9:38am) ``` Title: Monkey Maths Post by ThudanBlunder on Oct 30th, 2010, 9:38am 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, find their respective ages. Title: Re: Monkey Maths Post by ThudanBlunder on Oct 30th, 2010, 10:18am I managed to solve this puzzle, the longest one of its type I have seen. As nobody replied when I posted it last month, I am reposting it with my solution.It turns out to be solvable only when the father is older than the mother. So I think this may as well be stipulated in the puzzle statement.[hide]It is possible to derive three equations in three unknowns.Let m = present age of mother Let f = present age of fatherLet x = present age of monkeyFIRST EQUATIONIf, 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...Then f + (m/x) = x[x + (m/x)] which simplifies to x3 - (f - m)x - m = 0.......................(1)                                                                           ==============================SECOND EQUATION...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...If f > m,  m + a = (f - x) + 2(f - m) for some value of integer a. And monkey will be m + a years old in m + a - x yearsand m + a - x = m + (3f - 3m - x) - x                      = 3f - 2m - 2x yearsHence (if f > m) monkey will be m + a years old in 3f - 2m - 2x yearsORif m > f, m + a = (f - x) + 2(m - f) and m + a - x = m + (2m - f - x) - x                      = 2m - f - 2x yearsNow let b = 3f - 2m - 2x OR 2m - f - 2x...the monkey's mother will be five times as old as the monkey will be when... means LHS of equation is m + b for some value of integer b. And m + b = 3f - m - 2x (if f > m)                 = 3m - f - 2x (if m > f)...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...Then x + c = [(m + f)/7] - 4 for some value of integer c....when the monkey's father is one year more than ten times as old as the monkey is when...Then f + h = 10(x + c) + 1 for some value of integer h....as old as the monkey will be when the monkey's father is...When father is f + h years old, monkey is x + h years old. And x + h = [10(x + c) + 1] + x - f   ...the monkey's mother will be five times as old as the monkey will be when...Hence m + b = 5(x + h) = 50(x + c) + 5 + 5x - 5f                     = 50[{(m + f)/7} - 4] + 5 + 5x - 5fSo finally,  if f > m then3f - m - 2x = 50[{(m + f)/7} - 4] + 5 + 5x - 5fThis gives 57m - 6f + 49x = 1365 ................................(2a)ORif m > f then3m - f - 2x = 50[{(m + f)/7} - 4] + 5 + 5x - 5fThis gives 29m + 22f + 49x = 1365 .............................(2b)                                  ==============================THIRD EQUATION...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.                                               ...when the monkey's mother is eight times the age of the monkey.Then m + d = 8(x + d) for some value of integer d.Rearranging, x + d = (m - x)/7...when the monkey's father is one year less than twelve times as old as the monkey is when...Then f + e = 12(x + d) - 1 for some value of integer e.When father is f + e years old, monkey is x + e years old.And x + e = [12(x + d) - 1] - f + x  ...three times as old as the monkey will be...This equals 3(x + e) = 36(x + d) - 3 - 3f + 3x...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...So m + g = 3(x + e) - |f - m| for some value of integer g. ...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...When mother is m + g years old, monkey is x + g years old.And x + g = [3(x + e) - |f - m|] - m + xSo father will be 2(x + g)When father is 2(x + g) years old, monkey is 2(x + g) - f + x years old.Quite simply, a monkey's mother is twice as old as the monkey will be when the monkey's father is...Hence m = 4(x + g) - 2f + 2x               = [12(x + e) - 4|f - m| - 4m + 4x] - 2f + 2x               = 12(x + e) - 4f + 4m - 4m + 4x - 2f + 2x (if f > m)               = 12(x + e) - 6f + 6x               = [144(x + d) - 12 - 12f + 12x] - 6f + 6x               = 144(x + d) + 18x - 18f - 12               = [144(m - x)/7] + 18x - 18f - 12         7m = 144m - 144x + 126x - 126f - 84So (if f > m) 137m = 126f + 18x + 84...............................(3a)OR if m > f     m = 12(x + e) - 4m + 4f - 4m + 4x - 2f + 2x         = 12(x + e) - 8m + 2f + 6x         = [144(x + d) - 12 - 12f + 12x] - 8m + 2f + 6x         = 144(x + d) + 18x - 10f - 8m - 12         = [144(m - x)/7] + 18x - 10f - 8m - 12   7m = 144m - 144x + 126x - 70f - 56m - 84     So (if m > f) 81m - 70f - 18x  =  84..................................(3b)                                                      ==============================Hence the system of equations is:x3 - (f - m)x - m = 0.......................................(1) and if f > m57m - 6f + 49x = 1365 ................................(2a)137m - 126f - 18x = 84................................(3a)and if m > f29m + 22f + 49x = 1365 .............................(2b)81m - 70f - 18x  =  84..................................(3b)No real solutions exist when m http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gif f, but when f > m Mathematica produces the solution f = 25, m = 24, x = 3which forces a = 0b = 21c = 0d = 0e = 10g = 14h = 6 [/hide] Title: Re: Monkey Math Post by Noke Lieu on Oct 31st, 2010, 3:28pm It took me half an hour to read the puzzle statement! :D Powered by YaBB 1 Gold - SP 1.4! 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