Author 
Topic: Monkey Maths (Read 2564 times) 

ThudnBlunder
wu::riddles Moderator Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


Monkey Maths
« on: Oct 30^{th}, 2010, 9:38am » 
Quote Modify

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.

« Last Edit: Nov 26^{th}, 2010, 3:10am by ThudnBlunder » 
IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



ThudnBlunder
wu::riddles Moderator Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


Re: Monkey Maths
« Reply #1 on: Oct 30^{th}, 2010, 10:18am » 
Quote Modify

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. It is possible to derive three equations in three unknowns. Let m = present age of mother Let f = present age of father Let x = present age of monkey FIRST EQUATION 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... Then f + (m/x) = x[x + (m/x)] which simplifies to x^{3}  (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 years and m + a  x = m + (3f  3m  x)  x = 3f  2m  2x years Hence (if f > m) monkey will be m + a years old in 3f  2m  2x years OR if m > f, m + a = (f  x) + 2(m  f) and m + a  x = m + (2m  f  x)  x = 2m  f  2x years Now 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  5f So finally, if f > m then 3f  m  2x = 50[{(m + f)/7}  4] + 5 + 5x  5f This gives 57m  6f + 49x = 1365 ................................(2a) OR if m > f then 3m  f  2x = 50[{(m + f)/7}  4] + 5 + 5x  5f This 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 + x So 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)  4f  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  84 So (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: x^{3}  (f  m)x  m = 0.......................................(1) and if f > m 57m  6f + 49x = 1365 ................................(2a) 137m  126f  18x = 84................................(3a) and if m > f 29m + 22f + 49x = 1365 .............................(2b) 81m  70f  18x = 84..................................(3b) No real solutions exist when m f, but when f > m Mathematica produces the solution f = 25, m = 24, x = 3 which forces a = 0 b = 21 c = 0 d = 0 e = 10 g = 14 h = 6

« Last Edit: Nov 30^{th}, 2010, 8:27am by ThudnBlunder » 
IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



Noke Lieu
Uberpuzzler
pen... paper... let's go! (and bit of plastic)
Gender:
Posts: 1884


Re: Monkey Math
« Reply #2 on: Oct 31^{st}, 2010, 3:28pm » 
Quote Modify

It took me half an hour to read the puzzle statement!


IP Logged 
a shade of wit and the art of farce.



