Thread: greek maths puzzle

I found this maths riddle in an article and i can't work it out!!

somethings telling me this is easier than im making it and it should be quite simple??

His boyhood lasted 1/6 of his life, he maried after 1/7 more, his beard grew after 1/12 more and his son was born 5 years later, the son lived to half the fathers age and the father died 4 years after the son.

how long did the father live??



Any help would be great

cheers

Alex

The son lived to 1/2 his fathers age:

The fathers life proceeded as follows:



Sub the first equation into the second and solve for F, the father's age.

Originally Posted by alexc63

I found this maths riddle in an article and i can't work it out!!

somethings telling me this is easier than im making it and it should be quite simple??

His boyhood lasted 1/6 of his life, he maried after 1/7 more, his beard grew after 1/12 more and his son was born 5 years later, the son lived to half the fathers age and the father died 4 years after the son.

how long did the father live??



Any help would be great

cheers

Alex

Diophantus

CB

cheers guys

appreciated

" . . . . have you figured out Diophantus's son's age yet? No? Well - the problem says that the son lived to half his father's life. From the other half, if you take away 1/6th + 1/12th + 1/7th then you are left with with 5+4 = nine years. So 3/28th of his age was 9, so Diophantus turns out to have lived till 84, and his son till 42." -

Storytelling Science: Aryabhata and Diophantus' son

"This gives rise to a linear equation in Diophantus’ age x (much simpler than anything Diophantus has done) with x = 84 as the solution." -

http://www.fen.bilkent.edu.tr/~franz/M300/diopro.pdf




ok, this site is even better: http://mathworld.wolfram.com/DiophantussRiddle.html

k