After the death about 290A.D.of Diophantus a famous Greek mathematician someone !
Question:
After the death about 290A.D.of Diophantus a famous Greek mathematician someone described his life as a puzzle
he was a boy for 1/6 of his life
after 1/12 more, he aquired a beard
after another 1/7, he married
in the fifth year after marriage his son was born
the son lived half as many years as his father
diophantus died 4 years after his son
how old was diophantus when he died?
how did u get dat answer?
best answer plz
Answers:
This is a simultaneous equation question.
Firstly you know the guy dies in 290AD (i hope thats what it means). therefore you know he was born in 290AD-x (x is his age).
1/6 of his life is therefore 1/6x so he was a boy for 1/6x, and then after 1/12x he grew a beard. His age at this point was 1/6x+1/12x, and when he gets married his age is 1/6x+1/12x+1/7x which is simplified to 33/84x.
After 5 more years he has a child so he has a child at 33/84x+5.
The son lives for 1/2 of his fathers life so he lives for 1/2x.
We know that the father dies in 290AD so his son dies in 286AD.
Finally, the year in which the son dies is therefore:
The year of Diophantus birth+ his life up until he gives birth to his child+the lifespna of the child.
His life as explained is 290-x+33/84x+5 and the lifespan of the child is 1/2x.
290-x+33/84x+5+1/2x=286
4-x+33/84x+5+1/2x=0
9-x+33/84x+1/2x=0
9-x+75/84x=0
9=x-75/84x
9=9/84x
9*84=9x
x=84years
I hope this is right, also i got this first before the guy who said hes working on it, unless of course im wrong.