Here lies the remains of diophantus.he was a child for one sixth of his life.after one twelfth more,he became a man.after one seventh more,he married,and five years later his son was born.his son lived half as long as his father and died four years before his father.how old was diophantus when he died???

Question: Here lies the remains of diophantus.he was a child for one sixth of his life.after one twelfth more,he became a man.after one seventh more,he married,and five years later his son was born.his son lived half as long as his father and died four years before his father.how old was diophantus when he died???

Let D be the age of Diopahntus when he died (as an integer value).

Fractions are taken of his age, notably 1/6, 1/12, 1/7.

Since we are talking about Diophantus, and Diopahantine equations are equations with integer solutions, then it is reasonable to suppose that all numerical values used in the original statement involve integer values.

i.e. we can assume that when his age, D, is divided by 6, by 12, and by 7, then the results will be integers.

i.e. Diophantus' age is divisible by 6, 12 and 7

The only (integer) age, divisible by 6, 12 and 7, within normal lifespan is D = 84 years.

Check

1) he was a child for one sixth of his life. i.e. C = D/6 = 84/6 = 14 yrs

2) after one twelfth more,he became a man C + D/12 = 14 + 84/12 = 14 + 7 = 21 yrs

3) after one seventh more,he married. M = C + D/7 = 21 + 84/7 = 21 + 12 = 33 yrs

4) and five years later his son was born. F = M + 5 = 33 + 5 = 38 yrs

5) his son lived half as long as his father. S = D/2 = 84/2 = 42 yrs

6) and died four years before his father. Sd = F + S = 38 + 42 = 80 yrs

i.e. Diophantus was 80 yrs old when his son died, 4 yrs before his own death, which checks out!

Answer: Diophantus lived to the age of 84 yrs

answered Mar 16, 2014 by Level 11 User (80,500 points)