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Can you help me for a math problem?
Diophantus was a Greek mathematician who lived in the third century. He was one of the first mathematicians to use algebraic symbols.
Most of what is known about Diophantus's life comes from an algebraic riddle from around the early sixth century. The riddle states:
Diophantus's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and Diophantus died four years after his son.
How many years did Diophantus live?
(I do my own homework, i just dont get this)
1 Answer
- MathMan TGLv 71 decade agoFavorite Answer
Draw a time line of his life:
| - - - - - - - - - - - - - - - - - - - - - - - - - - - - - |- - - < n/2 > - - - - | - - - - - - - | ← n years life
Youth (gap) Grew beard (gap) Married (gap) Son born (gap) Son died
1/6 .... 1/12 . . . . . . .... 1/7 ... . . . . . . . 5 . . . . . . . .. . . .... | - - - 4 - - - |
n = n/6 + n/12 + n/7 + 5 + n/2 + 4
n = n/4 + n/7 + 5 + n/2 + 4
n = 3n/4 + n/7 + 9
1/4 n - 1/7 n = 9
7n - 4n = 9 * 28
n = 3 * 28 = 84
84/6 = 14 = youth
84/12 = 7 years gap, grew beard at 21
84/7 = 12 years gap, married at 33
5 years gap, had son at 38
Son lived 42 years - to his age 80
4 years later he died at 84.