Math is old. Diophantus of Alexandria (now Egypt, then under a Greek rule) who lived 18 centuries ago is frequently called the father of algebra or a satan of algebra. He wrote 13 books that he titled "Arithmetica" containing various numerical problems and their solutions. Diophantus was the first Greek mathematician who recognized fractions as numbers. In a tribute to him one of his admirers described his life as a fraction riddle:

*Diophantus' youth laster 1/6 of his life. He grew a beard after 1/12 more of his life. After 1/7 more of his life, Diophantus married. Five years later he had a son. The son lived exactly 1/2 as long as his father, and Diophantus died just four years after his son's death. All of this totals the years Diophantus lived.*

The question is whether all this info is enough to determine how long Diophantus lived? Prove it!

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Answer ideas accepted any time until midnight on Saturday September 10th (EST), on our Family Puzzle Marathon. They will be hidden till then and everyone who submitted a valid solution will get a puzzle point.

## 10 comments:

Yes, as it's a pretty straightforward algebra problem:

1/6x + 1/12x + 1/7x + 5 + 1/2x + 4 = x

Solve for x, and you get 84.

1/6x+1/12x+1/7x+5=1/2x-4

=84 years old

The answer is 84

1/6x+1/12x+1/7x+5+1/2x+4=x

9=x(1-1/6-1/12-1/7/-1/2)

9=3/28x

3x=9*28

x=84

1/6*x + 1/12*x + 1/7*x + 5 + 1/2*x + 4 = x

14/84*x + 7/84*x + 12/84*x + 42/84*x + 9 = x

75/84*x + 9 = x

9/84*x = 9

x = 84

Yes, there is enough information. Diophantus lived to be 84.

1/6x +1/7x + 1/12x + 5 + 1/2x + 4 = x, where x = length of D's life

42/252x + 36/252x + 21/252x + 5 + 126/252x + 4 = 252/252x

225/252x +9 = 252/252x

9 = 27/252x

252/27 x 9 = x

84 = x

14 (1/6 = youth) + 7 (1/12 = beard) + 12 (1/7 = married) + 5 (had son) + 42 (1/2x = son's life) + 4 = 84

Yes there are enough clues here. Using L (or could be x or whatever) for Diophantus Lifespan....several ways to approach it.

One can nose around (iterate) and see what number fits, probably something between 50 and 100. That is quite often a good approach to puzzles, but maybe not respected as a proof.

Or play the algebra,

1/6L + 1/12L + 1/7L + 5 + 1/2L + 4 = L

That does seem to be a clumsy sequence of numbers....so

L/6 + L/12 + L/7 + L/2 = L - 9

find a common denominator in those fractions, like 84, bingo, nice,

just see if it works

84/6 is 14, 84/12 is 7, 84/7 is 12, 84/2 is 42 ....75....L is indeed 84.

Alternatively we could do it with some decimals, get pretty darn close and round it off:

1/6 = .1666

1/12= .0833

1/7= .1429

½ = .50

subtotal about .893

.893L = L-9

L - .893L = 9

.107L = 9

L = 9 / .107 = 84.112

D was (1/6 +1/12 +1/7)x +5 years old when he get his son.

x being his age at his death.

his son lived y= 1/2 x years

D died 4 years after his son's death so we have to take into account the time before he get a son plus the time that his son lived plus four which is:

(1/6 + 1/12 + 1/7)x + 5 + 1/2x + 4 = x

solving for x,

x=84

anne-Marie

Diophantus lived to be 84 yers old.

If his youth was 1/6 of his life (L), then the rest of his life can be written as:

1/12L + 1/7L + 5 + 1/2L + 4 = 5/6L

L=84

Yes, there is enough information.

1/6x + 1/12x + 1/7x + 9 = 1/2x

Solve for x which equals 84.

84 years old is an amazingly long life span for a 2nd century AD, but it does explain 13 Arithmetica books. Congratulations on exact answer and another puzzle point to Bean, icmdiaz, Marina M., Ilya, Annie, Tom, anne-marie, Susan and Lisa.

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