Friday, September 9, 2011

Life, in Fractions


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!

Top image by freeasfreedom, distributed under the CCL.

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:

Bean said...

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.

icmdiaz said...

1/6x+1/12x+1/7x+5=1/2x-4
=84 years old

Marina Marshak said...

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

Ilya said...

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

Annie said...

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

Tom said...

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

Anonymous said...

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

Susan said...

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

Lisa said...

Yes, there is enough information.

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

Solve for x which equals 84.

Maria said...

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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