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# Diophantus

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Diophantus - Back to the Cool Math Games

We know little about this Greek mathematician from Alexandria, called the father of algebra, except that he lived around 3rd century A.D. Thanks to an admirer of his, who described his life by means of an algebraic riddle, we know at least something about his life.

Diophantus's youth lasted 1/6 of his life. He had his first beard in the next 1/12 of his life. At the end of the following 1/7 of his life Diophantus got married. Five years from then his son was born. His son lived exactly 1/2 of Diophantus's life. Diophantus died 4 years after the death of his son.

How long did Diophantus live?

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Diophantus - solution

There is an easy equation to reflect the several ages of Diophantus:

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

So the solution (x) is 84 years.

We know little about this Helen mathematician from Alexandria (called the father of algebra) except that he lived about the year 250 B. C. Due to one admirer of his, who described his life by the means of an algebraic riddle (math brain teaser), we know at least something about his life.

Diophantus's youth lasted 1/6 of his life. He had his first beard in the next 1/12 of his life. At the end of the following 1/7 of his life Diophantus got married. Five years from then his son was born. His son lived exactly 1/2 of Diophantus's life. Diophantus died 4 years after the death of his son.

How long did Diophantus live?

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• 2 months later...

i'm pleased to say that i figured this one out all by myself

here's the long way around it (or in my eyes, an organized way to see what's going on).

1. Y = (1/6)L

2. B = Y + (1/12)L

3. M = B + (1/7)L

4. S = M + 5

5. P = S + (1/2)L

6. L = P + 4

Diophantus' age at the point of...

L = he died

Y = the end of his youth

B = the end of his beard

M = the day he got married

S = the day his son was born

P = the day his son died

6 equations, 6 unknowns, cake!

Substitute P from 5 into 6 and solve for S

Substitute S from 6 into 4 and solve for M

Substitute M from 4 into 3 and solve for B

Substitute B from 3 into 2 and solve for Y

Substitute Y from 3 into 2 and solve for L

...and presto!

(don't be afraid if, as you're working it out, you get some weird fractions)

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I basically found all the multiples of 12 up to the resonable dying age. Then I took all of those numbers and tried to find the answer. It is much easier this way.

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I think the easiest way to approach it is to realize that, unless they were rounding, two numbers were needing to diving into the age evenly: 7 and 12. Thus, multiply 7 and 12 to get the age at which he died. The rest works out nicely from there.

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1/6+1/12=1/7+5+1/2x+4=x 1/6=14,1/12=7,1/7=12,14+7+12+5+4=x-1/2x,14+7+12+5+4+=1/2x,42=1/2x,x=84

he lived to 84

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• 4 weeks later...

Knowing that ancient Greeks did not use floating point operations, (yes they use portioning like 1/2 etc.) we can figure out that all the parts of life are integers or better natural numbers.

So from this i make:

1/2,

1/6,

1/7,

1/12

Finding the Lowest Common Multiple:

So 7*12 = 84.

Diophatius lived 84 years.

his life was:

Youth 14 years,

Beard 14 + 7 = 21

He got married 21 + 12 = 33 Years

His son was born on 33+5=38 years old

his son died when he was 38+42=80 years old

He died 84 years old.

I believe that solving this using X is not a puzzle but an equation.

This is the puzzle solution ;-)

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• 4 weeks later...

i agree with all of this except for the people that think the LCD gives up his age. just because you can split his life into 84ths does not mean whatsoever that he is 84. it just happens to work out that way. the easiest way to set it up is this way in my opinion:

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

x = 14/84x

7/84x

12/84x

42/84x

------

x = 75/84x + 9

-75/84x -75/84x

9/84x = 9

X84 X84

9x = 756

/9 /9

x= 84

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• 1 month later...
i agree with all of this except for the people that think the LCD gives up his age. just because you can split his life into 84ths does not mean whatsoever that he is 84. it just happens to work out that way.

That's true, except that we need to keep in mind that we're talking about the Greeks. So we certainly wouldn't expect a solution relatively prime to 84. Further, we're talking about a puzzle that the Greeks meant one another to be able to solve readiy; so we're probably looking for a factor or multiple of 84. That makes the possible solutions 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84, and (at the outside of possibility) 168. Of these, only 42 and 84 are at all reasonable, and we need only check them.

Of course we can solve it easily using algebra... but using any tools not available to the Greeks is cheating.

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• 3 weeks later...

The Greeks were using Algerbra at least by 600BC and the invention of Algerbra predates that by over 1000 years. Needless to say Greeks could have used algerbra to solve the problem.

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• 2 months later...

Some of the periods of his life are given in fractions of his total life and some are given in actual years. 75/84 of his life is accounted for in the fractions, and the rest of his life (the remaining 9/84) is given in years (9).

(9/84)x = 9

x = 84

Edited by Kend0g187
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• 2 weeks later...
1/6+1/12=1/7+5+1/2x+4=x 1/6=14,1/12=7,1/7=12,14+7+12+5+4=x-1/2x,14+7+12+5+4+=1/2x,42=1/2x,x=84

he lived to 84

Hmm... thats excactly coorect but lf we would've simply taken the lowst common multiple of 6,7,12 ie. 84 it would become simpler

Edited by Mihir
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• 1 month later...

Since his son's life is half of his life

so whatever information given there is equal to the other half of his lifetime

so let L = his whole life

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

(3/28) L = 9

L = 9 x 28 / 3

= 84

His life = 84 years old

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The text of the riddle states,

"Due to one admirer of his, who described his life by the means of an algebraic riddle" so it seems fair to use algebra to solve this puzzle.

My solution follows what others have done. I started by finding a common base for the constraints, which set me up to create two equations. Plugging one into the other . . .

11/28d + 9 = s

2s = d

11/28d + 9 = 1/2d

11/28d + 9 = 14/28d

9 = 3/28d

9X28 = 3d

3X28 = d

d=84

There's probably a hundred ways to prove this one. Kind of fun though.

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• 2 weeks later...

You would have to consider we are talking about someone who was living around 250 BC and people didn't live as long back then. 2 reasonable ages would be 28 or 42.

Edited by Enigma Time
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Talk about smart people, but he lived to be 84 years old.

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• 1 month later...

I as usual I have not read any previous post, it corrupts your thinking.

I figure the best answer for this is that he lived 1/1 of his life.

Also I notice he had an interesting life, not only did he get married as a youth but he got his first beard half way through his youth . . . this must be the new modern definition of youth, it goes 'til your thirty. Maybe in another ten years it will go 'til we're forty, that way we can stay young forever.

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• 4 months later...

I solved this quite a different way.. No equations with variables or finding the LCM at all.

Youth=1/6

First beard=1/6+1/12 (1/4)

Married=(1/4)+1/7 (11/28)

Son=(11/28)+5yrs

Son lived=14/28

Son died= 25/28+5 yrs

He died=25/28+9yrs

3/28 of his life=9yrs, so 1/28=3 years, and 28/28=84 years.

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