## May 11, 2009

### Maths Challenge #4 – How Old Was Diophantus?

Diophantus’ childhood lasted one sixth of his life. His beard grew after one-twelfth more. He married after one-seventh more. His son was born five years later. The son lived to half his father’s age. Diophantus died four years after his son. How old was Diophantus when he died?

###### Who was Diophantus?

"Diophantus was probably Greek and he lived in ancient Alexandria. Some time around AD 250 he wrote a book about solving algebraic equations - with a slight twist: the solutions were required to be fractions, or better still, whole numbers. Such equations are called Diophantine equations to this day."

Find out more in Professor Stewart's Cabinet of Mathematical Curiosities by Ian Stewart

### 6 comments by 2 or more people 1. #### Jon Lewis

You can kind of guess this because 84 is the lowest common multiple of 6, 12, 7.
Alternatively you can work it out the long way….

11 May 2009, 13:46

2. #### Eleanor Lovell

Diophantus was 84 when he died. Let $x$ be his age. Then; $\frac{x}6+\frac{x}{12}+\frac{x}7+5+\frac{x}2+4=x$

So $\frac9{84} x=9$

Therefore $x$ = 84

12 May 2009, 12:53

3. #### Mathew Mannion

For those people who don’t really understand how you can just pluck those fractions out of thin air…

Taking the length of Diophantus’ life as $D_L$...

Diophantus’s childhood lasted one sixth of his life.

So, $D_C=\frac16D_L$

His beard grew after one-twelfth more.

So, $D_B=D_C+\frac1{12}D_L=\frac3{12}D_L$

He married after one-seventh more.

So, $D_M=D_B+\frac17D_L=\frac3{12}D_L+\frac17D_L=\frac{21}{84}D_L+\frac{12}{84}D_L=\frac{33}{84}D_L$

His son was born five years later.

So, $D_{SB}=D_M+5=\frac{33}{84}D_L+5$

The son lived to half his father’s age.

So $S_L=\frac12D_L$

Diophantus died four years after his son.

The date of Diophantus’ death is the date of his son’s birth plus his son’s life plus 4, so: $D_L=D_{SB} + S_L+4=\frac{33}{84}D_L+5+\frac12D_L+4=\frac{33}{84}D_L+\frac{42}{84}D_L+9=\frac{75}{84}+9$

So we can solve for $D_L$ $D_L-\frac{75}{84}D_L=9$ $\frac{9}{84}D_L=9$ $D_L=84$

12 May 2009, 12:55

4. #### David

interesting to know that history is relying on this problem / puzzle (which is from that time indeed) to state the man’s age!

31 May 2009, 00:06

5. #### Asif

x = age of Diophantus at death

1/6x + 1/12x + 1/7x + 5 + 1/2x + 4 = x… Read more

Re-arrange divide by x
1/6 + 1/12 + 1/7 + 1/2 + 9/x = 1

Combine some fractions
9/12 + 1/7 + 9/x = 1

Combine more fraction
63/84 + 12/84 + 9/x = 1

Combine even more fractions
75/84 + 9/x = 1

Subtract 75/84 from both sides
9/x = 84/84 – 75/84

Summate
9/x = 9/84

Equate
x = 84

01 Jun 2009, 22:54

6. #### zaipong

I think it should be 65.333 years old haha.

coz his son only lived to half Diophantus’s age.
which means,

SL = 1/2 (x-4)

**quoting Mathew Manion

I think it will be clearer if we do a time scale for this problem

01 Jun 2009, 23:39

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Warwick Challenges are mini academic challenges from University of Warwick professors, set via the micro-blogging service Twitter (and also via this blog).
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