**The Math Adventurer’s Rule: Figure It Out for Yourself!**

Whenever I give a problem in an Alexandria Jones story, I will try to post the answer soon afterwards. But don’t peek! If I tell you the answer, you miss out on the fun of solving the puzzle. Figure it out for yourself — and then check the answer just to prove that you got it right.

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Remember, don’t peek until you’ve given the problem a good workout on your own!

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**For Diophantus:** Did you try algebra? Try to write each part of his life as a fraction of the whole number of years.

**For Fibonacci:** On January 1st, you have a pair of baby rabbits. On February 1st, your rabbits are half grown. On March 1st, your original pair of rabbits is mature, and they produce one new pair of babies. Continue the pattern…

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## The Epitaph of Diophantus

Let’s make the letter D stand for the number of years that Diophantus lived. He was a boy for 1/6 of his life, which would be D/6 years. Then he was a youth for 1/12 of his life, until he got a beard — that’s D/12 years. After another seventh, he got married, so that’s D/7 more years as a bachelor. Then, when he had been married 5 years, his son was born. That’s just a plain 5 years, not a fraction.

This is where the riddle gets tricky. The phrase “when he had reached the measure of half his father’s life” could mean the son was half as old as Diophantus was when the son died, or that the son lived half as long as Diophantus lived TOTAL. Take the second interpretation. (The first will lead to a fractional age, but Diophantine problems must have whole number solutions.) So the son lived D/2 years.

Finally, Diophantus lived 4 more years after his son died. Again, that’s a plain 4 years, not a fraction of anything.

Put all of these parts together to make an algebraic equation for how long Diophantus lived:

D = (D/6) + (D/12) + (D/7) + 5 + (D/2) + 4

And then imagine solving that with Roman numerals!

Find a common denominator and solve the equation:

D = (75/84)D + 9

D = (25/28 )D + 9

(3/28 )D = 9

(1/28 )D = 3

D = 3 x 28 = 84 years

## Fibonacci’s Rabbits

January: Baby rabbits = 1 pair.

February: The bunnies are half grown = 1 pair.

March: One pair of adults produces one pair of babies = 2 pair.

April: One pair of adults, one new set of babies, and the first babies are half grown = 3 pair.

May: Two pairs of adults produce two pairs of babies, and last month’s babies are half grown = 5 pair.

June: 8 pairs.

July: 13 pairs.

August: 21 pairs.

September: 34 pairs.

October: 55 pairs.

November: 89 pairs.

December: 144 pairs.

So on Christmas day, you are being overrun by a total of 144 x 2 = 288 rabbits. Sounds to me like it’s time for rabbit stew!

## To Be Continued…

Read all the posts from the May/June 1998 issue of my ** Mathematical Adventures of Alexandria Jones** newsletter.

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This Diophantine word problems are really interesting, great contribution

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