In the future equations

Originally posted by lemon lime
[b]Derek will be 3 times his age today

Three times whose age?[/b]

Originally posted by JS357
If d =3d is not a typo, d=0 (a newborn)

Originally posted by AThousandYoung
It's supposed to read D=3d where D=d+y

Originally posted by AThousandYoung
In some future time, Derek will be 3 times as old as he is today. So if Derek is 15, then in 30 years he will be 45 - three times 15.

The main problem seems to be that we have six variables abcdey but five equations.

Originally posted by lemon lime
The question asks [b]Who is the youngest and how old is he?
This might be one of those brain teasers where most of the information is irrelevant. So I will say the answer is...

Derek is the youngest, and he is zero years old on his birthday.
(newborn, day of birth)

I'm assuming d =3d is not a typo,[/b]

Im so confused right now...

Apparently the important bit of the problem is the fact that all the ages are under 50 and what i din't mention was that it's to do with drinking age laws in the UK which means all the 5 people are between 18 and 50.
Knowing this you have to use something called a diophantine equation which,as has been said involves introducing a 6th unknown.
I've not fully researched it yet, but it implies you then have to try a few numbers and see if they work.
The answer published in the latest Sunday times was c is the youngest and he is 20.
Thanks for all your replies and I hope it's been interesting for you all.
I'd never heard of a Diophantine equation so when I've time it's yet another area of maths I'm going to look into.
I detested Maths at school.
How things change over the years!!

Originally posted by venda
Apparently the important bit of the problem is the fact that all the ages are under 50 and what i din't mention was that it's to do with drinking age laws in the UK which means all the 5 people are between 18 and 50.
Knowing this you have to use something called a diophantine equation which,as has been said involves introducing a 6th unknown.
I've not fully ...[text shortened]... f maths I'm going to look into.
I detested Maths at school.
How things change over the years!!

A Diophantine equation is basically just a polynomial equation where you seek only integer solutions.

I didn't care much for math either until I entered college to study computer science. Then the logic and math started clicking.

"The Monkey and the Coconuts" is a classic Diophantine problem.

Three sailors and a monkey were shipwrecked on a desert island. The sailors worked all day to collect coconuts, but were too tired that night to count them. They agreed to divide them equally the next morning. During the night, one sailor woke up and decided to get his share. He found that he could make three equal piles, with one coconut left over, which he tossed to the monkey. Thereupon he had his own share and left the remainder in a single pile,

Later that night, the second sailor awoke and, likewise, decided to get his share of coconuts. He also was able to make three equal piles, with one coconut left over, which he tossed to the monkey. Somewhat later, the third sailor awoke and did exactly the same thing with the remaining coconuts, making three equal piles, taking his share and tossing the leftover coconut to the monkey.

In the morning, all three sailors noticed that the pile was considerably smaller, but each thought that he knew why and said nothing. When they then divided the remaining coconuts equally, each sailor received seven and one was left over, which they tossed to the monkey.

How many coconuts were in the original pile?

79?

Originally posted by Blood On The Tracks
79?

I'm not sure it quite equates to the ages problem, in that you give the solution '7' at the end.

Anyway, easy to see there were 22 in last pile

That is the other 2 sailors shares after sailor 3 took his and gave monkey a coconut, realising this 'pile' is 2 x share at any one time is the key

So, he was originally looking at a pile of 34 (3 x 11, plus 1)

By same logic, previous pile was 52 (3 x 17, plus 1) - 17 being half the 34

And one more time, original pile is 79 (3 x 26, plus 1)

You say 79 works . .any other solutions? Your seem to suggest there may be

Originally posted by Blood On The Tracks
I'm not sure it quite equates to the ages problem, in that you give the solution '7' at the end.

Anyway, easy to see there were 22 in last pile

That is the other 2 sailors shares after sailor 3 took his and gave monkey a coconut, realising this 'pile' is 2 x share at any one time is the key

So, he was originally looking at a pile of 34 ...[text shortened]... 9 (3 x 26, plus 1)

You say 79 works . .any other solutions? Your seem to suggest there may be