OK, I get it. Still 2 or more possible solutions following the second clue, meaning there must be 2 or more solutions, i.e. the sum must be 13.

so 1,6,6, or 2,2,9

so it must be 1,6,6, again assuming that 2,2,9 can't have a youngest. Which is bollocks, as I'm a twin and I'm younger by a minute. As my twin delights in reminding me.

Yes, that's been extensively covered.

Where is this pig problem?

So here is my problem. I'll arrange the pigs as:

0,19,2,3

great, every pen I go to gets closer to 10, until I go round again. My final pen is 7 "off" the target value and the first goes back to being 10 off the target value and surely that will never change?

This isn't a mathematical problem I reckon.

As you've already said, it would not be possible.

Question from a Year 9 (!) maths paper...
A student in a maths class was trying to get some information from her teacher.
Sally: How many children have you got, Mrs Smith?
Mrs Smith: Three
Sally: How old are they?
Mrs Smith: If you multiply their ages together it comes to 36
Sally: That doesn’t give us enough information
Mrs Smith: If you add the ages, you get the number on the door over there
Sally: That still doesn’t give us enough information, Mrs Smith
Mrs Smith: The youngest is called Amanda
Sally: Thank you, now we know how old they are
How old are her children?

AND

Problem. — Place twenty-four pigs in four sties so that, as you go round and round you may always find the number in each sty nearer to ten than the number in the last.

Edited by mr. apollo (11 Mar 2016 3.55pm)

How would 3 be closer to 10 than 9?

Definitely not a numerical/maths only answer.