Oink !

You have 9 pigs.

Build 4 pens to house the pigs.

Every pen must hold an odd number of pigs.

All pens must be used, and all pigs accommodated.

How?

Originally posted by Mathurine
You have 9 pigs.

Build 4 pens to house the pigs.

Every pen must hold an odd number of pigs.

All pens must be used, and all pigs accommodated.

How?

Three pens with three pigs each, one large pen surrounding the other pens. The small pens hold three pigs - an odd number - and the large pen holds nine pigs - an odd number - and three pens.

Richard

Originally posted by Shallow Blue
Three pens with three pigs each, one large pen surrounding the other pens. The small pens hold three pigs - an odd number - and the large pen holds nine pigs - an odd number - and three pens.

Richard

Splendid! 😀

very good problem (and solution)

Originally posted by Shallow Blue
Three pens with three pigs each, one large pen surrounding the other pens. The small pens hold three pigs - an odd number - and the large pen holds nine pigs - an odd number - and three pens.

Richard

Alternatively, you can build normal pens and have two pigs in three of the pens and three in the fourth. Two is an odd number, because it's the only even prime. 😉

You can actualy make 4 perfect square pens and Place them in a PLUS shape so one side from each pen makes 4 sides for a new one.

..[]
[] []
..[]


kinda like that (of course the corners would touch). so the 4 pens in that position make a new one in the middle.

put 1 pig on left 1 pig in middle and 1 pig in right.
3 in top and 3 in bottom.

But that's "actually" constructing five pens!

... nice variant, notwithstanding!

🙂

Oink!

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