All the kings are coming together for a celebration and they are all bringing their elephants with them. Our task is to design holding pens for these elephants. Here are the king’s directions.



1. The pens have to hold exactly 42 elephants - no more, no less.



2. You will have exactly 100 logs – no more, no less. The sides of each pen are only one log in length. Pens may not share a side.



3. You must use ALL the logs, but the pens do not have to be the same shape.



4. Each pen holds exactly two less elephants than its number of sides. For example: a square pen will hold two elephants (4-2) and an octagon pen will hold six elephants (8-2).

All the kings are coming together for a celebration and they are all bringing their elephants with them. Our task is to design holding pens for these elephants. Here are the king's directions.



1. The pens have to hold exactly 42 elephants - no more, no less.



2. You will have exactly 100 logs – no more, no less. The sides of each pen are only one log in length. Pens may not share a side.



3. You must use ALL the logs, but the pens do not have to be the same shape.



4. Each pen holds exactly two less elephants than its number of sides. For example: a square pen will hold two elephants (4-2) and an octagon pen will hold six elephants (8-2).

I'm not much of a mathematician but was able to solve it in about 5 minutes by trial and error.
It's pretty clear that to use 100 logs for only 42 elephants, the pens have to be 'wasteful.'
In other words, a lot of 3 sided pens for only 1 elephant.
That's all I will say for now.

(If your teacher requires equations I can't help.)

There are multiple answers to this problem - I've found three so far using pens of 3 sides or more. And yes, you must use at least some 3-sided pens.

However I suggest you consider the use of 2-sided pens, which can hold zero elephants each. Then the problem is real simple to solve.