You have 42 parrots and 7 cages of various sizes. Lock all the parrots inside the cages so that no cage is empty and each cage contains an odd number of parrots.

A hint: this puzzle requires some unusual thinking.

Warning: the solution is below.

At first it seems simple: 42 divide by 7, 6 parrots in each cage. But the number of parrots in each cage should be odd. OK, let's try 1 parrot in each of the first 6 cages. How many parrots are left? 42-6=36. Put 36 in the last cage. Again, not an odd number.

Now we note that in general, there is no way that 7 odd numbers could be added to produce an even number. So, there must be a trick! A trick that does not involve the numbers.

And the trick is to place cages inside each other. There was a note of cages being of different sizes. So, place one parrot inside a small cage, place this cage inside a large cage. Both cages have odd number of parrots. You are left with 41 parrot and 6 cages. Now, it is easy. Anyway you do it, it works. Just keep numbers odd.

You can try this puzzle now on your friends and family if you are not afraid to be beaten. A simpler version of it is to pack 6 presents inside 7 boxes of various sizes so that no box is empty.

You can try this puzzle now on your friends and family if you are not afraid to be beaten. A simpler version of it is to pack 6 presents inside 7 boxes of various sizes so that no box is empty.

Enjoy!

## 3 comments:

I'll start the comments off. I think that this is an interesting example of having to put everything you know on the back burner and just think the problem out. You need an odd number of birds to go into an odd number of cages, but it can't be done.

So you have to be innovative. Because you are dealing with real objects that have to be confined, you can put one cage containing birds into an empty cage. That gets around the problem. Or does it?

The cage inside the cage begs the question of what is inside the larger cage. Is it a cage, or is it the birds? What would happen if you let the birds out and they occupied the larger cage. What would you have then? You'd have an empty cage which means that you have disobeyed the conditions of the problem.

If you leave the birds in the smaller cage aren't you counting them twice: Once for the small cage and once for the larger one?

I don't know much about group theory, but if you put a group of 7 inside a larger group container, what have you done according to group theory. My guess is that you have cheated somehow. It's an interesting problem to try and think out, but is it math?

Okay, I had a different out-of-the-box answer. Arrange the cages in a circle, with 6 around the edge and one in the center. Put 7 birds in each of the outer ring of cages. The center cage will quickly get lots of bird poop kicked into it; thus, it is not "empty". I have witnessed the messiness of owning pet birds at a friend's house. Yuck!

Although I am a little late, I want to answer the Parrot puzzle. After shifting all the birds around from cage to cage I could not come up with odd numbers for each cage, but...when one of the birds flew away, it was possible to meet the requirement for odd numbers.

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