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!





