[Guest]

Hi,

I did a quick search in the forum but i am not able to see this puzzle. Of course this puzzle is simple but just wanted to share with you.

There are 50 coins on table A, out of which 30 are showing heads and 20 are showing tails (I mean you can 30 heads up and 20 tails up). And you are blind folded.

Now you have to pick some number of coins from table A and keep them on table B (table B is empty initially) such that both the tables have equal number of heads up.

Thanks

Shravan

[Guest]

I've heard a variation of this problem given with cards instead of coins.

Take any 30 coins from table A, and place them opposite side up on table B. You're done.

Explanation:

Let's say there were n heads in that group of 30 before flipping. Therefore, the remaining pile (of 20) must contain (30-n) heads, since we know there are 30 heads total.

Before flipping, the pile of 30 has that same number of tails (since in a group of 30 coins with n heads, there must be (30-n) tails).

So by flipping that pile of 30, we end up with (30-n) now heads-up, which matches the heads-up count of the pile of 20.

[Guest]

I've heard a variation of this problem given with cards instead of coins.

Take any 30 coins from table A, and place them opposite side up on table B. You're done.

Explanation:

Let's say there were n heads in that group of 30 before flipping. Therefore, the remaining pile (of 20) must contain (30-n) heads, since we know there are 30 heads total.

Before flipping, the pile of 30 has that same number of tails (since in a group of 30 coins with n heads, there must be (30-n) tails).

So by flipping that pile of 30, we end up with (30-n) now heads-up, which matches the heads-up count of the pile of 20.

Nice and quick reply.... and yes ... this is the answer.... a simple puzzle