Wednesday, May 27, 2009

Solution

If you like puzzles and you haven’t seen my post on Singapore where I shared a puzzle that some clients gave me, please take a look and read no further!

For those of you who have already been thinking about it, here’s the solution:

There are 30 coins in front of the blind man. 18 have heads facing up, 12 have tails facing up. His aim is to separate the coins into two groups such that there is an equal number of coins with heads facing up in both groups.

The blind man takes 18 of the coins and puts them in a group to his left. He takes the remaining 12 and puts them in a group to his right. He doesn’t flip any coins over yet. The trouble is that he doesn’t know how many coins on the left have heads facing up. It could be all 18, it could be 12, and it could be a minimum of 6. What he does know is that the heads that don’t go to the left pile, must be in the right pile. So if there are 18 heads on the left, there are zero on the right. If there are 12 heads on the left, there are 6 on the right etc. To insert some letters: if there are x heads on the left, then there are 18-x heads on the right.

The blind man then flips over all the coins in the left pile. Problem solved. If there were 18 heads on the left then they all turn to tails. If there were 12 heads on the left, then those 12 turn to tails and the remaining 6 turn to heads. To use letters again, if there were x heads on the left, then after flipping all of the coins on the left, there are now 18-x heads on the left. This is the same number as the number of heads on the right.

I like puzzles and I thought that was a good ‘un. Hope you had fun thinking about it.

2 comments:

  1. Nice. It really does work. I couldn't break out of the box of thinking that you must split the coins into two equal groups and was fliping all the coins over, instead of just one side. Epic fail on my part. I'll post the world's hardest logic puzzle sometime soon, which I did manage to solve (sort of).

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  2. Very nice. Post moar!

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