This is not a new puzzle, but it popped up in the Introduction to Logic class I am taking online.
I thought the solution was very interesting. I'll put the solution in the comments later, so you don't have to look it up until you are done thinking about it.
You are shown 100 coins arranged on a table, with 40 showing heads up and the other 60 showing tails. You are then blindfolded and told that you can move the coins around on the table and flip them over all you want, but you can't remove any coins and you can't see through the blindfold. When you are done, the coins should be separated into two groups (they don't have to be the same size) and there should be exactly the same number of heads showing in each group.
The solution to this does not involve any tricks (ask a friend to tell you when you've got it) or imaginative thinking (turn each coin so it is standing on its edge), just deduction. The original solution I came up with, in fact, was an algebraic one. Forty years of teaching math tends to make me attack problems by writing equations. The first comment contains a hint if you get stuck. Good luck.
A common technique in mathematics to finding patterns is to do a simpler problem. So experiment with just ten coins on your desk and see what works. Change the number of heads up every so often to help you see a pattern. So do 4 heads and 6 tails for a while, then 3 and 7, and so on.
ReplyDeleteRemember there are 40 heads and 60 tails. First, separate them (at random; remember you are blindfolded) into a group of 40 and a group of 60, probably mixed heads and tails. You can’t count them, but, suppose there were 15 heads and 25 tails in the 40 group. That would leave 25 heads and 35 tails in the 60 group.
ReplyDeleteoriginal
40 Heads 60 Tails
40 group-----------------60 group
15 Heads 25 Tails-----25 Heads 35 Tails
If I now flip over every coin in the 40 group, I would have 25 heads in each group. Let’s do it a little more generically. I am going to separate into a group of 40 (group A)and a group of 60 (group B) . Suppose there are X heads in group A. Since there are 40 coins altogether in that group, the other 40 – X coins must be tails. Since you had X heads in group A and 40 heads when you started out, the other 40 – X heads must be in group B. We don’t really care how many tails are in group B.
original
40 Heads 60 Tails
40 group------------------------60 group
X Heads (40 – X) Tails-----(40 – X) Heads ? Tails
If I now flip over all the coins in group A, I have 40 – X heads in each group.
To be even more generic, you can assume that you don’t know how many coins to put in each group to start, say, Y in group A and the remaining 100 – Y in group B.
original
40 H 60 T
Y coins in group A-------(100 – Y) coins in group B
X Heads (Y – X) Tails---(40 – X) Heads ? Tails
Which becomes, after you flip all the coins in group A,
Y coins in group A-----(100 – Y) coins in group B
X Tails (Y – X) Heads-----(40 – X) Heads ? Tails
Set Y – X = 40 – X and get Y = 40. So group A has 40 and group B has the remaining 60.
So if the original set of coins had 28 heads and 72 tails, move the coins into two groups, one of 28 and one of 72 and flip all the coins in the 28 group. Now the two groups have exactly the same number of heads. I don't know ahead of time how many heads there will be. It depends on how many of the heads got pulled into the 28 group. But the number of heads will be the same. Pretty cool, huh?