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Penny Problem (Posted on 2003-10-14) Difficulty: 3 of 5
You are blindfolded before a table. On the table are a very large number of pennies. You are told 128 of the pennies are heads up and the rest are tails up.

How can you create two subgroups of pennies, each with the same number of heads facing up?

See The Solution Submitted by Ravi Raja    
Rating: 3.7500 (16 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Heads I win, tails you lose. | Comment 9 of 18 |
Let Z be the original mass of coins, consisting of 128 heads and x tails. Select 128 coins arbitrarily (by feeling, since you are blindfolded). Call this group A. A consists of m heads and n tails. m + n = 128. Z is now reduced to 128 - m heads and x - n tails. Now flip all the coins in A. A will then consist of n heads and m tails. But since m + n = 128, it follows that 128 - m = n. So the selected group A and the revised original pile Z now contain the same number of heads.
  Posted by Dan on 2003-10-15 04:07:22
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