Puzzle: 50 coins
With your eyes closed, somebody puts 50 coins on the table: 40 heads and 10 tails. Let's assume that you cannot tell the side of the coin from touching or any other mechanism. Your job is to divide up the coins in two groups. All the 50 coins must be in one of the two groups, but the number of coins in each group may be different. Your aim is to have the same number of tails in each group. What do you do?
3 Comments:
If you can turn the coins over - this one was easy
Just divide the coins into a group of 10 (G10) and a group of 40(G40). Let us assume that G40 has x tails, therefore G10 has 10-x tails and therefore x heads.
Just flip over each of the coins in G10 and it will now have x tails and 10-x heads.
I am wondering if there is a solution without flipping coins over.
Abhi must have a more mathematically inclined brain than I :)
I think arrived at the same solution as him, but with a different methodolgy.
I paired the ratio down (2 tails, 8 heads) and put 10 quarters on the table in front me.
Pile one contains 2 coins, pile two contains 8. As long as you flip all the coins in pile one, you will have the same number of tails in both groups (whatever the equivalent ratio).
what s the solution?
Post a Comment
<< Home