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 Penny Problem (Posted on 2003-10-14)
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)

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 Got it... | Comment 10 of 18 |
First, make a subgroup of 128 pennies, dun matter which ones really. Flip all of them over. All of the pennies should now have equal face up and face down.

The ones that have heads will be flip to tail. The tails will be flip to head so even amount. So have a number of pennies in group A. Take 128 pennies from it to form group B. Now you have some number of heads (call it H). So group A has 128 - H heads. By flipping the pennies in group B you will get 128 - H heads as well. Ex: 10 pennies, 4 heads. Take 4 pennies and have 1 head and 3 tails in group B. Group A now has 3 heads. Flip pennies in group B and get 3 heads now. So same number.

 Posted by PWP on 2003-10-17 12:49:35

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