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?
Separate out 128 pennies from the large group (call these 128 group A, and the remainder group B)
Turn all of Group A over, one by one. Then you have the same number of heads-up-pennies in both groups.
Proof:
When you separate group A, you have from 0 to 128 heads-up-pennies there, lets call the number X.
Therefore, you have 128-x heads-up-pennies in group B.
If you flip over all the pennies in Group A, you now have 128-x heads-up pennies in Group A as well.
(e.g. If you had 5 heads-up-pennies in group A when you first separated the groups, then you must have had 123 tails-up in group A, and 123 heads-up in Group B. When you flip over *ALL* in group A, you will have 123 heads-up in Group A as well.... this satisfies the problem.)
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