Subj:.....A Pile Of Pennies (S659)           From: MathNexus.wwu.edu           on 1/25/2009 Drawing from MathNexus.wwu.edu...

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Source: http://mathnexus.wwu.edu/Archive/problem/detail.asp?ID=152   You are blindfolded, then asked to sit down at a table.  On the table is a large number of pennies.  You are told that ten of the pennies show HEADS up, while the rest show TAILS. You cannot feel the difference between a HEADS or TAILS being up. Your Task: Arrange the pennies into two disjoint groups so that each group shows an equal number of HEADS up. Note: Though blindfolded, you are still able to count the pennies and turn any penny over while sorting them into groups. . . Drawing from tom on 8/21/2009 . . Hint: Without a blindfold, try to solve the problem by experimenting with a pile of pennies.  Be sure to always start with exactly ten pennies showing HEADS. . . Finger pointing down from darrell94590 on 1/2/2006 .       Drawing from Ripleys-Believe It Or Not ..               THE SOLUTION   If you have not solved it, you will be surprised by the simplicity of the solution.  While blindfolded, move any ten pennies from the original pile into a new group.  Turn over all ten of these pennies.  You have now solved the problem. Again, if you do not believe it, try it!  Now why does it work.  In the pile of ten pennies separated out, N of those showed HEADS (where N is a value from 0 to 10) and 10-N showing TAILS, with the remaining 10-N HEADS being left in the original group.  When you flipped all ten pennies, you now have N showing TAILS and 10-N showing HEADS, just as was desired. As a generalization, if the original pile had Y pennies showing HEADS, you would....?

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