**Mathematical Amazements and Surprises**":

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"You are seated at a table in a dark room. On the table there are twelve pennies, five of which are heads up and seven of which are tails up. (You know where the coins are, so you can move or flip any coin, but because it is dark you will not know if the coin you are touching was originally heads up or tails up.) You are to separate the coins into two piles (possibly flipping some of them) so that when the lights are turned on there will be an equal number of heads in each pile."

"Your first reaction is 'you must be kidding!' How can anyone do this task without seeing which coins are heads or tails up? This is where a most clever (yet incredibly simple) use of algebra will be the key to the solution."

Posamentier Continues:

"Let's 'cut to the chase' (You might actually want to try it with 12 coins.) Separate the coins into two piles, of 5 and 7 coins each. Then flip over the coins in the smaller pile. Now both piles will have the same number of heads! That's all! You will think this is magic. How did this happen. Well, this is where algebra helps us understand what was actually done."

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Explanation: ...At the start, there are 5 heads showing among 12 coins. After separating into piles, let's say the 7-pile now has "h" heads. The 5-pile then has "5 - h" heads, and "5 - (5 - h)" tails (or, just "h" tails). Once you flip the entire smaller pile, all the tails ("h" of them) become heads, and all the heads become tails. Thus you are left with "h" heads in the "5-pile," the same as the number in the "7-pile." Whaaa-laaahhhh! Math is a beautiful thang!!

## 3 comments:

I followed Sol's link here. Thanks for suggesting the use of just 12 coins. I've seen a problem like this, with bigger numbers, and never played with it enough to 'get it'.

I didn't use algebra just now to get it, though. I made a table with 5 on one column (for the pile with 5 in it) and 7 on the other. That first pile can have from 0 to 5 Heads, and the other will have what I think of as the complement (what you called 5-h). Flipping the whole smaller pile makes them have the same. Very cool.

I'd like to do a magic show soon with tricks like these. (Challenge post.)

This can't be right.

Imagine that there are 7 heads total.

You divide them up into two groups: 5 (all heads) and 7 (with 2 heads). If you flip over the group of 5, you'll have 5 tails in that group, compared with 2 heads in the other group.

There are many other examples where this strategy doesn't work.

Hi Stuart, Posamentier is postulating a specific set-up where heads are fewer than tails (sorry if this wasn't clear from the post). In your instance of 7 heads, tails are the smaller group(5) and it is tails that come out equal.

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