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Let the number of coins with heads facing upwards in the smaller pile be x.

The number of coins with heads facing upwards in the larger pile will be 5-x.

The number of coins with tails facing upwards in the smaller pile will also be 5-x. Once the coins in the smaller pile are flipped, the number of coins with heads facing upwards in the smaller pile will be 5-x, which is equal to the number of coins with heads facing upwards in the larger pile.

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let the number of heads in Small pile (5) be H

then, there will be

(5-H) heads in Big pile  (since there are total of 5 heads) and

(5-H) tails in Small pile (since total coins in small pile is 5)

So, when you flip the coins in the small pile, all the tails will become heads -> (5-H) heads in small pile after flipping.

 

 

cheers.

 

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Thank you all for your reply.  Greatly appreciate it.  Will get ds to try.

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This is a Bofore After question.

We have 5 Head and 7 Tails.

Small 5 Big 7 After Small Big 7 Result
5 Head 7 tail 5 Tail 7 tial No head
4 H 1 Tail 6 T 1 H 4 T 1 H 6 T 1 H 1 Head
3 H 2 T 5 T 2 H 3 T 2 H 6 T 2 H 2 Head

Just repeat until all Tails.

 

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This is interesting and I can try this “magic trick” too.

To think about it, it is only logical so. But I don’t know how to put these in Mathematical terms…

There are only 5 heads in the original group of 5 (“High-Five” group). So if 1 head left High-Five group to join the bigger group (“Lucky-Seven” group), 1 tail must come to fill the gap in High-Five group and get flipped (into 1 head) with the other 4 heads (flipped into 4 tails) in the High-Five group. Naturally as a result, there is only 1 head in Lucky-Seven and 1 head in High-Five group.

Likewise if N head(s) left High-Five group to join the bigger group (“Lucky-Seven” group), N tail(s) must come to fill the gap in High-Five group and get flipped into N head(s), with the other 5-N head(s) (flipped into 5-N tail(s)) in the High-Five group. Naturally as a result, there is only N head(s) in Lucky-Seven group and N head(s) in High-Five group.  

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