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Originally Posted by Mike Betts
it works for all combinations
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Here's the algebraic proof that the solution of grabbing 32 coins and flipping them works.
Let's call our pile of 18 Pile I
Let's call our pile of 32 Pile II
Let a be the number of heads in Pile I
Let b be the number of tails in Pile I
Let x be the number of heads in Pile II
Let y be the number of tails in Pile II
Therefore,
eq1: a + b = 18 (From Pile I)
eq2: x + y = 32 (From Pile II)
eq3: a + x = 32 (From the total number of heads)
eq4: b + y = 18 (From the total number of tails)
Combining eq2 & eq3 we see that,
y - a = 0 -->
y = a
Therefore the number of heads in Pile I is the same as the number of tails in Pile II
Now, we flip all 32 coins in Pile II. So all the heads become tails and vice-versa. So now we have y heads and x tails in pile II.
Since a = y, we see that the number of heads in both piles I & II is equal!