Penny Wise

A friend told me this riddle once upon a time and I liked it for its elegance. Goes like this.pennies

There are 50 pennies spread out on a table, tails-up. I blindfold you and flip over 17 of them at random. While blinded, can you divide the pennies up into two piles such that you can be sure there are an equal number of coins lying heads-up in each pile? You may turn over whichever coins you wish, as many times as you wish.

No stupid play-on-words trick. Nor do you have fine-enough sensory resolution on your fingertips to determine which side of the penny is facing up based on the imprint. The solution is purely logical.

Take your time..

Figured it out?  It’s pretty easy once you realize that both piles don’t have to be the same size, hey? Just make one pile of 17 pennies and flip each coin in that pile, et voila.  (Alternatively you could stand all the pennies up on edge and make two piles of them, resulting in no heads-up coins in both…..but that’s cheap.) Most people that I’ve shared this with end up intuiting the answer, satisfied that it seems to make sense, without actually testing every possible case. Those people have better things to do. I, however, can be helplessly OCD about things like this, so here’s the  general solution I worked out.

N = total # of coins on the table

X = # of coins lying heads up

p1 = # of coins in first pile

h1 = # of heads in the first pile

t1 = # of tails in the first pile

h2 = # of heads in the second pile

Solution: First let p1 = X.  Making this assignment allows us to make the following statements:

(1) h1 + t1 = p1 = X (all coins in first pile)

(2) h1 + h2 = X (total number of heads-up coins on the table)

Now flip all coins in the first pile and call the new number of heads and tails in that pile h1′ and t1′ respectively. Therefore:

(3) h1′=t1

(4) t1′ =h1

Substituting eq’n (3) and (4) into (1), and (4) into (2) gives us:

(5) t1′ + h1′ = X

(6) t1′ + h2 = X

Substituting (6) into (5):

t1′ + h1′ = t1′ + h2

h1′ = h2

Note that the solution does not depend on the values of X or N.

And thats my two cents.

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