SOLUTION: 100 coins are on a table showing 10 heads and 90 tails. You are blindfolded. Is it possible to divide the coins into 2 piles with an equal number of heads?

Question 1210390: 100 coins are on a table showing 10 heads and 90 tails. You are blindfolded. Is it possible to divide the coins into 2 piles with an equal number of heads?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
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100 coins are on a table showing 10 heads and 90 tails. You are blindfolded.
Is it possible to divide the coins into 2 piles with an equal number of heads?
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Yes, of course - if you ask another person, who is sighted, to do it for you :)

Another way for a blindfolded person to perform this assignment on his or on her own
is to select 5 head coins from the entire collection of 100 coins by touch.

In this case, you use your tactile senses instead of your sight.

To me, it remains unclear how this post and your question do relate to Math . . .

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Label the piles A and B.
Randomly select 10 coins to form pile A.
The remaining 90 coins are pile B.

Let x be a whole number in the set {0,1,2,...,9,10}
x is between 0 and 10
If x coins in pile A are heads, then 10-x coins in pile A are tails since pile A has 10 coins.
Further we know there are 10-x heads in pile B since there are 10 heads total.

Turn over all coins in pile A.
x heads ---> x tails
10-x tails ---> 10-x heads
We'll have 10-x heads in pile A and 10-x heads in pile B.
At this point we have guaranteed both piles have the same number of heads.
Unfortunately we won't be able to determine how many heads are in each pile.

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Let's look at an example.
Let's say there are x = 3 heads in pile A and 10-x = 10-3 = 7 tails in pile A.
The remaining 10-x = 10-3 = 7 heads are in pile B.

Now let's turn over every coin in pile A.
3 heads ----> 3 tails
7 tails ----> 7 heads
We will have 7 heads in each pile.

I encourage you to try other values of x in the set {0,1,2,...,9,10} so you can get a sense of how this is working.

Note carefully I say "turn over" rather than "flip" because I want to avoid the idea of randomly flipping a coin.
When I say "turn over" I simply mean "any heads becomes tails or vice versa".

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Quick Recap: Select any 10 random coins. Turn those 10 coins over (leave the rest as they are).

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