Blindfold coin partition
A pile of n coins lies on a table, some heads up and some tails up. You are blindfolded but may touch and move coins.
You are told exactly k coins are heads up (0 ≤ k ≤ n).
You may split the coins into two piles (any sizes), then choose one pile and flip every coin in that pile. The other pile is untouched.
How can you make the two piles contain the same number of heads? (When n is even, equal heads in both piles also means equal heads and tails on the whole table.)
Take any k coins as one pile and flip that entire pile; leave the other pile unchanged.
Why it works
Suppose x of the chosen k coins were heads. After flipping, that pile has k − x heads.
The remaining n − k coins already had k − x heads (since there are k heads in total), so the other pile also has k − x heads.
Note
Without knowing k, one split-and-flip step cannot guarantee equal head counts in both piles.