100 coins, 10 heads up, blindfolded.
The blindfolded coins puzzle: 100 coins on a table, exactly 10 showing heads, and you cannot see or feel which. Split them into two piles with the same number of heads. It sounds impossible with zero information. It is a one-move solution, and the proof fits in a sentence.
A hundred coins lie flat on a table. Exactly ten of them show heads; the other ninety show tails. You are blindfolded, wearing gloves, and cannot tell heads from tails by any means. You may move coins around and flip any coin over as often as you like.
Divide the coins into two piles so that both piles contain the same number of heads.
The piles do not need to be the same size. They only need equal head counts.
A common first reaction is that this cannot be done, because you receive no information at all. That reaction is correct about the information and wrong about the conclusion: you do not need to know where the heads are, you only need an operation whose result is guaranteed regardless. Before the hints, try describing any pile split in terms of an unknown k, the number of heads that happened to land in the first pile, and see what flipping can do to k.
Stuck? Open hints one at a time
Each hint gives away one more layer. Stop the moment something clicks.
Hint 1 a nudge
Take any 10 coins into a separate pile. Call the number of heads in it k. How many heads are in the other 90?
Hint 2 the key move
The small pile has k heads and 10 minus k tails. Flipping every coin in it swaps those counts. Compare with the big pile.
The solution
Reveal the full solution
Take any ten coins and flip every one of them. Those ten are one pile; the other ninety are the second.
The proof is a single accounting step. Suppose the ten you picked contain k heads, for some k you do not know. Then the remaining ninety contain 10 minus k heads. Your ten also contain 10 minus k tails. Flipping all ten turns its heads into tails and its tails into heads, so afterwards it shows exactly 10 minus k heads: the same as the other pile. Every possible value of k, from 0 to 10, gives equality, so the result holds no matter what you picked.
What makes this feel like a trick is that the answer never determines k. Both piles end with 10 minus k heads, an unknown number, and the puzzle only asked for them to be equal. Solving for the unknown was never required; arranging for it to cancel was.
What this puzzle is really testing
Whether you can act correctly under total uncertainty by choosing an operation that is invariant to the thing you do not know. Engineers meet this constantly: a rebalancing step that works whichever side is heavier, a swap that fixes an ordering whichever way it was wrong. The Dutch national flag partition is the same idea in code, moving each element to the right region without ever inspecting the global arrangement, and the in-place reversal patterns rely on flip-style operations whose effect is guaranteed by counting rather than by looking.
Follow-ups to expect
Generalise to n coins with h heads: take any h coins and flip them, same argument. What if you want the piles to have equal numbers of tails instead? Equal heads in piles of size 10 and 90 does not give equal tails, so the move changes: take any 90 coins and flip them, and the identical argument runs with tails in the role of heads. And the sharp one: can you do it if you do not know how many heads there are? No, and being able to say why (the operation needs the count to pick the pile size) shows you understood the proof rather than the trick.