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You have 100 coins laying flat on a table, each with a head side and a tail side. 10 of them are heads up, 90 are tails up. You can't feel, see or in any other way find out which side is up. You cannot flip any of the coins or stand them on their edges. Split the coins into two piles such that there are the same number of heads facing up in each pile.

Can this problem be solved? If so, show your solution.

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    $\begingroup$ This is a variant of a common logic puzzle, where the usual solution would involve flipping. $\endgroup$ – Rand al'Thor May 11 at 6:20
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    $\begingroup$ Ted-ed has also a riddle on this. Here is the link :- youtube.com/watch?v=pnSw8g3DPHw $\endgroup$ – Anonymous May 11 at 7:02
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    $\begingroup$ If you puzzle is sufficiently unconstrained, and accepting of wild lateral-thinking answers, that's not acceptable here. Puzzles should ideally have a single obviously-best answer, or at the very least a very small set. We're not idea-generation. See: puzzling.meta.stackexchange.com/q/1254 $\endgroup$ – bobble May 11 at 18:21
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    $\begingroup$ It seems that the puzzle doesn't have a proper answer without crazy lateral ideas. I am giving you -1. $\endgroup$ – Dmitry Kamenetsky May 11 at 23:13
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    $\begingroup$ I think there may be a misunderstanding in your concept of "puzzle". Puzzles (as requested by PSE) have a problem, an intended solution, and logical and reasonable steps to get there - even if there is a lateral thinking (for some) step. If anything in this list is missing, then it isn't a puzzle. For example, using this question, if intended answers could include "Use magic to give the coins enough intelligence to sort themselves into piles for themselves", this could be a creative answer to an open-ended question, but it would not be a puzzle with a puzzle solution. :) $\endgroup$ – Graylocke May 12 at 3:44
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If you are unsure about the puzzle, you are probably misquoting a classical puzzle as Rand Al'Thor mentions.

If you cannot distinguish the type of the coins, regardless of how you make two piles, there is always a way to swap a heads-up and a tails-up coin between the two piles, changing the heads-up counts. So you cannot guarantee the counts are the same.

I could think of an out-of-the-box solution however.

Split each coin in half and make 2 piles of 1/2 coins.

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You should:

Take the coins on a rocket ship ride into deep space, and turn off your engine. In a zero gravity environment far from any massive body, there is no perceptible gravitational field, and no concept of "up" or "down". Split the coins however you want - both piles have zero coins that are heads up, and zero coins that are tails up, since the direction "up" does not even exist where you are!

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    $\begingroup$ Or along those lines, you could save the trip, place all coins into two heat-resistant ceramic buckets on the launchpad under the engines; as soon as you turn the engine on, you'll reduce them all into two puddles of molten slag, zero coins heads or tails up. $\endgroup$ – Darrel Hoffman May 11 at 18:14
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A solution to this problem exists because there are an even number of both heads-up coins and tails-up coins, and thus they can be split evenly into two piles. If a solution exists, then all you would need to do is arrange the coins into two piles, and randomly swap one coin from each pile, over and over. Given enough time, you would generate all possible arrangements of coins, and at least one is a solution to the problem. Since the original question doesn't specify that you need to leave the coins in a solved state, you would have found the solution, you just wouldn't know when it happened. But, it can be done.

There are something like 9.332E157 total permutations of arrangements, but I believe there are only 11 discrete states, so it's likely that no human could ever do this, but an un-described player definitely could!

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  • $\begingroup$ When I first encountered this puzzle, my immediate solution was to pick the coins in a ordered pattern such that the two piles would have the same number of head and tails. After some deliberation, the coins are random and no matter what order you use, the two piles will also be random. $\endgroup$ – ATL_DEV May 11 at 16:27
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    $\begingroup$ Since we only need the number of heads to be the same in both piles, you don't need two piles of 50 coins each. If you split into two piles of 5 coins and 95 coins, you only need to try 100 choose 5 configurations (about 75M). At one configuration per second working 24/7, you could do this in just a couple of years. Also, don't swap randomly, or it's possible (although infinitesimally likely) that you never finish. $\endgroup$ – Nuclear Hoagie May 11 at 17:10
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    $\begingroup$ @NuclearHoagie An even faster solution is just to start with all the coins in one pile and move them one by one to the other pile. At the beginning all the coins - and therefore all 10 heads-up coins - are in the first pile, and at the end, zero heads will be in the second pile. Since you only move one at a time, the number of heads changes by at most one at each step, so you must reach every number between 10 and zero. Therefore there must be at least one step where the number of coins in the first pile is 5 - and is therefore equal to the number of coins in the second pile. $\endgroup$ – 2012rcampion May 11 at 18:23
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    $\begingroup$ @2012rcampion Now that's a good answer that deserves its own write-up - it's the only feasible solution so far. $\endgroup$ – Nuclear Hoagie May 11 at 18:27
  • $\begingroup$ @2012rcampion Almost. You can start with 5 in one and 95 in the other, and ones by one move one coin, until you are at 95 and 5. This is a total of 91 permutations, and you are guaranteed that at least one will have 5 heads in each. $\endgroup$ – David G. May 12 at 5:40
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Based on @Florian-F's out-of-the-box approach, I would

bring a belt sander, put it on the table, and sand off both sides of all coins.

Then, I would randomly split the coins into two piles. Any two piles will have the same number of heads facing up, which would be exactly

zero (since all heads have been sanded off)

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I don't see a solution without some kind of 'cheating'.

My first idea was to stand the coins on their edge, however this is not allowed but perhaps one of the following might be allowed:

1.) Fix coins and turn the table by 90 degrees upwards Make two piles. It doesn't matter how many coins are in the first or second pile.

Now use some sticky tape (scotch) to glue all coins on the table

Then turn the table by 90 degrees. Now no coin in either pile points upwards or downwards

Or

2.) You do not turn any coin you ask somebody else to flip 10 coins

Like original solution but you ask somebody else to flip the coins.

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Hire someone to sort the coins so that all heads and all tails together, but make sure they do not tell you which way they did it (to prevent "in any other way find out which side is up"). Take five coins from the front and five from the back and put them into a separate pile. Now the new pile has 5 heads and 5 tails and the original pile has 5 heads and 85 tails.

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