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Detect my chosen coin (Posted on 2016-01-19) Difficulty: 4 of 5
This is what I’ve told my two mathematician friends:

“Imagine a 64-square chessboard with a coin on each square.
Each of the coins has either head or tails facing up, chosen at random.
I check the board and decide which coin will be my favorite one.
One of you (say A) will be with me, see the chessboard and I will reveal to him (only to him) which coin is my favorite. He then must flip over exactly one of the coins on the chessboard, while the other mathematician (B) is in another room not looking.

Once the coin is flipped over, the uninformed mathematician (B) is summoned into the room and must deduce which coin is my favorite only by examining the coins on the chessboard.
To secure absence of any other hints A is escorted out of the room.

Clearly, prior to the procedure, you are free to discuss the problem between the two of you and establish its solving strategy. You have no time limit, you are free to use any kind of calculator, but any communication between you two is strictly prohibited”

What strategy can the two mathematicians devise to ensure that my favorite coin can always be correctly identified?

No Solution Yet Submitted by Ady TZIDON    
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A way of looking at it | Comment 15 of 19 |
A similar (but not so technical) way of looking at it.
What is the effect of flipping a coin?
Let write on each coin the value of its square-number from 1-64.
But we write it express as sum of its two power components or packets (f. es. coin 37 as sum of 32 + 4 + 1).
Imagine the chessboard full of coins in head position. The chessboard is in order with all the numbers and packets: 64 numbers written with theirs 32 pks 32, 32 pks 16, ... 32 pks value 1.

Now I flip the coin 37 and broke the harmony. One pk 32, one pk 4, one pk 1 are singles now. If I flip other coins I will add single packets but other packets will regain parity. Single packets are significative in a enviroment dominate by parities, as them indicate singularities, something as disorder.

We can use them to solve the problem. Just we need to force that the only single packets in the whole set of tails would be those composing the chosen number. So we need to flip a coin able to parify the others.
======
In Ady's example
3=2+1
7=4+2+1
20=16+4
61=32+16+8+4+1
Singles= 32,8,4,1 composing number 45
To have 8 (the chosen number) I have to parify 32,4,1. I flips coin 37 and leave a single 8 pk.

Math B just check that there is only a single pk 8 value.
Edited on February 4, 2016, 6:30 pm
  Posted by armando on 2016-02-04 15:31:32
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