A disappearing coin trick uses “lucky” numbers to solve puzzles. The trick can be used to prove a problem is insoluble by presenting a single number which must vanish for that particular puzzle. In this way, the trick allows you to solve various problems which were previously not solvable.
In my paper I show, for instance, that all the problems which cannot be solved by any finite series are insoluble by the disappearing coin trick. The trick is especially important for problems involving multiplication tables, which, among other things, require that the numbers being multiplied remain in one piece.
What is the vanishing coin trick and how does it work?
The vanishing coin trick can be thought of as a form of game theory. Games of chance take place, for example, when the total number of balls are given at the beginning and then are distributed to players at random. In a game of chance, the number of outcomes can be thought of as the number of ways in which the coin can land on a particular number. This means that any number can be a solution. For such problems, we want to find a solution, otherwise some odd numbers can win out. And, of course, for such puzzles we want the solution to be a simple addition. This means that we need the answer to be at most a single number, so that there is an infinite number of ways in which it can be obtained.
We can now ask who among us has a “great” number. This was an idea of Bertrand Russell, and I am now in the process of refining his suggestion. I suggest that each of us has a great number in the neighbourhood of 10. We may call these numbers p and q. If the total number of p’s among 10’s is a certain value, then our number is greater than that number. If p’s are so very rare that they cannot be found by means of ordinary analysis, then the number of p’s is less than the total number of p’s out there: the total number is less than 10.
So, with the vanishing coin trick, we can choose any number p, as the number to which the coin must vanish, and choose the number q such that:
(p-q)(p-m): P ≤ q ≤ m
If we choose p, we end up with p, p, i, i = …i, j, …j, k. Let us say that we chose p. This means that p ≤ q
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