Consider the following 'trick' (WARNING: very lame)
Think of a number. Multiply this number by two. Add four. Divide the number by two. Subtract the number you were originally thinking of. I guess that the number now in your head is two.
This trick might stump non-mathematicians for a few minutes, but most people will eventually figure it out. The trick relies on the distributive law of numbers, which is not entirely trivial, but most people will find the trick rather lame once they figure it out.
I was wondering if it is possible to do a "I know what number you're thinking of" that relies on less trivial results, from number theory for example, tricks that require some heavier results to explain and are thus more 'magical'. Too avoid being flagged as too subjective or vague, I will try to specify this.
Consider two people, the mathematician $M$ and the non-mathematician $N$. $M$ asks $N$ to think of a number ($M$ may restrict the numbers from which to choose to a sufficiently large set, like the numbers between 1 and 1000 or the prime numbers) and then lets $N$ apply a finite sequence of functions to this number. These functions should preferably be easy and well-known, i.e. basic arithmetic operations, division with residue, decimal representation of a number (e.g. "take the fifth digit"), à la limite prime factorisation, but no functions which require a calculator like trigonometric functions. The following two conditions are imposed:
- The function must have a finite image, i.e., $M$ should be able to give a finite, preferably small, set in which the image of the function always lies. For example "Either you're thinking of 5 or you're thinking of 29" would still impress $N$. On the other hand, the trick bellow relies on some serious number theory, but gives an infinite image, which I think is less impressing to most people.
Think of a natural number which cannot be written as a sum of three squares of natural numbers. If the number you're thinking of is divisible by four, divide it by four. Keep dividing by four untill you can no longer do so in $\mathbb{N}$. Add one to the number you now have. The number you end up with is divisible by eight.
- The function uses specific properties of the natural numbers or integers, not just algebraic manipulations. This is what I mean by non-trivial. In particular, the same trick should not work in any commutative ring $A$. For example, tricks relying on the fact that $\mathbb{Z}/p\mathbb{Z}$ is a field for any $p$ prime could satisfy this.
I'd be interested in knowing what's possible here.