Have Your Mind Read

Do you think it's possible to have your mind read?

(try the following)

(explanations below)

Come now, there is ABSOLUTELY NO EVIDENCE that mind reading is possible - so there has to be a more down-to-earth explanation:

Fido Puzzle

In this trick, you think of a number, make an anagram of it, and subtract the smaller number from the larger. You then circle one non-zero digit of the answer, and enter the remaining digits into the computer. The computer tells you what the digit you circled is.

The trick works because the result of the subtraction is guaranteed to be divisible by 9. This implies that the sum of the digits of the answer must be a multiple of 9. So, for example, if you type into the computer 3, 5 and 7 (totalling 15) the computer knows that the missing digit must be 3, to make the total up to 18, the next integer multiple of 9.

But why does subtracting an anagram of a number from the number itself always yield a multiple of 9?

Suppose the original subtraction that you came up with was this:

5	7	3
3	5	7	-

2	1	6

We can decompose each of these numbers, so that for example 573 becomes 500+ 70+3. The subtraction can then be written like this:

5	0	0	+
	7	0	+
		3	+
3	0	0	-
	5	0	-
		7	-

2	1	6

which we can rearrange so that similar digits are grouped in pairs.

5	0	0	+
	5	0	-
	7	0	+
		7	-
		3	+
3	0	0	-

2	1	6

or (500-50) + (70-7) + (3-300) = 216

Now each term of this sum is of the form i10^m – i10ⁿ where I is the digit and m and n are whole numbers. For example the first term is 5´10² - 5´10¹.

We can take I out as a factor and rewrite the term as

I(10^m – 10ⁿ)

and then, assuming that m > n*, we can take out 10ⁿ as a factor as well (noting that 10^m/10ⁿ = 10^(m-n)) giving

i10ⁿ(10^(m-n) – 1)

Now 10^(m-n) is an integer power of 10 (10, 100, 1000 etc) and if we subtract 1 from any integer power of 10 we get a string of nines (0, 9, 99, 999 etc), which of course is divisible by 9. So the contents of the bracket must be divisible by 9, and as the factor outside the bracket is an integer, the entire expression must be divisible by 9.

So we can see that when we subtract an anagram of a number from the number itself, we can write the answer as a sum of terms each of which is divisible by 9. It follows that the whole answer must also be divisible by 9.

* If m < n, you just take out 10^m as a factor and end up with the expression i10^m(1 - 10^(n-m) ). In this case the contents of the bracket always come out as –9, -99, -999 etc and the argument still holds.