In the above mentioned book, there's a trick that won't work as explained (page 171-2). I'd like to know if someone has experimented with it. The rule I came up with is to double the first digit of the number chosen by the spectator, and use that result as the total number of numbers written beneath the first one. For instance, in Gardner's example, 5 times 2 is ten. If the chosen number was 389, the numbers beneath the first one would be six. Hope it makes sense. Will this rule always work?

Hello federico,

Gardner's description works for me okay. Here's an example, using his name number of 81885 (and the prediction number is 21885).

You write 81887.

Spectator writes any number, say, 20991.

You write the 9's-complement of this: 79008.

Spectator writes any number, say, 17345.

You write the 9's-complement of this: 82654.

The sum is 281885, as predicted.

Hope this helps,

Thomas Henry

Thomas! Thanks for responding. I am referring to the trick that uses 2 digit numbers. Last line on page 171, continued on page 172.

Hi again,

Oops, silly me; I was looking at the wrong page.

Anyway, this other procedure also works. Taking your number 389 as an example, the start number will be 3+89=92. And the 3 tells us there will be three pairs of numbers written below it. So for example,

92
16 (written by spectator)
83
22 (written by spectator)
77
35 (written by spectator)
64

When added up, the result is 389, as expected.

Incidentally, as to the name of this method, this is the principle behind "M#nt#l Logs." (I've put in # to disguise this from search engines; but you'll recognize that first word).

Thomas Henry