Skull Shakers No

By PETER A. McDONALD

"Right Out of My Depth"

To be honest I am throwing this into the series for two reasons. In the first place I know from the boys that when George Blake has delivered up some little known facts in the past . . . (e.g. that Singleton deal business) he has aroused much interest. In the second place, I lack the necessary mathematical training and ability to work out (a) whether what I am about to describe ALWAYS works and (b) if so, why. In other words, I hope that someone will chew over it and let us have the benefit of their experience if they crack the problem. I can only say that AS FAR AS I CAN DISCOVER ... and I have tried to take representative selections ... the thing always works if confined to the limits I am going to mark out.

By now you will all, I hope, be agog. First let's look at the thing as though it were a telephone test. You ring up a pal and ask him if he will join in a little experiment in thought-reading. Say that minds tend to work in certain weird grooves and provided he allows you to set his mind working in this way you should be able to read his mind for him. Tell him to write down any number of three figures (e.g. 321). The only stipulation is that they must all be different figures and none of them must be a nought. Then he is to write out this number again but this time in reverse order ... in our example 123. Thus he has in front of him two three-figure numbers. He then subtracts the smaller of these three-figure numbers from the larger of them ... in our case he would take 123 from 321 and be left with 198. (Yes, I know you've heard of this one before, but hear me out). So far so good.

He has arrived at a three figure number entirely by chance and since he has told you none of the figures at any stage in the proceedings you can't possibly tell what it is. Fair enough? You then tell him to multiply this three-figure number (it might JUST be a two-figure one but this is catered for below . . . assume it is three--figure for the time being) by any number between 1 and 10. He does not even tell you this number.

This leaves him (say) with a four figure number and you tell him to write the letters A, B, C and D above each of the figures in the order they come. Now (if he hasn't become utterly wearied) he multiplies A by B and jots down the answer; then he multiplies B by C. and jots down the answer; then C by D and, finally, D by A. Thus he has four figures arrived at in the chanciest of chancy manners. There isn't a possibility of your knowing any of them. Now the thought-reading starts.

You tell him to strike out three of the four "answers", calling them out to you as he does so. Immediately he has called out these three figures you tell him to concentrate on the fourth and after a few seconds you announce the number he is thinking of.

Now I know that the routine is pretty intricate and if any-one can find a slicker way of using the formula no-one will be more pleased than I shall be. But though it sounds long-winded on paper it doesn't really take such an awfully long time in practice and the result is startling. After all, how CAN the magician know what the figure is ? He hasn't been told ONE figure all the way through the calculation and the three figures he is told at the end don't seem to bear any relation to the one which is being thought about. Unless there really is something in this thinking in a groove business what solution is there to offer?

And now I have to break the sad news that this doesn't always work with every set of numbers Let's start at the beginning. It is a well-known fact (to conjurers) that if a three-figure number (e.g. 321) is written down then re-written in reverse order (123) and the smaller number (123) taken from the larger number (321) the result (in our case, 198) will be a number whose digits total 18. It must, therefore, either be a three-figure number or the figure 99. Most usually it is a three-figure number as in our example. When you are doing the trick, however, stop at this stage and tell him that if there are only two figures in his answer he is to add a nought to the end as you want to ensure that there are plenty of figures at the end of the experiment.

Now tell him to multiply the three-figure number by any number between 1 and 10 (i.e. not including 1 and 10). This is where it goes one stage further than the old "18" dodge which, of course, forms the basis of the effect, for it would seem (I have not been through every one of the dozens of possible combinations but I've never had a "dud" yet) that no matter what number he chooses to multiply the other by, the digits in the answer will also add up to 18.

At this point you have to ask him how many figures there are in his answer. If there are four, you are all set to go ahead as set out above . . . i.e. the A, B, C. D stunt. If there are only three figures tell him to add a nought to the end. (Note: This is not so fishy as it sounds for if it was necessary to add a nought earlier on . . . i.e. to the 99 ... it will not be necessary to add another; therefore you only have to ask him to add a nought at ONE stage at the most). Remind him before he starts his multiplications of A to B etc. that 3 times O is O . . . some people forget. Now here is the secret; the product of A and B plus the product of B and C plus the product of C and D plus the product of D and A ALWAYS (so it seems to me after many experiments) total 81. So, when he strikes out tl ree of the products and reads them out io you you just have to jot them down, add them, and deduct the result from 81 to find out the fourth figure . . . i.e. the one he ¡s thinking of!

An example is the best way of showing how the thing woiks. Using the previous figures again wo had leached an answer of 198 whe:i our imaginary v ctim had deducted 123 from 321. Now suppose that, unknown to the magician, he multiplies this number by 8. His calculation will read : 198 x 8

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