Forcing a particular number on an audience member, though apparently fair and random means, has a long and venerable history in magic. One popular way of forcing is using magic squares, as explored in the last Card Colm.
Applications considered there include counting down to a card carefully placed in a particular position in a deck in advance—perhaps also predicted in writing earlier—or having a spectator go to the corresponding page of a certain book and focus on the first word or sentence, while you "read their mind." Below we consider several related number forces, and refer the interested reader back to the February 2007 Card Colm for potential applications.
The year 1843 saw the birth of William Rowan Hamilton's quaternions in Dublin, Ireland, as well as the birth of John Venn in Hull, England. Consider the "convex hull" below, which consists of eight numbers:
Note that the numbers 1, 8, 3, 4, 9, 2, 7, 6 above, wrapped in an infinite circular loop, are John Venn's birth year followed by each respective digit's complement in 10 (so 9 matches with 1, etc.).
If we take any three consecutive numbers starting and finishing with evens, e.g., 8, 3, 4, or 6, 1, 8, those numbers sum to 15. The reason is simple: the circle above forms the perimeter of the standard 3 by 3 magic square (which also has 5 in the middle). The evens form the corners of that square, so the suggested sums are the magic square top, bottom and two side sums.
The other constant sums of interest in the usual magic square (forming the cross and the diagonals) correspond to the central option, namely, supplementing the circle with a central 5 and summing each even number with both the 5 and the (even) number directly across from it, e.g., 4 plus 5 plus 6.
Imagine tightening up this second circle by having the odd numbers move closer to the center: the magic circle becomes the standard magic square. There are also four other varying perimeter sums of three consecutive numbers above, starting and finishing with odds, namely 12, 16, 18 and 14. You may wish to draw attention to a couple of these totals in performance, to play up the random sums that can result. Regardless, here's a simple way to force 15 every time.
First arrange a circle of cards (face up or face down, your choice) as above, with an Ace in place of 1. Ask a spectator to select any card and use its value (peeking at the card face if the circle is face down) to count in either direction to another card. Regardless of whether an even or odd card is first selected, an even card results. Furthermore, the counting does not even have to be done in one fixed direction: the original selected value merely determines the number of "one-card moves" to be carried out. Finally have the even value arrived at added to its two neighbours on one side (not on each side). We refer to this as "end summing." Another way to get the correct total is to allow for "one last move (in either direction) to get to the final card," and then do "central summing:" namely, add that number to its two immediate neighbours, one on each side. The central option can be worked in as appropriate here too.
Suitably directed, this can all take place while your back is turned so that you seem to have zero information.
A presentational alternative is to have the spectator hold the packet of eight (face down) cards arranged in the hand, and request that the cards be cut several times before using the resulting top card to determine a count, moving cards one by one from top to bottom, finally summing the values of the top three cards after the card transferring is complete.
Let's modify the numbers in the sequence to give different triple sums. We could just add 1 to each number to get 2, 9, 4, 5, 10, 3, 8, 7, with sums 18 when adding three consecutive numbers starting and ending with even positions (but odd values this time). Here there is a parity switch: the odds now occupy the even positions, and vice versa. The odd card arrived at after the first count must be summed with its two immediate (even) neighbours, i.e., we do central summing. Similarly, by adding 2, 3 or 4 to each number, we can get triple sums of 21, 24 or 27, using the J, Q and K to represent the numbers 11, 12 and 13, and using central summing as appropriate.
We can also get a triple sum of 12 if we subtract 1 from each value of the original sequence to get: 0, 7, 2, 3, 8, A, 6, 5, where we use a valueless Joker to represent 0.
The following table summarizes these options, with the original 1834 base case in bold. Here, and throughout what follows, even positions are italicized.
There are other parity preserving shift possibilities which require end summing in all cases. For instance, we can add 2 to just the even numbers, getting 1, 10, 3, 6, 9, 4, 7, 8, with triple sums 19, or subtract 2 instead to get 1, 6, 3, 2, 9, 0, 7, 4, with triple sums 11. We could also add 4 to the evens, getting 1, 12, 3, 8, 9, 6, 7, 10, with triple sums 23. Or we could add 2 or 4 to just the odd numbers, getting 3, 8, 5, 4, 11, 2, 9, 6, with triple sums 17, or 5, 8, 7, 4, 13, 2, 11, 6, with triple sums 19. We can add 2 to the evens and 4 to the odds, yielding 5, 10, 7, 6, 13, 4, 11, 8, with triple sums 23, or add 4 to the evens and 2 to the odds, yielding 3, 12, 5, 8, 11, 6, 9, 10, with triple sums 25. The next table summarizes all of these:
There are also other parity switching possibilities which all require central summing. E.g., we can add 1 to the evens and add 3 to the odds, getting 4, 9, 6, 5, 12, 3, 10, 7, with triple sums 20, or we can add 3 to the evens and add 1 to the odds, getting 2, 11, 4, 7, 10, 5, 8, 9, with triple sums 22. Alternatively, we can subtract 1 from the evens and add 3 to the odds, getting 4, 7, 6, 3, 12, 1, 10, 5, with triple sums 16, or we can add 3 to the evens and subtract 1 from the odds, getting 0, 11, 2, 7, 8, 5, 6, 9, with triple sums 20. (In this last case you'll need to use a Joker and a Jack.) We can even add 5 to the evens and subtract 1 from the odds, getting 0, 13, 2, 9, 8, 7, 6, 11, with triple sums 24. And of course, we can also subtract 1 from the evens and add 1 to the odds, getting 2, 7, 4, 3, 10, 1, 8, 5, with triple sums 14, or we can add 1 to the evens and subtract 1 from the odds, getting 0, 9, 2, 5, 8, 3, 6, 7, with triple sums 16. The next table summarizes all of these:
Note that the forced sum is odd if and only if we use end summing, since everything is set up to force landing on an even. The table below subsumes all of the earlier ones.
Curiously, 13 is missing from this list of forces, as is 26!
In cases where the same triple sum (e.g., 16, 19, 20, 23, 24) can be forced with different circles of eight, we can start with sixteen cards, arranged appropriately, deal into two piles, show how different they are, and then give the spectator a choice of which one to work with.
Above we have been careful to keep evens and odds separate, and also to avoid the use of duplicate numbers. Furthermore, we've steered clear of negative numbers. However, we can relax these restrictions with interesting results, as we did in similar circumstances last time around. There, we accommodated having some values used twice by using cards of different suits but the same colour, and we attempted to dispel the negative associations associated with negative numbers in magic squares by associating negative values with red cards. Rather than adding all of the card values as we did earlier, we add credits for black card values and subtract debits for red card values, to get predictable overall balances. In the February 2007 Card Colm, we also obtained new magic squares from the standard three by three one by other means, such as adding constants to three of the eight perimeter elements. Similar adjustments can be made here to generate new magic circles of eight in addition to the ones listed above. We leave it to the interested reader to pursue these options.
Another direction to explore is to start with the perimeters of four by four or five by five magic squares, to get magic circles of twelve or sixteen respectively!