Copyright ©2006 by Paul Niquette, all rights reserved. 
Before reading Miss Silverwood’s explanation, you may want to allow yourself to be mystified still further by a clever animation that applies paranormal numbers. Or so it seems.t the conclusion of her mindreading demonstration Miss Silverwood shook her head. “Identifying each deleted digit was merely a trick,” she said. “I don’t need paranormal powers to know exactly what you are thinking. Right now, for example, you are thinking that I am going to show you how the trick is done. You are right to think that. Let me begin by commenting that there is something rather special about the integer nine.” Readers may think Miss Silverwood had in mind a rather facetiously discovery in another puzzle  that nine is the smallest nonprime odd integer. You are not right to think that.“Throughout the decimal world,” she said, “the cipher nine enjoys a unique place among digits: nine is exactly one less than ten. Thus, 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1,…” She reminded her students that every decimal integer X, with its digits symbolized by x_{i}, looks like this… X = x_{0} + 10 x_{1} + 100 x_{2} + 1000 x_{3} +… …and can be expanded as follows: X = x_{0} + (9 + 1) x_{1} + (99 + 1) x_{2} + (999 + 1) x_{3} +… Using elementary algebraic operations, Miss Silverwood wrote on the blackboard… X = x_{0} + x_{1} + x_{2 }+ x_{3} +… + 9 x_{1} + 99 x_{2} + 999 x_{3} +… “By inspection,” said she, “you will see that the underlined terms are each divisible by nine. Meanwhile, all the others represent what?” She nodded to a student with an upraised hand, “I know what you are thinking – that x_{0} + x_{1} + x_{2 }+ x_{3} +… merely adds up all the digits in the integer X, and you are right. Something else, too: If that sum happens to be divisible by nine, the integer X will also be divisible by nine. All those underlined terms have already been shown to be divisible by nine. So, you see why I think the cipher nine is quite special.” or an illustration, Miss Silverwood wrote a large decimal integer on the blackboard, 127,689,354 and said, “This integer is divisible by nine. Go ahead and check that.” She then calculated the sum of its digits (45). “Of course, 45 is indeed divisible by nine – and, by the way, so too is the sum of four and five.” Most puzzle solvers will see immediately that the results are the same for every decimal integer comprising those same digits, thus 123,456,789 and 987,654,321 are each divisible by 9, along with all possible arrangements of those ten digits, which totals 10! Punctuation concluding the previous sentence is not exclamatory but factorial (10! = 3,628,800).“It does not take a mindreader,” said Miss Silverwood, “to know what all of you have been thinking up to this point: ‘Who cares if a given integer is divisible by nine?’ Well, the answer to a more general question does have practical interest – particularly for reading minds: ‘When a given decimal integer X is divided by nine, what is the remainder?’ ” The mathematical expression in Miss Silverwood’s day was rem_{9}(X), and it would be pronounced as follows: “the remainder modulo 9 of X.” Today’s spreadsheet notation is “mod(X, 9).” “We found that rem_{9} (127,689,345) = 0, she said. “By changing, say, the last digit from five to six, we will see that rem_{9} (127,689,346) = 1.” Using a modern calculator to find that out is not especially convenient: 127,689,346 divided by 9 equals 14,187,705.1111…1, from which we must subtract the integer portion 14,187,705 to isolate the repeating decimal 0.111…1; finally, we must compute the remainder by multiplying 9 times 0.111…1 = 1. Another way is to multiply the integer portion of the quotient by nine and subtract the result from the original integer. Either way, taking the sum of the digits is easier.“The sum of the digits in 127,689,346 is 46,” said Miss Silverwood, “and the sum of those two digits, four and six, equals ten. We see straight away that rem_{9} (10) = 1. Other digits can be substituted: for example, try rem_{9} (237,779,855) = 8. Naturally, any remainder modulo nine is limited to the ciphers 0, 1, 2, 3, 4, 5, 6, 7, and 8, inasmuch as rem_{9} (9) = 0.” Puzzle solvers will observe that, given a decimal integer N, where rem_{9}(N) = 0, we may insert an arbitrary integer x into N at any location, creating thereby an integer N'and immediately discover that rem_{9}(N') = x. Thus, rem_{9}(1,276,8x9,345) = x; the sum of the digits will be 45 + x, and by inspection, rem_{9}(45+ x) = x.“There is a shortcut for modulo9s,” said Miss Silverwood. “It is called 'casting out nines', and it goes like this.” She wrote N = 1,276,8x9,345 on the blackboard. “You simply cross out any combinations of digits that total nine.” Her procedure went like this… 1 2 7 6 8 x“This confirms that rem_{9} (N) = x,” she said. Some solvers will argue that, for certain integers (8256, for example), 'casting out nines' will not be a shortcut at all. Nevertheless, let’s keep the procedure in mind for the mindreading demonstration.“Modulo9 comes in handy for checking your arithmetic,” said Miss Silverwood. “For summing, where X plus Y = Z, you will see that rem_{9} (X) plus rem_{9} (Y) must always be equal to rem_{9} (Z). And for multiplying, where X times Y = Z, then rem_{9} (X) times rem_{9} (Y) must always be equal to rem_{9} (Z).” It is not difficult to confirm Miss Silverwood’s assertions. Her own proofs were quite elegant, each based on the observation that every integer X can be represented by 9 times some integer quotient plus a remainder (X = 9Q + R).“Such calculations will not catch every number blunder,” she said, “but only extremely unlikely errors escape detection, and each will have plenty to do with the cipher nine.” The students began looking intently at the clock on the wall  but not with their usual afternoon restlessness. “Some of you,” said Miss Silverwood, “are thinking that I am running out of time – that I must hurry up and finish telling you how the mindreading trick worked. Well, the job is almost done. You all recall that at the beginning, I invited you to choose an arbitrary decimal integer. Let’s call it A. I then asked you to form a new integer B by simply rearranging all the digits in A. You found the difference by subtracting one from the other, A  B. Be prepared for a surprise.” Miss Silverwood raised her eyebrows and exclaimed, “For any value of A and any rearrangement of its digits B, rem_{9} (A  B) = 0!” he glanced at the clock and quickened her pace. “Take any particular digit d in A. Let’s say it is located at the fifth position from the decimal point, so that its value in A will be 10^{5} d. Suppose you move d two positions to the right in B, making its value there 10^{3} d. Using the special features of the cipher nine, we have the digit d multiplied by one  plus the digit d times a number comprising five nines, as 99999 d. In B, we have the same digit d multiplied by one  plus the digit d times a number comprising three nines, as 999 d.” Miss Silverwood lined up these terms on the blackboard, ready for the subtraction of A – B. “We see that d – d = 0!” Her exclamation point stood a foot tall. “Look at that other coefficient of d,” she exclaimed. “The difference 99999 – 999 = 99000, which is divisible by nine. The same outcome is assured for all other digits in A as each gets relocated in B, which confirms that rem_{9} (A  B) = 0.” The school bell rang. “You now have everything you need to perform the Paranormal Number trick,” said Miss Silverwood. She grasped up her purse and clomped toward the door. “Have a good weekend.” The students continued to sit at their desks, scribbling in their notebooks. And discovering. On Monday, Mr. Johnson returned to his algebra class and was astounded by the paranormal powers of his students. Some of them. 




