Search This Blog

Monday, September 24, 2012

Freaky facts about 9 and 11
Scroll down to Force 1089 for a fun psychic power trick

It's easy to come up with strange coincidences regarding the numbers 9 and 11.  See, for example,

http://www.unexplained-mysteries.com/forum/index.php?showtopic=56447

How seriously you take such pecularities depends on your philosophical point of view. A typical scientist would respond that such coincidences are fairly likely by the fact that one can, with p/q the probability of an event, write (1-p/q)^n, meaning that if n is large enough the probability is fairly high of "bizarre" classically independent coincidences.

But you might also think about Schroedinger's notorious cat, whose live-dead iffy state has yet to be accounted for by Einsteinian classical thinking, as I argue in this longish article:

http://www.angelfire.com/ult/znewz1/qball.html


Elsewhere I give a mathematical explanation of why any integer can be quickly tested to determine whether 9 or 11 is an aliquot divisor.

http://www.angelfire.com/az3/nfold/iJk.html

Here are some fun facts about divisibility by 9 or 11.

# If integers k and j both divide by 9, then the integer formed by stringing k and j together also divides by 9. One can string together as many integers divisible by 9 as one wishes to obtain that result.

Example:

27, 36, 45, 81 all divide by 9

In that case, 27364581 divides by 9 (and equals 3040509)

# If k divides by 9, then all the permutations of k's digit string form integers that divide by 9.

Example:

819/9 = 91

891/9 = 99

198/9 = 22

189/9 =21

918/9 = 102

981/9 = 109

# If an integer does not divide by 9, it is easy to form a new integer that does so by a simple addition of a digit.

This follows from the method of checking for factorability by 9. To wit, we add all the numerals, to see if they add to 9. If the sum exceeds 9, then those numerals are again added and this process is repeated as many times as necessary to obtain a single digit.

Example a.:

72936.    7 + 2 + 9 + 3 + 6 = 27.  2 + 7 = 9

Example b.:

Number chosen by random number generator:

37969.  3 + 7 + 9 + 6 + 9 = 34.  3 + 4 = 7

Hence, all we need do is include a 2 somewhere in the digit string.

372969/9 = 4144

Mystify your friends. Have them pick any string of digits (say 4) and then you silently calculate (it looks better if you don't use a calculator) to see whether the number divides by 9. If so, announce, "This number divides by 9." If not, announce the digit needed to make an integer divisible by 9 (2 in the case above) and then have your friend place that digit anywhere in the integer. Then announce, "This number divides by 9."

In the case of 11, doing tricks isn't quite so easy, but possible.

We check if a number divides by 11 by adding alternate digits as positive and negative. If the sum is zero, the number divides by 11. If the sum exceeds 9, we add the numerals with alternating signs, so that a sum 11 or 77 or the like, will zero out.

Let's check 5863.

We sum 5 - 8 + 6 - 3 = 0


So we can't scramble 5863 any way and have it divide by 11.

However, we can scramble the positively signed numbers or the negatively signed numbers how we please and find that the number divides by 11.

6358 = 11*578

We can also string numbers divisible by 11 together and the resulting integer is also divisible by 11.

253 = 11*23, 143 = 11*13

143253 = 11*13023

Now let's test this pseudorandom number:

70517. The sum of digits is 18 (making it divisible by 9).

We need to get a -18. So any digit string that sums to -18 will do. The easiest way to do that in this case is to replicate the integer and append it since each positive numeral is paired to its negative.

7051770517/11 = 641070047

Now let's do a pseudorandom 4-digit number:

4556. 4 - 5 + 5 - 6 = - 2. Hence 45562 must divide by 11 (obtaining 4142).

Sometimes another trick works.

5894. 5 - 8 + 9 - 4 = 2. So we need a -2, which, in this case can be had by appending 02, ensuring that 2 is found in the negative sum.

Check: 589402/11 = 53582

Let's play with 157311.

Positive digits are 1,7,1
Negative digits are 5, 3, 1

Positive permutations are

117, 711, 171

Negative permutations are

531, 513, 315, 351, 153, 135

So integers divisible by 11 are, for example:

137115 = 11*12465

711315 = 11*64665

Sizzlin' symmetry
There's just something about symmetry...

To form a number divisible by both 9 and 11, we play around thus:

Take a number, say 18279, divisible by 9. Note that it has an odd number of digits, meaning that its copy can be appended such that the resulting number 1827918279 yields a pattern pairing each positive digit with its negative, meaning we'll obtain a 0. Hence 1827918279/11 = 166174389 and that integer divided by 9 equals 20312031. Note that 18279/9 = 2031,

We can also write 1827997281/11 = 166181571 and that number divided by 9 equals 203110809.

Suppose the string contains an even number of digits. In that case, we can write say 18271827 and find it divisible by 9 (equaling 2030203). But it won't divide by 11 in that the positives pair with positive clones and so for negatives. This is resolved by using a 0 for the midpoint.

Thence 182701827/11 = 16609257. And, by the rules given above, 182701827 is divisible by 9, that number being 20300203.

Force 1089
An amusing property of the numbers 9 and 11 in base 10 arithmetic is demonstrated by Force 1089, the name of a trick you can use to show off your "mentalist" powers.

"It is important that you clear your mind," you say to your target. "Psychic researchers have found that a bit of simple arithmetic will help your subconscious mind to focus, and create a paranormal atmosphere."

You then instruct the person to choose any three-digit number where the first and last digit differ by at least 2. "Please do not show me the number or your calculations," you insruct, handing the person a calculator to help assure they don't flub the arithmetic.

The target is then told to reverse the order and subtract the smaller from the larger number. Take that difference and do a reversal of order on that. Then add those two numbers.

You toss a couple of books over to the target and say, "Be careful to conceal your number. Now use the first three digits to find a page in one of the books you choose." Once the page is found, you instruct the person to use the last digit to count words along the top line and stop once reaching that number.

"Now, PLEASE, help me, and concentrate hard on the word you read."

After a few moments, you announce -- of course -- the exact word your target is looking at.

Your secret is that the algorithm always yields the constant 1089, meaning you only have two words to memorize.

These symmetries always, in base 10, produce numbers divisible by 9 and 11 (or 99), even though the first number may not be so factorized.

Consider 854-458 = 396. 396+693 = 1089.
Further, 396 = 4*99; 693 = 7*99; 1089 = 9*11^2.

If we differ the first and last digit by 0, of course, the algorithm zeros out on the next step. If we differ them by 1, the subtraction yields 99, as in 100 - 001 = 99, which turns out to be a constant at this step.

Consider 433-334 = 99.

In the case of a number with 2n digits, we discover that the algorithm always yields numbers divisible by 9. However, for 2n+1 digits, the algorithm always yields a set of numbers divisible by 99. BUT, it does not yield a constant.

(I have not taken the trouble to prove rigorously the material on Force 1089.)

Ah, wonderful symmetry.

No comments:

Post a Comment