Samuel's Blog

Numbers, numbers, lots of numbers

June 12th, 2008 at 02:02am

I’m not entirely sure how this train of thought started, but I think it was a couple nights ago when I was thinking about the benefits and disadvantages of the AM radio frequency stepping in Australia and the US from a marketing perspective. For those of you that aren’t aware, Australia uses a 9kHz stepping system whereas the US uses a 10kHz stepping system, so in the US radio stations on the AM band have frequencies which are divisible by 10kHz such as 1040, 920 and 730, whilst in Australia they are divisible by 9kHz, so you have frequencies like 873, 954, 1206 and 1494 (first three stations are obvious, if anybody wants to guess the fourth I’d be quite obliged…as far as I can tell there is only one station in Australia on that frequency, but ACMA keep changing the layout of their website so I can’t be sure).

Anyway, it was around 3am when I was thinking about all of this and my mind wandered on to the way numbers which are divisible by nine and ten work in a base 10 numbering system such as the one we use. I’ll take a step back for a moment here to explain the base ten numbering system for people who are not familiar with the term. Basically it means a numbering system with ten distinct characters available for use in any given number, so in the case of our usual numbering system the available characters are 0123456789. Effectively the way it works is that you count up (eg. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and when you run out of characters you increment the leading digit by one character and start again with the following digit(s) (eg, you would carry on with 10, 11, 12 etc). To make it blatantly obvious what I mean by “leading digit”, it would be possible to repeat those original numbers as “00, 01, 02, 03, 04, 05, 06, 07, 08, 09” or even as “000, 001, 002, 003, 004, 005, 006, 007, 008, 009”. The “leading digit” in this case is the one immediately preceding the digit which has run out of characters. In the case of “0099″ the bold zero would be the leading digit as both digits behind it have reached their highest digit and so to continue we must increment the leading digit by one and reset the digits behind it to the lowest available character which in our case is zero.

Other numbering systems also exist, such as base 2 (0, 1, 10, 11, 100, 101, 110, 111, etc) and base 16 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D, 1E, 1F, 20 etc). It starts to get a tad more confusing if you’re converting between numbering systems but it’s helpful to be able to do so. As an example, the number 91 from the base 10 numbering system would be represented as “1011011” in base 2 and “5B” in base 16. If it still doesn’t make sense to you, perhaps reading the wikipedia article on the subject might be helpful.

Anyway, on with my story (and returning to base 10 for the moment). My thought was specifically about numbers which are divisible by 9 and 10. In base 10 there are certain exclusive rules about these numbers which make them very easy to spot. All numbers which are divisible by ten have a zero as their final digit (eg 10, 50, 730), whilst the sum of the digits in a number which is divisible by nine, is equal to nine or a number which, if you continue the process of adding the digits until you reach a single digit number, will equal nine (eg. 18 (1+8=9), 72 (7+2=9), 99 (9+9=18 1+8=9), 67833 (6+7+8+3+3=27 2+7=9). As implied by the term “exclusive rules”, numbers which are not divisible by nine or ten will not fit these rules.

I was interested in working out whether or not these rules would still apply in numbering systems other than base 10.

It’s quite clear that the rule for numbers which are divisible by ten would not continue to hold true (as 20 in base 9 is “22” and in base 11 is “19”) but I was too tired to work out whether the “divisible by nine rule” would continue to work in other numbering systems. My theory was that, seeing as “divisible by ten” didn’t work, neither would “divisible by nine”, however the rules would work in the form of “divisible by n where “n” is derived from base n” as a substitute for “divisible by ten”, and to the same extent a rule for “divisible by n-1 where “n” is derived from base n” as a substitute for “divisible by nine”. In both cases, if my theory was accurate, the rules I was originally working with would just be the base 10 equivalent of the new rules anyway.

So, to test this, I’m going to work with examples from base 9, base 11, and base 12. For the sake of clarity, the symbols available in the versions of these numbering systems that I’m using for this theory are displayed below (it’s also so that once I confuse myself I will be able to just count along the symbols).
Base 9: 012345678
Base 11: 0123456789A
Base 12: 0123456789AB

In the case of “divisible by n where “n” is derived from base n”, all numbers which are divisible by “n” should end in zero which is the lowest available symbol.

For base 9, we can use the example numbers (as expressed in base 10) 9, 27 and 99, all of which are divisible by nine.
9: 10
27: 30
99: 110

All of these, when expressed in base 9, end in zero, fitting the theory.

For base 11, we can use the example numbers (as expressed in base 10) 11, 55 and 132, all of which are divisible by eleven.
11: 10
55: 50
132: 120

All of these, when expressed in base 11, end in zero, again fitting the theory.

And for base 12, we can use the example numbers (as expressed in base 10) 12, 72 and 240, all of which are divisible by eleven.
12: 10
72: 60
240: 200

All of these, when expressed in base 12, end in zero, again fitting the theory.

The “divisible by n where “n” is derived from base n” rule, where numbers which meet the criteria exclusively end in zero seems to work, although it is a fairly straight-forward rule to prove.

It gets a bit more interesting with the “divisible by n-1 where “n” is derived from base n” rule where the sum of the digits in the number need to add up to “n” in base n as the addition needs to be performed in base n, making the whole process so much more confusing.

Working in base 9, where the available symbols are 012345678, and therefore the digits of numbers divisible by 8 need to add up to 8. To make this slightly easier to follow, I’m just going to work up in multiples of 8 (so 8, 16, 24 etc) converted to base 9 (which would be 8, 17, 26 etc, as you will find if you use the example symbol set and count up in groups of 8):
8 (8=8)
17 (1+7=8)
26 (2+6=8)
35 (3+5=8)
44 (4+4=8)
53 (5+3=8)
62 (6+2=8)
71 (7+1=8)
80 (8+0=8)
88 (8+8=17 1+7=8)
etc

With this number set it works. With Base 11:
Available characters: 0123456789A
A (A=A)
19 (1+9=A)
28 (2+8=A)
37 (3+7=A)
46 (4+6=A)
55 (5+5=A)
64 (6+4=A)
73 (7+3=A)
82 (8+2=A)
91 (9+1=A)
A0 (A+0=A)
AA (A+A=19 1+9=A)

Another number set with which the theory works.

Finally Base 12
Available characters 0123456789AB
B (B=B)
1A (1+A=B)
29 (2+9=B)
38 (3+8=B)
47 (4+7=B)
56 (5+6=B)
65 (6+5=B)
74 (7+4=B)
83 (8+3=B)
92 (9+2=B)
A1 (A+1=B)
B0 (B+0=B)
BB (B+B=1A 1+A=B)

I suppose if I stop and think about it, the whole thing makes perfect sense. Unfortunately to convince myself that it made sense, I had to write it down and prove that the theories were correct.

I have another, far less coherent, theory about zero being an even number and the possible erosion of the entire numbering system if it isn’t, but I’ll save that for another day. I can’t see much point in tarnishing a perfectly good theory with a theory which is probably completely nuts…and I’m sure you’ve seen enough numbers for today.

Samuel

Entry Filed under: Bizarreness

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