[I should point out that only after
writing this whole thing out did it occur to me to look online to see
what others have said about base twelve.]
When I was writing up
the division rules I was reminded of something someone once said about different
bases. Specifically, they thought it would have been better if we
used base twelve because twelve has more divisors. Because I feel
like using random symbols that probably won't even show up for most
people, I'm going to introduce two symbols that are new to me instead of using the
traditional method of dealing with new digits (which is to use Latin
letters.) So Ñ¦ means ten and ÏŸ means eleven.
Thus 1ÏŸÑ¦ base twelve would be 1 gross
(144 base ten) plus twelve times eleven, plus ten. Which is the same
as 286 base ten.
Given that we're all already trained in
base ten it will obviously be harder for us to use base twelve regardless. For example, when I wanted to work out what 1/7th would
be in base 12 (.186Ñ¦35 repeating, if I did it right) most of what I
did was using base ten.
Also, the person who said base twelve
would be better was in no way referring to the division rules.
Fractions possibly*, but not division rules. All of which is to say,
what I'm doing here is completely pointless.

0
This is exactly the same as base ten,
but there was an error in my thing for base ten. Zero isn't a divisor
of anything, by definition. I honestly don't know what that is. The
only number it could be a divisor of is zero, and apparently saying
that zero divides zero would be problematic for something. Thus
zero is the only integer that doesn't divide itself (for any other
integer X, X is a divisor of X) and also the only integer that
doesn't divide zero (for any other integer X, X is a divisor of
zero.)
It's not as if saying zero was a
divisor of zero would make zero over zero stop being indeterminate,
0=c*0 for an infinite number of integers c, so 0/0 would still be
impossible to solve, and you'd still have to resort to limits if you
wanted some kind of answer. Still, there must be a reason of some
sort.
The division rule for zero will be the
same in every base.

1
Exactly the same as base ten: is it an
integer?
The division rule for 1 will be the
same in every base.

2
Exactly the same as base ten: is the
last digit divisible by two?
Two is a divisor of
ÏŸÑ¦03495Ñ¦349ÏŸ3Ñ¦091ÏŸ3495159Ñ¦ because two is a divisor of Ñ¦.
This still works because 2 divides
twelve. So if the last digit is 0, 2, 4, 6, 8, or Ñ¦, the number is
divisible by 2.

3
This is the first thing that is
different. Is the last digit divisible by three?
Three is a divisor of 81Ñ¦ÏŸ125ÏŸ7ÏŸ934589
because three is a divisor of 9.
This works because 3 divides twelve.
So if the last digit is 0, 3, 6, or 9, the number is divisible by 3.

4
Is the last digit divisible by four?
Four is a divisor of
Ñ¦0466208ÏŸ490Ñ¦34Ñ¦090 because four is a divisor of 0.
This works because 4 divides twelve.
So if the last digit is 0, 4, or 8, the number is divisible by 4.

5
This is the first hard one, it's sort
of like it's somewhere between the rules for eleven and seven in base
ten. You break up the digits into groups of two, add the lowest,
subtract the next lowest, add the next, subtract the one after, and
so on.
Of course, you can't do that when
there are only two digits, but in that case there's another rule.
Actually, there's more than one. Use which ever one is easier. You
can either multiply the twelves digit by 2 and add the ones digit, or you can multiply the ones digit by 2
and add the twelves digit.
So take a number, say:
ÏŸÑ¦0943530048Ñ¦513, then break it up:
ÏŸÑ¦+0943+5300+48Ñ¦5+13
then cry because you don't have a
calculator that operates in base twelve which means that you have to
work out the addition and subtraction yourself. Of course, if you
did have a base twelve calculator handy, you could just use that to
see if it's divisible by five without a division rule.
I may very well be getting all of this
wrong and I'm not going to do it in one step, but I think that,
combing terms pairwsie I get this:
ÏŸ1+10+4892= Ñ¦146 = 127
Then repeat:
+127= 26
Now we use one of the other rules:
We could say 2*2 + (6) = 4 6 = Ñ¦
which is divisible by five.
Or we could say 2 + 6*(2) = 2 +10
= Ñ¦ which is likewise divisible by five.
So, in theory at least, it is
divisible by five. Is that actually true? Do I have to check? Ok,
I'll check (says the idiot who typed in a random 15 digit number.) My calculator says that that's equal
to 15199987645464555 base ten, so yes, it is divisible by five.
Lucky random number punching I guess, it has to happen one out of
five times.
The first rule works because five
divides 101 base twelve, meaning 100 base twelve is 1 mod five. So
it's just like the sevens rule in base ten, but with one less digit.
The first of the other rules is because
10 = 2 (mod 5) base twelve. So we can replace every instance of 10
with 2, as in 26 = 2*10 6= 2*2 6 (mod 5).
The second is another case of this
being like seven in base ten. You see 5 divides 21 base 12, so we
can use the same trick we used with seven in base ten. 20 = 1 (mod
5) base twelve. So that means that we can take the 2*10 6 from
before and multiply it by 2 to get 2*(20) – 6*(2) = 2 –
6*(2) (mod 5).
So basically it comes down to which
digit you'd rather multiply by two. If things would be easier
multiplying the twelves digit, multiply it by two, if it would be
easier to multiply the ones digit, multiply it by negative two.
Here, with the digits being 2 and 6, either way is easy, but maybe
another time you don't remember what ÏŸ times two is, so if that
shows up you'd rather use a different digit. Also, generally, the
math will be simpler if you multiply the smaller number.
It's possible my brain may have
exploded in the process of making this description.

6
Is the last digit divisible by six?
6 is a divisor of
Ñ¦1ÏŸ2034931841ÏŸ04984203491040340 because six is a divisor of 0.
This works because 6 divides twelve.
So if the last digit is 0 or 6, the number is divisible by 6.

7
The first part of the rule is the same
as in base ten, break things into groups of three, alternate adding
and subtracting them.
The second part isn't all that far
from what it was in base ten. Instead of subtracting twice the last
digit from the rest, you add three times it to the rest.
In rare circumstances it might be
useful to know that you can add the last digit to five times the
rest as an alternate test.
ÏŸÑ¦0943530048Ñ¦510
ÏŸ+Ñ¦09435+30048Ñ¦+510 =
9ÏŸÑ¦135+042 = 907
90+7*3 = 90+19 = Ñ¦9 I have no idea if that is divisible by
seven, so we repeat the second rule:
Ñ¦+9*3 = Ñ¦+23 = 103
10+3*3 = 10+9 = 19 we already
discovered this is divisible by seven, but this feels like a good
time to use the last rule.
1*5+9 = 12 That's just 7*2, but say
we didn't know that. Either the second or the third rule would do.
For the second rule 1+2*3 = 1+6 = 7
For the third rule 1*5+2 = 5+2 = 7
I already checked that this was indeed
divisible by seven (which is why it's so close to the number I used
for 5.) So our rule has worked.
Ok, so why do these things work?
The first thing is because seven is a
divisor of 1001 base twelve just like it's a divisor of 1001 base
ten. So the same rule applies. The second thing is because 30 = 1
(mod 7) base twelve. We've made use of this sort of thing a lot by
this point, so I'm thinking I don't need to explain it again. The
third thing is because 10 = 5 (mod 7) base twelve. It's not going to
be useful most of the time, but if you've got a two digit number and
the twelves digit is a lot smaller than the ones digit, as in the
case of the 19 we bumped into above, it might be useful.

8
Are the last two digits divisible by
8?
Of course, we've got a question of how
to tell. You take the ones digit and add it to four times the
twelves digit.
If we're faced with
1Ñ¦3963ÏŸÑ¦07Ñ¦9071938451Ñ¦49853403ÏŸ4, we only have to look at ÏŸ4.
How would one know whether
ÏŸ4 is divisible by eight?
(Other than the fact I made sure it
is, I mean.)
So we go to rule two, ÏŸ*4 + 4 = 38 +
4 = 40. If that's not obvious we repeat.
4*4+0 = 14. 1*4 + 4 = 8. So it's
divisible by eight.
The first rule works because eight
divides 100 (which is, I guess, pronounced, “one gross.”) It's
the same reason that we only have to look at the last two digits when
looking at 4 in base ten.
The second rule is because 10 = 4 (mod
8) base 12.

9
Are the last two digits divisible by
9?
To deal with the two digit number we
add the ones digit to 3 times the twelves.
So we look at
1Ñ¦3963ÏŸÑ¦07Ñ¦9071938451Ñ¦49853403ÏŸ3 and ignore everything but the
ÏŸ3.
ÏŸ*3 + 3 = 39 + 3 = 30, and now it's
occurring to me that maybe things of the form ÏŸX are bad examples for this kind of rule.
Anyway, repeat to get 3*3 + 0 = 9.
Done.
Lets do another example. Consider
nine squared. That's 69 in base 12, but nothing about it jumps out
at me as saying it's divisible by nine, so if we just randomly came
across it we'd probably need to check instead of instantly knowing it
was 9 squared and thus divisible by nine.
6*3 + 9 = 16 + 9 = 23.
2*3+3 = 9
This works for pretty much the exact
same reasons as the eight rules. Nine divides a gross, so we only
have to look at the last two digits. 10 = 3 (mod 9) base twelve,
hence the multiplying the twelves digit by three.

Ñ¦ (ten)
The lazy rule: check two and five.
The complex annoying rule?
This is where we flashback to the
rules we used with twelve in base ten. Except worse.
We can add the last digit two twice
what remains when we remove it. Or, if we don't want to multiply
such a big number, we can do something really screwy with the number.
Take the first digit, add to it twice the twelves, digit, four times
the gross digit, negative two times the next digit, negative four
times the next digit, two times the next digit, and so on.
The sequence is 2, 4, 2, 4, and
repeat. Starting with the twelves digits and moving up the higher
digits.
Do I have to do an example?
41ÏŸ4Ñ¦ÏŸ459529784
4*4+1*2+ÏŸ*(4)+4(2)+Ñ¦*4+ÏŸ*2+4*(4)+5(2)+9*4+5*2+2*(4)+9*(2)+7*4+8*2+4
Which is a massively intimidating
problem. Hopefully it works out like this:
14+2–38–8+34+1Ñ¦–14–Ñ¦+30+Ñ¦–8–16+24+14+4
= 5Ñ¦
5*2+Ñ¦ = Ñ¦+Ñ¦ = 18, though really we
can see it's divisible by Ñ¦ at the Ñ¦+Ñ¦ step.
1*2 + 8 = Ñ¦ and we are done.
That wasn't nearly as bad as I
expected it to be (at least in part because my random numbers aren't
as random as they could be, so there was some useful canceling.) It
would have been better if I hadn't misread a sign and been forced to
redo the addition and subtraction as a result.
Anyway, I'm thinking that it's not
worse than the rule for 12 base ten.
The explanation is, entirely about
twelve being 2 modulo ten. So twelve squared is 4, and twelve cubed
is 8, but 8 = 2 (mod Ñ¦) and I would much rather multiply something
by two than eight. Also, having signs alternate can keep the sum
from getting too high. In the above I added up 15 things, many of
them two digit numbers, and the result was only a two digit number
because six of those things were negative. Returning from the
digression, twelve to the fourth power is 2*(2) = 4 (mod Ñ¦). To
get twelve to the fifth power we multiply by two again, but remember
what I said about not liking to multiply by eight when I can multiply
by two? Well that goes for negative eight as well, and 8 = 2 (mod
Ñ¦).
So that's where the 2, 4, 2, 4,
pattern comes from.

ÏŸ (eleven)
Add up all the digits, is the result
divisible by ÏŸ?
41ÏŸ4Ñ¦ÏŸ459529789
4+1+ÏŸ+4+Ñ¦+ÏŸ+4+5+9+5+2+9+7+8+9 = 83
8+3 = ÏŸ. So eleven does divide it.
This works for the exact same reason
the rule for nine works in base ten. It will work for 101
regardless of what base you happen to be in. In any given base 10 =
1 (mod 101).

10 (twelve)
Is the last digit zero?
41ÏŸ4ÏŸ459529780 is clearly
divisible by 10.
Again, this is a rule that works for
all bases. That's basically what being in base whatever means.
14514X = 14514*10 + X, and X is always less than 10, so the only way
10 divides the number is if X=0.

11 (thirteen)
Take the ones digit, subtract the
twelves digit, add the gross digit, subtract the next digit, and so
on. Is the result divisible by 11?
41ÏŸ4Ñ¦ÏŸ459529789
41+ÏŸ4+Ñ¦ÏŸ+45+95+29+78+9 = 11
So it is divisible by 11 (thirteen.)
This is exactly the same as the rule
for 11 base 10 and will be the same for 11 in any base. 10 = 1 (mod
11) in every base (that's what 11 means after all) so the rule will
always work.

So, to address the question of whether
or not it's easier, here's what we've got:
In any base the rules for 0, 1, (10 –
1), 10, and (10 + 1) will be the same. (Though in base two 101=1
and so the rule for 101 is redundant.) That means that for any base from three to twelve more than a third of the fourteen numbers we looked at were going to follow those rules. They are neither easier
nor more difficult.
In base ten, 2 and 5 have extremely
easy rules, you just look at the last digit. (The same is true of 10
but we already covered that.) In base twelve 2, 3, 4, and 6 fall into
this extremely easy category. We've doubled the size of this category. This is an example of more divisors of the base
making things easier since the reason these are easy is because they are divisors.
In base ten, 4 has you look at the last
two digits. In base twelve the same rule applies to 8 and 9. The
size of this category has likewise doubled.
Again, more divisors makes things easier. In this case it's because they're multiple of divisors.
All that's left are five, seven and
ten.
Seven is a lot like how it was in base ten. Five and ten are where things turn out harder, because five is no longer a factor of the base, and ten is no longer the base. So the change isn't all making things easier. Still, ten in base twelve seems to fall in the same category as things like six and twelve in base ten, so it doesn't seem to be making things harder so much as shifting where the difficulty lies.
The rule for five in base twelve doesn't have a good analog in base ten. (Because 101 base ten is a prime number.) It's easier than the seven rule in base ten and harder than 11 rule in any given base, but similar to both, and that's probably about as precise a comparison as we can make.
So overall it seems like things are
quite a bit easier. Several things have been made easier, other things haven't changed much, only two things have been made harder, and in the case of one of those it's arguable that the difficulty has been shifted onto it rather than created.
I have no idea what difference any of
this would make in practice, though.
I do wonder about the importance of
even and odd numbers. What if it weren't easy to figure them out?
Consider, for example, any odd numbered base. To know that a number
is even you'd have to add up all the digits and see if that was even.
For a large number you'd probably have to repeat the process
multiple times. If that were the case, would we still think in terms
of even and odd? Would we still have names for them? What if, instead of being able to tell whether
something was even or odd at a glance, you could tell that it was
divisible by three at a glance? Would that have changed our
thinking?
In base 12, both would be things you
could tell immediately upon seeing a number. Would that make a
difference in the way one thinks?
Might we have a trichotomy of divisible by three, one more than divisible by three, and one less than divisible by three to go alongside the existing dichotomy of divisible by two and one moreorless than divisible by two? If we did, would it, in any way, matter?
Consider the fact that people actually
say, “Two is the only even prime,” as if that's any more
meaningful than saying, “X is the only prime divisible by X,”
where X is any given prime. Well yes, two is the only prime number
divisible by two, and three is the only prime number divisible by
three.
Anyway, there really was no point to
this, and given that I kept slipping from base twelve to base ten and
back again, there's a fair chance the math is wrong in places.



* So, fractions. If I knew how to make
a table this would be easier to communicate. Consider fractions of
the form 1/X where X is a single digit number.
In base ten three such fractions
terminate after a single digit. 1/2 = .5, 1/5 = .2, and 1/10 = .1
In base twelve five such fractions do.
1/2 = .6, 1/3 = .4, 1/4 = .3, 1/6 = .2, 1/10 = .1
I don't know about anyone else, but I
find myself facing thirds more often than I'm facing fifths. (I have
yet to meet a recipe that calls for a fifth of a cup, for example.)
I definitely come across quarters more than fifths. So base twelve
seems to be winning there.
In base ten there is one such fraction
that terminates after two digits: 1/4 = .25
In base twelve two such fractions do:
1/8 = .16, 1/9 = .14
In base ten 1/3, 1/6, 1/7, 1/8, and 1/9
fall into neither of these categories.
In base twelve 1/5, 1/7, 1/Ñ¦, and 1/ÏŸ
fall into neither.
Which is to say, even though it has
fewer numbers to deal with, base ten has more things that fail to be
nice. Of course 1/8 base ten does
terminate, after only three digits at that**. So if we count that as being
nice then they both have the same rate of failing to be nice, though
base twelve still has more nice things.
Base ten has two things that repeat the
same digit infinitely: 1/3 and 1/9.
Base twelve has but one, that being
1/ÏŸ.
Anyway, just looking at 1/2, 1/3, 1/4 you can see why someone might want to do math in base 12. .4 is a much nicer thing to work with than .33333333333333andsoon. .3 is nicer than .25 as well.
** By now you've probably noticed a similarity between the number of digits we have to look at in a division rule for X and the number of digits in 1/X. I'm not sure how obvious it is though. It sort of makes me wonder if I had to list the one digit fractions and two digit fractions or if everyone already knew.