## Tuesday, August 28, 2007

### Base 13 - Part II: The secret

Normally we do our calculations in base 10, so the number 489 can be written as:

$489= 4 \times 10^2 +8 \times 10 +9$

In general, for integers a, b, c from zero to nine we have that,

$abc = (a \times 10^2) + (b \times 10) +c$.

This can be extended for larger integers, but we can completely generalise it by using a single formula. I hadn't originally meant to blog about this, but the comic has given me an opportunity to do so. If we let m denote the integer we have, then m can be written in base 10 by the following formula, where $a_n, a_{n-1},... a_1,a_0$ are integers from zero to nine and n is any positive integer,

$m = (a_n \times 10^n) + (a_{n-1} \times 10^{n-1}) + ... + (a_1 \times 10) + a_0$

In other words $m = a_n a_{n-1} a_{n-2}... a_1a_0$ (compare with example above).

This is also know as the decimal system which we use on most occasions; and for the non-mathematician, probably all the time!! (By non-mathematician I don't mean Physicists or Chemists, but the people who don't like maths... )

However, interestingly enough we can work in other bases like 18, 3, 7 etc. The choice of base is up to you, but according to the book I'm reading (from which I'm getting most of this!!), "the choice of base 12 has been advocated, since twelve is exactly divisible by two, three, four and six, and, as a result, work involving division and fractions would often be simplified."1 Since I mentioned base 13 \sout{above} below, I will use that as an example of a different base.

A reminder for those who didn't catch the previous post. A mysterious somebody in a red t-shirt has claimed that they're '21 in base 13 only'. You join us at .... the moment of truth. :D (Anyone remember that show!!)

We have the number 21, so we're going to work backwards (I think). I.e if we were in base 10 then 21 would mean, $2 \times 10 + 1$. Since this is base 13, an integer k can be written using the following formula,

$k= (b_n \times 13^n) + (b_{n-1}\times 13^{n-1})+ ... +(b_1 \times 13) +b_0$,

where k=$b_nb_{n-1}...b_1b_0$ and $b_n, b_{n-1},...b_1, b_0$ are integers from zero to 12 (and as above, n is any positive integer).

In our case k=21, which tells us that $b_1=2, \;b_0=1$. Hence if we plug that into the formula we should get the correct age:

$k= 21= (2 \times 13) +1$.

So if I haven't stupidly misunderstood this whole base business, then unfortunately the age of the person in the red t-shirt is no longer a secret. If I have done things correctly, the age should be 27. (If I haven't understood this properly and this post is a load of rubbish, then it'll teach me to not read books at weird times like 4am! That's an excuse BTW.) Did you get that answer? :o

1. What is mathematics? Richard Courant and Herbet Robbins

Anonymous said...

you've understood it to a nicety.
i wouldn't count this as "algebra",
by the way ... "number theory", rather.
VME

beans said...

Hi,

It is actually quite a 'sweet' topic! :o I remember, before starting my degree, naively saying to a university student that I enjoyed calculus (i.e. integrating and differentiating to me then!), and something about algebra. He'd laughed and said that I had no idea what algebra I'd be getting into. A year on and I still have no idea!!

Well a short idea, but not a concise one. I need shorting out in the mathematical sense. (BTW, I'm agreeing - just felt the urge to waffle about something!)