Basic Bases!

Yes... it's how bases work.

Just as you can say 56 (base 10) is 5 * 10 + 6, you can then convert 31 in base A to 3 * A + 1. :)

Well, when I see something like $xy_{A}$ I translate this to

$x*A^{1} + y*A^{0} = xA + y$

to know what value in base $10$ I'm dealing with.

If it was something like $xyz_{A}$ then the base $10$ value would be

$x*A^{2} + y*A^{1} + z*A^{0} = xA^{2} + yA + z$ .

If the base is unknown and we are given an equation like $31_{A} = 2_{A} * 17_{A}$ then I think the approach I've taken to determine $A$ is as efficient as any. If we know the base and we have to determine the value of, say, $2_{13} * 17_{13}$ , then we can work in base $13$ directly without converting to base $10$ first.

So with $2_{13} * 17_{13}$ , we start by noting that

$2_{13} * 7_{13} = 11_{13}$ and $2_{13} * 10_{13} = 20_{13}$ ,

and so $2_{13} * 17^{13} = 11_{13} + 20_{13} = 31_{13}$ .

It's a bit confusing when dealing with a two-digit base such as $13$ , because something like $10_{13}$ could be interpreted as either $10$ or $1*13 + 0 = 13$ in base $10$ . I'm not sure what is normally done to distinguish between the two, but I myself would write $(10)_{13}$ to indicate $10$ in base $10$ , and $10_{13}$ to indicate $13$ in base $10$ .

Hopefully I haven't made this even more confusing. when I'm in doubt I convert everything into base $10$ first so that I'm in familiar territory, but with practice you can begin to work in different bases directly without first converting to base $10$ .