Hi Jade.

I'd start by remembering that we use place-value notation for our numbers. This means that each "place" represents a different value. With base 10 numbers, the 4 in the number 3421 is in the "hundreds" place. When you count, starting with zero, from the right to the left, you see that the four is in position 2, so that place represents the 10² value (base^position). It is easy to verify that

3421 = ( 3 * 10³ ) + ( 4 * 10² ) + ( 2 * 10¹ ) + ( 1 * 10⁰ ).

The principal is the same when working in other bases.

Let's say we have 453₆ (we use subscripts to denote the base) and we want to convert to base 11. The original base here is called "6" if we agree to always write the base itself in base 10. This is the usual practice, but it is important to remember that the base is really always 10 when written in the base itself. What I mean is that 10₆ = 6₁₀. 453₆ is composed of three digits each in its place, so expressed in expanded base 6 notation this is:

453₆ = ( 4₆ * 10₆² ) + ( 5₆ * 10₆¹ ) + ( 3₆ * 10₆⁰ ).

Now we convert these small numbers directly from base 6 to base 11 and substitute them in:

4₆ = 4₁₁
5₆ = 5₁₁
3₆ = 3₁₁ (all these are easy because we are going from a small base to a larger base).
10₆ = 6₁₁ (you have to think this one out a bit more).

Let's leave the exponents as is, since 0, 1, 2 are all smaller than both bases anyway. So the substitution gives us this:

453₆ = ( 4₁₁ * 6₁₁² ) + ( 5₁₁ * 6₁₁¹ ) + ( 3₁₁ * 6₁₁⁰ ).

It really helps to write the multiplication table for 6₁₁ in base 11 if you cannot do it in your head. I'll solve this in steps:

453₆
= ( 4₁₁ * 6₁₁² ) + ( 5₁₁ * 6₁₁¹ ) + ( 3₁₁ * 6₁₁⁰ )
= ( 4₁₁ * 33₁₁ ) + ( 5₁₁ * 6₁₁ ) + ( 3₁₁ )
= 121₁₁ + 28₁₁ + 3₁₁

And finally we just add in base 11:

453₆ = 151₁₁.

You'll notice that I had to ensure I was "carrying" on elevens instead of tens, but that is just working in base 11, so we haven't really used base 10 through this whole conversion. However, I want to check my work in base 10:

453₆ = 4 * 36 + 5 * 6 + 3 = 144 + 30 + 3 = 177.

151₁₁ = 1 * 121 + 5 * 11 + 1 = 121 + 55 + 1 = 177.

Now you try this approach with your base 6 to 12 conversion, Jade.

Cheers,
Stephen La Rocque.

Jade,

Here is another technique that is similar to the method I would use to put 177 in base 6. Suppose you have 177 hockey pucks and you want to express this number in base 6. Arrange the pucks in stacks of 6 pucks each. You get 29 stacks of 6 pucks each with 3 left over. Now arrange the 29 stacks of 6 pucks into groups of 6 stacks each. You get 4 groups of 6 stacks each and 5 stacks of six pucks left over. Thus you have

4 groups of 36 pucks
5 stacks of 6 pucks and
3 more pucks.

That's is 4 × 6² + 5 × 6 + 3 = 453₆ pucks.

Arithmetically what is did was

177 ÷ 6 = 29 with remainder 3
29 ÷ 6 = 4 with remainder 5
4 ÷ 6 = 0 with remainder 4

Read the remainders from bottom to top to get 453₆.

Now suppose you want to convert 453₆ to base eleven. I can use the same technique, repeatedly divide by eleven and record the remainder. The main difference is that you have to express eleven in base 6 and divide in base 6. Eleven is 15₆ so here is my calculation (without the details)

453₆ ÷ 15₆ = 24₆ with a remainder of 1
24₆ ÷ 15₆ = 1₆ with a remainder of 5
1₆ ÷ 15₆ = 0 with a remainder of 1

Thus 453₆ = 151₁₁

Penny