# Thread: Converting Bases

1. ## Converting Bases

1.) Convert (547)_10 to the _6 notation

2.) Convert (6A3)_12 to the decimal notation.

Note: _10 and _6 mean base 10 and base 6, respectively.

2. Originally Posted by fifthrapiers
1.) Convert (547)_10 to the _6 notation.
Note 6^4 is too much.

6^3=216 is the starting point.
This divides the number 2 times remainder 115.

Now 6^2=36 divides this number 3 times, remainder 7.

Now 6^1=6 divides this number 1 times, remainder 1.

Thus,
(2311)_6 is the required number.

3. Originally Posted by ThePerfectHacker
Note 6^4 is too much.

6^3=216 is the starting point.
This divides the number 2 times remainder 115.

Now 6^2=36 divides this number 3 times, remainder 7.

Now 6^1=6 divides this number 1 times, remainder 1.

Thus,
(2311)_6 is the required number.

Where's the 2311 from, and I thought using this algorithm that you go until you have remainder 0

EDIT: Oh, I see where it's from. Thanks.

4. I don't get what the A is. In my notes, I have this as an example:

(28A)_12:

12^2 = 1
12^1 = 20
units = 10

I'm assuming when it goes over the base?

5. Hello, fifthrapiers!

There is a "trick" to converting to another base . . .

1) Convert 547 to base-6.

1. Divide the number by 6, note the remainder. .547 ÷ 6 = 91, rem. 1

2. Divide the quotient by 6, note the remainder: .91 ÷ 6 = 15, rem. 1

3. Repeat step 2 until a 0 quotient is reached:. . . 15 ÷ 6 = 2, rem. 3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 ÷ 3 = 0, rem. 2

Now read up the remainders.

Answer: . 547 .= .2311_6

2) Convert (6A3)_12 to base-10.

(6A3)_12 .= .6·12² + 10·12 + 3·1 .= .987

6. Originally Posted by Soroban
[size=3]

(6A3)_12 .= .6·12² + 10·12 + 3·1 .= .987

What exactly is the A? Mind explaining this a bit more?

7. Originally Posted by fifthrapiers
What exactly is the A? Mind explaining this a bit more?
When using a base higher than 10 we need symbols for numbers greater than 9 but less than 10. So the typical count goes:
1 2 3 4 5 6 7 8 9 A B C D ... 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D ... 20 etc.

For base 12 we get:
1 2 3 4 5 6 7 8 9 A B 10 11 12 13 14 15 16 17 18 19 1A 1B 20 etc.

-Dan