How to convert Hexadecimal to Duodecimal?

Question

B4A9 = 22921

What is the solution?

Please use the 7-3-2-1 code.

I don't know if the 7-3-2-1 code is correct.
(for example binary has 8-4-2-1 code to convert)

I need the solution. Thanks!

Answer 1

I don't know what a "7-3-2-1" code is, but base conversion any base (X) to another base (Y) is easy if you know how to multiply and add using base-Y arithmetic. I'll use uppercase letters, as usual, for extra digits so that A is ten, B is eleven, and so forth, and lowercase suffixes of "h" for hexadecimal and "t" for duodecimal (base twelve). Let's use your example of B4A9h:

Take the first digit of the input number (Bh) expressed in the target base (Bt) as the initial result. (For larger digits, you would have: Ch=10t, Dh=11t, Eh=12t, or Fh=13t--but those won't be needed in this example.) Call this value r1.

For each additional digit, multiply the previous result by sixteen (14t) and add the base twelve representation of that digit. So:
r1 = Bt (the first digit, see above)
r2 = r1*14t + 4t
r3 = r2*14t + At
r4 = r3*14t + 9t

There are no more digits, so r4 is the answer.

I don't have a base twelve calculator, and don't have the base-twelve times table memorized, so with pencil and paper I use a shortcut based on 10h*x = 14t*x = (2*2 + 10t)*x.
Take the number x (in duodecimal), double it, double it again, then add the value of x with a trailing zero appended. With x=r1, you get
r1 = Bt
r1*2 = 1At
r1*2*2 = 38t
38t + B0t = 128t = Bt * 14t
So r2 = 128t + 4t = 130t
Similarly, I get r3 = 180At, then r4 = 22921t, which agrees with your result.

Answer 2

Convert it first to base 10, then convert to base 12.

9 = 9 x 16^0 = 9 x 1 = 9
A = 10 x 16^1 = 10 x 16 = 160
4 = 4 x 16^2 = 4 x 256 = 1024
B = 11 x 16^3 = 11 x 4096 = 45056
Add them up, B4A9 = 46249

Now to convert 46249 to duodecimal, divide by 12 successively until the quotient is smaller than 12. List down all the remainders and the final quotient, and thats your answer.

Lets do it for 46249:
46249/12 = 3854 r 1
3854/12 = 321 r 2
321/12 = 26 r 9
26/12 = 2 r 2

There you have it. Start with the final quotient, which is 2, then affix the remainders from bottom to top: 2921
answer = 22921

Answer 3

"Duodecimal"? Isn't that base 12? Not a very useful system. Or do you mean binary (base 2)? I'm going to assume you mean binary...

It's very easy, provided you have the following table at hand -

Hex . Bin.
0. . . .0000
1 . . . 0001
2 . . . 0010
3 . . . 0011
4 . . . 0100
5 . . . 0101
6 . . . 0110
7 . . . 0111
8 . . . 1000
9 . . . 1001
A . . . 1010
B . . . 1011
C . . . 1100
D . . . 1101
E . . . 1110
F . . . 1111

Now all you have to do is substitute the hex digit with the 4 corresponding binary digits. This works because the base of hex, 16, is the 4th power of the base of binary, 2.

So say you have the hex number 4A09. Converted to binary it would be

. . 4 . . A . . . 0 . . 9
0100 1010 0000 1001

You can drop the leading zero, but it is common to leave a space every 4 digits of a binary number to make it easier to read, much like putting in a comma every 3 digits of a decimal number. So then you'd have

4A09(b16) = 100 1010 0000 1001(b2)


...and if you actually DID mean base 12, the way I would do it is to first convert the hex number to decimal because I don't have a calculator that works in base 12. Then I would divide out the largest power of 12 that would go into it, then the next largest, etc.

Answer 4

This Site Might Help You.

RE:
How to convert Hexadecimal to Duodecimal?
B4A9 = 22921

What is the solution?

Please use the 7-3-2-1 code.

Answer 5

Try using 5-3-2-1 rather than 7-3-2-1

Answer 6

These may help :-)