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I am a new student of basic Mathematics and I am learning number systems, so this is a basic question. This exercise gives me the following duodecimal number:

9AX (where A=11, X=10)

And it requires to be converted to binary, octal and decimal. Of course I am doing this by hand so the answer could not be using WolframAlpha...

First, I am dividing the number to get the binary form, getting: (10011101001). Then, I group them in groups of four: (1001 1101 0010). Then, in groups of three: (100 111 010 010). This gives the octal number (4722). Final decimal number would be (2514).

My results are wrong according to the answer sheet. It should be (1011001010)2, (2625)8 and (1429)10.

Can you tell me what am I doing wrong? Thank you!

2 Answers 2

1

To check your answer we could compare the values in the normal decimal system: $(9AX)_{12}=9\times 12^2+11\times 12^1+10\times 12^0=1438$

$(10011101001)_2=1\times2^{10}+0\times2^{9}+0\times2^{8}+1\times2^{7}+1\times2^{6}+1\times2^{5}+0\times2^{4}+1\times2^{3}+0\times2^{2}+0\times2^{1}+1\times2^{0}=1257$

However it seems the answer sheet also has it wrong:

$(1011001010)_2=1\times2^{9}+0\times2^{8}+1\times2^{7}+1\times2^{6}+0\times2^{5}+0\times2^{4}+1\times2^{3}+0\times2^{2}+1\times2^{1}+0\times2^{0}=714$

To translate to binary I find it easiest to start from decimal $(1438)_{10}$ and use the following procedure: If the number is even the rightmost digit is 0, else it is 1. If the number is odd subtract by 1. Then divide by two. Then repeat the above, adding digits to the left of your result, until you reach 1. In this case:

$1438\ \ \ \ 0$

$719\ \ \ \ \ 10$

$359\ \ \ \ \ 110$

$179\ \ \ \ \ 1110$

$89\ \ \ \ \ 11110$

$44\ \ \ \ \ 011110$

$22\ \ \ \ \ 0011110$

$11\ \ \ \ \ 10011110$

$5\ \ \ \ \ 110011110$

$2\ \ \ \ \ 0110011110$

$1\ \ \ \ \ 10110011110$

You can check in the same way as above that $(10110011110)_2=(1438)_{10}$.

  • 0
    No problem, glad to help2012-11-03
1

You are clearly doing the conversion to binary wrong: Your last duodecimal digit is $X=10$, so the number must be even, yet your binary form ends with a $1$.

If I were you, I'd convert it to decimal first:

$9 \times 12^2 + 11 \times 12^1 + 10 \times 12^0 = 1296 + 132 + 10 = 1438$ (your answer sheet is wrong, $1429$ is represented in duodecimal as $9A1$)

And then convert it do binary:

$\begin{array}{rcll} 1438 &>=& 1024 &:& 1 \\ 414 &<& 512 &:& 10 \\ 414 &>=& 256 &:& 101\\ 158 &>=& 128 &:& 1011\\ 30 &<& 64 &:& 10110\\ 30 &<& 32 &:& 101100\\ 30 &>=& 16 &:& 1011001\\ 14 &>=& 8 &:& 10110011\\ 6 &>=& 4 &:& 101100111\\ 2 &>=& 2 &:& 1011001111\\ 0 &<& 1 &:& 10110011110\\ \end{array}$

After this you can separate it in groups of 3 to obtain the octal representation.