Hexadecimal dividing $\frac{17a}{12}$ with remainder

Question

How much is it $\frac{17a}{12}$ (in hexadecimal) I got $15$ with remainder $a$. Can someone confirm its true?

Thank you!

Answer 1

No, it's false.

Since the claim is equivalent to $$17A_{16}=12_{16}\cdot15_{16}+A$$

One can confirm as the following way:

$$17A_{16}=16^2+7\cdot16+10=378$$

While

$$15_{16}\cdot12_{16}+A=(16+5)(16+2)+10=388\neq378$$

Obviously, by observation, we have

$$15_{16}\cdot12_{16}+A=17A_{16}+A$$ $$15_{16}\cdot12_{16}=17A_{16}$$

So it should be:

$$\frac{17A_{16}}{12_{16}}=15_{16}$$

Answer 2

Just do long division...

12 / 17A \ 15
     12
     --
      5A
      5A
      --
       0

Answer 3

Just like in the case of the division algorithm in the base then system, first you take a look at $$17a:12$$

and you realize that $17/12=1$ and the remainder is $5$. So you have after the first step $$ \begin{matrix} 17a&:12&=&1\\ \color{white}{|}5 \end{matrix} $$ then you "bring down" the next digit of the dividend: $$ \begin{matrix} 17a&:12&=&1\\ \color{white}{|}\color{blue}{5a} \end{matrix} $$

then you guess the result of the division $\color{blue}{5a}:12$. My first guess is $5$. Indeed $12\times5=5a$.

You right this down the following way

$$ \begin{matrix} 17a&:12&=&1\color{red}5\\ \color{white}{|}\color{blue}{5a}\\ \ \color{red}{ 5a} \end{matrix} $$

now, you subtract $\color{red}{5a}$ from $\color{blue}{5a}$. The result would be the remainder, zero this time, because there is no more numbers in the divident to be brought down.