Hexadecimal dividing $\frac{17a}{12}$ with remainder How much is it $\frac{17a}{12}$ (in hexadecimal)
I got $15$ with remainder $a$.
Can someone confirm its true?
Thank you!
 A: 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:
$$\frac{17A_{16}}{12_{16}}=15_{16}$$
A: 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.
A: Just do long division...
12 / 17A \ 15
     12
     --
      5A
      5A
      --
       0

