Does anybody have a guide to division of hexadecimal base numbers?
For example divide 57F6D by B9.... without first converting to decimal...
I need to do it using long division... on paper... without converting it to any other base... I've been given an example, but when I've been revising, I can't understand how to do it...
57F6D is divided by B9 to give 79B as quotient and 6A as remainder... that's the example...
Hello, yuud!
Does anybody have a guide to division of hexadecimal base numbers?
For example: .$\displaystyle 57F6D_{16} \div B9_{16}\quad\hdots$ without converting to decimal.
My advice is to learn to "think" in hexidecimal
. . or learn to quickly covert from hexidecimal to decimal and back.
In hexidecimal, the division looks like this:
. . $\displaystyle \begin{array}{cccccccc}
&& & & 7 & 9 & B\\
&& -- &--&--&--&-- \\
B\,\;9 & ) & 5 & 7 & F & 6 & D \\
& & 5 & 0 & F \\
& & -- & -- & -- \\
&&& 7 & 0 & 6 \\
&&& 6 & 8 & 1 \\
&&& --&--&-- \\
&&&& 8 & 5 & D \\
&&&& 7 & F & 3 \\
&&&& --&--&-- \\
&&&&& 6 & A
\end{array}$
As everyone has said, exactly like you divide in decimal.
Of course, you need to know some multiplication and division "facts" first- again just like you learned for decimals:
Multiplication tables:
1 times 1= 1, times 2= 2, times 3= 3, ..., to 1 times F= F.
2 times 1= 2, times 2= 4, times 3= 6, times 4= 8, time 5= A, times 6= C, times 7= E, times 8= 10, times 9= 12, times A= 14. times B= 16, time C= 18, times D= 1A, times E= 1C, times F= 1E.
3 times 1= 3, times 2= 6, times 3= 9, times 4= C, times 5= F, times 6= 12, times 7= 15, times 8= 18, times 9= 1B, times A= 1E, times B= 21, times C= 24, times D= 27, time E= 2A, times F= 2D
etc.
continuing in that way, you would eventually get to
7 times 1= 7, times 2= E, times 3= 15, times 4= 1C, times 5= 23, times 6= 2A, times 7= 31, times 8= 38, times 9= 3F, times A= 46, times B= 4E, times C= 54
it was that last one, that 7 times B= 4E and 7 times C= 54 that caused Soroban to try B as the first "digit" in the quotient, just as if, in base 10, you had 79 divided into 4883, you would try a trial divisor of 7 or 8 because you know that 7 tmes 7 is 49 and 7 times 8 is 56 (it would be that "9" in the one's place of the divisor that would cause you to try 8 rather than 7). Once you have an idea of where a trial divisor should be, it's just "try and check".