A New Method to divide Numbers by 8 (Eight)!

Basic Knowledge: Any Three digit number will be in the form of $100x+10y+z$ where x is it's hundred's digit, y is the tens digit and z is the units digit. E.g. 341= $100 \times 3+ 10 \times 4+ 1$

$\overline {xyz}$ represents the number as a value. If we write that the number = $xyz$ it might be percieved for $x \times y \times z$

Let the number be here $\overline {xyz} = 100x+10y+z$

According to the procedure, The number formed by the first two digits = $\overline {xy}=10x+y$

Multiply by two to get $20x+2y$

Adding the units digit, we get $20x+2y+z$

Divide it by Eight. $\dfrac {20x+2y+z}{8}$

Let us ignore the step where we get the remainder and the quotient and we add them to the original two digit number, because it makes no difference whether we divide and add or we add first and divide. (E.g. We break a number 341 into 100+241. We divide them individually by 8 getting remainders 4 and 1. We add them, and we get 5 as the remainder which is the remainder when 341 is divided by 8).

Instead, we directly add our number i.e. $\dfrac {20x+2y+z}{8}$ to our original two digit number.

Adding, $\dfrac {80x+8y+20x+2y+z}{8}$

$\Rightarrow \dfrac {100x+10y+z}{8}$

$\Rightarrow \dfrac {\overline {xyz}}{8}$ Thus this would give us the remainder and the quotient when the original there digit number is divided by 8

Woot! Thus we get the desired result and the method works!. :)

Let the number be $\overline{abc}$ . The successive steps have us calculate: $\overline{ab};$ $2\overline{ab};$ $2\overline{ab}+c;$ $\frac{2\overline{ab}+c}8;$ $\overline{ab} + \frac{2\overline{ab}+c}8 = \frac{10\overline{ab}+c}8 = \frac{\overline{abc}}8.$