New Formula For Divisibility Of 8 Discovered By Me

We can find whether a number is divisible by 8 or not by following steps:

1# first multiply the no. Leaving ones digit by 2 or leaving the last 2 digit by 4 ( leaving last n digits by 2^n)and add the leaves no. , if the result is divisible by 8 the original no. Is divisible by 8 , to check whether the resulting no. Is divisible by 8 or not repeat the process.

Ex. Let original no. Be 34512 now leaving last digit , 3451x 2 +2( 1 digit is leave(2^n))

3451×2 +2= 6904 , 690×2+4= 1384, 138×2+4= 280, 28×2+0=56is divisible by 8 so 34512 is divisible by 8

#NumberTheory

Easy Math Editor

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Comments

Can you use LaTeX and paragraphs, please?

Because it's big and bulky...

Yajat Shamji - 7 months ago

I'll rephrase it in a more straightforward way:

$8|10a + b \rightarrow 8|2a + b$ for positive integers $a$ and $b$ . The "new divisibility rule" takes advantage of that, and iterates using that.

I personally don't see the point in applying this, the conventional method is much faster.

Elijah L - 7 months ago

Take an ex. 6948, 69×4+48= 324, 3×4+24= 36 not divisible by 8 so 6948 is not divisible by 8 . Isn't this easy

Hemant Kaushik - 7 months ago

@Hemant Kaushik – Sure, but the conventional rule is to take the last $3$ digits of the number and see if that is divisible by $8$ . That way, I only need to do $1$ total operation, as compared to your $2$ for $6948$ . I don't see a point in using a less efficient method to test for the divisibility rule of $8$ .

Elijah L - 7 months ago

Yeah, but my rule also describes the conventional rule although the conventional rule is a part or simplification of my rule as , let we leave last three digits than the no. Left we have to multiply by 8 as of 2^3(formula for n digits leaved is 2^n)and add the leaves no. , Now if the leaved no. ( The last three digits of original no. ) Is divisible by 8 then adding the multiplied no will be divisible by 8 that why we need just last three digits , it also describes how last two digits are taken in case of 4. It is not just a Divisibility rule for 8 , it is for all powers of 2(2^n) . Seems like it has gone very lengthy , sorry as I am not a good explainer .

Hemant Kaushik - 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$