# Divisibility by 8

A month or so ago, when I started learning exploring Number Theory, I looked more deeply into divisibility rules. The one that I got from Google for 8 was "A number is divisible by 8 if and only if its last three digits are divisible by 8".

This method is obviously accurate and can be proved really easily. However, it's not that useful... When I see a 3-digit number, I can not tell immediately if it is a multiple of 8. So I thought about how I can verify if 3-digit number is a multiple of 8 very quickly without a calculator, and came up with this.

A number $$\overline{...abc}$$, (meaning a number ending in the three digits a, b and c) is a multiple of 8 if...

(1) $$a$$ is even and $$\overline{bc}$$ is a multiple of 8

OR

(2) $$a$$ is odd and $$\overline{bc}$$ is a multiple of 4 and not 8.

This method is much simpler than the one I found from Google, because you need only remember 2-digit multiples of 4 and 8 instead of 3-digit multiples.

I wrote this proof for my method. It's my first attempt at writing a proof, so my format probably won't be very nice, but the mathematics is valid.

Note by Nicolas Bryenton
3 years, 11 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$