# Divisibility rule

Example: $$2^1$$=2 --> $$2\mid2$$

If a number has their last digit divisible by 2, than the number is divisible by 2

$$2^2$$=4--> $$4\mid2$$, $$4\mid4$$

If a number has their last two digit divisible by 4, than the number is divisible by 4

$$2^3$$=8--> $$8\mid2$$, $$8\mid4$$ , $$8\mid8$$

If a number has their last three digit divisible by 8, than the number is divisible by 8

and so on...

How would you word this into a conjecture?

• do you mean word the conjecture?, or the question title – Saketh Malyala Oct 27 '18 at 19:04
• @SakethMalyala you are right. I worded that wrong. – user608997 Oct 27 '18 at 19:06
• I mean, you can just say that for any natural $k$, if the last $k$ digits of a number are divisible by $2^k$, so is the whole number. – Don Thousand Oct 27 '18 at 19:07

Your conjuncture is essentially a true statement. The formulation of the statement is in @Ethan Bolker's answer. Here is also a prove for the statement:

Each number having $$n+1$$ digits can be written as follows:

$$10^na_n + 10^{n-1}a_{n-1} + \cdots+10a_1 + a_0 \tag 1$$

• Notice that we can extract $$10$$ from the first $$n$$ terms and the number will be $$10m+a_0$$ where $$m = (10^{n-1}a_n + 10^{n-2}a_{n-1} +\cdots+a_1)$$. Now, if $$2$$ divides this number, $$2$$ shall divide its form, $$10m+a_0$$. However $$2|10$$, hence, it will divide $$10m$$. Therefore $$a_0$$ must be divisible by $$2$$ so that the number is divisible by $$2$$.

• For the general case where $$2^k$$ divides $$n$$ if and only if $$2^k$$ divides the last $$k$$ digits, extract $$10^k$$ from the first $$n-k+1$$ terms, and follow the previous argument given that $$2^k$$ divides $$10^k$$ as $$10^k = 2^k\times5^k$$.

Your conjecture is (essentially) correct. You should be able to fill in the blank in

If a number has their last ?? digit divisible by ??, than the number is divisible by $$2^k$$.

But you are using the "$$|$$" symbol incorrectly. It means "divides", not "is divisible by", so $$4 \mid8$$ but $$8 \nmid 4$$ .

Finally, the statements you write after the arrows are true when you turn them around, but they don't prove the conjecture. Were you supposed to do that?

• we have been using "$|$" meaning "factor of". That's how I was taught. Maybe I was give the wrong information. – user608997 Oct 27 '18 at 19:54
• @DeNel If that's what you were taught then the source of your information is out of sync with the standard usage. – Ethan Bolker Oct 27 '18 at 20:16
• Well, $2$ is a factor of $10$, and $2|10$. I guess it is a matter of @DeNel’s application of the definition. – Lubin Oct 27 '18 at 21:44
• @Lubin I hurried. THe reading "is a factor of" is correct, of course. But then $8 | 4$, as written, is obviously wrong. – Ethan Bolker Oct 28 '18 at 2:29

Hint $$\ \ 2^k\mid a\!+\!10^k b\iff 2^k\mid a\,\$$ by $$\,\ 2^k\mid 10^k = 2^k 5^k$$

Better $$\,\ a\bmod 2^k = (\underbrace{a\bmod 10^k}_{\large {\rm first}\ k\ {\rm digits}})\bmod{ 2^k},\$$ an example of the simpler multiple method.

Better $$\,\ a\equiv b\pmod{\!10^k}\,\Rightarrow\, a\equiv b\pmod{\!2^k}$$

Better $$\,\ a\equiv b\pmod{\!mn}\,\Rightarrow\, a\equiv b\pmod{\!n}\,\$$ by $$\,\ n\mid mn\mid a-b$$

e.g. $$\bmod 1001\!:\ \color{#c00}{10^{\large 3}}\!\equiv -1 \,\Rightarrow\, a=12,013,002\equiv 12(\color{#c00}{10^{\large 3}})^{\large 2}\!+13(\color{#c00}{10^{\large 3}})+2\equiv 12\!-\!13\!+\!2\equiv 1$$

so $$\ 7\!\cdot\!13=10^2\!-\!10\!+\!1\mid 10^3\!+\!1\,\Rightarrow\, a \bmod 13 = (a \bmod 1001)\bmod 13 = 1\bmod 13 = 1$$.

In congruence language: $$\ a\equiv 1\pmod{\!13j\!=\!\!10^3\!+\!1}\,\Rightarrow\, a\equiv 1\pmod{\!13}$$

That's the idea behind one divisibility test for $$13$$.