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?

  • $\begingroup$ do you mean word the conjecture?, or the question title $\endgroup$ – Saketh Malyala Oct 27 '18 at 19:04
  • $\begingroup$ @SakethMalyala you are right. I worded that wrong. $\endgroup$ – user608997 Oct 27 '18 at 19:06
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    $\begingroup$ 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. $\endgroup$ – 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 $10^k = 2^k\times5^k$.

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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?

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  • $\begingroup$ we have been using "$|$" meaning "factor of". That's how I was taught. Maybe I was give the wrong information. $\endgroup$ – user608997 Oct 27 '18 at 19:54
  • $\begingroup$ @DeNel If that's what you were taught then the source of your information is out of sync with the standard usage. $\endgroup$ – Ethan Bolker Oct 27 '18 at 20:16
  • $\begingroup$ Well, $2$ is a factor of $10$, and $2|10$. I guess it is a matter of @DeNel’s application of the definition. $\endgroup$ – Lubin Oct 27 '18 at 21:44
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    $\begingroup$ @Lubin I hurried. THe reading "is a factor of" is correct, of course. But then $ 8 | 4$, as written, is obviously wrong. $\endgroup$ – 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 $13$.

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