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This may be a simple question but I would like to know the theory behind it.

Assume I have two different integer numbers of any digit length and both are divisible by the prime number 23.

When I concatenate them they seem to be always divisible by 23.

Exampe: a and b is divisible by 23. Then why ab is also divisible by 23? a = 460, b = 69, ba= 69460 $\div$ 23 = 3020

Please answer only If you are kind enough to answer simple questions kindly.

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4 Answers 4

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Concatenating a number $a$ with another number $b$ can be expressed in terms of addition and multiplication; in particular, if $b$ has $d$ digits, then to form the concatenation $ab$, you must shift the digits of $a$ by $d$ digits to the left - meaning you multiply it by $10^d$ - and then add $b$ to that.

In particular $$ab = a\times 10^d+ b.$$ This gives you all you need to prove it from two properties: first, if $n$ is divisible by $c$, then every multiple of $n$ is also divisible of $c$. Also, if $n_1$ and $n_2$ are both divisible by $c$, then so is $n_1+n_2$. Thus, since $a$ and $b$ are divisible by some divisor $c$, so is $a\times 10^d$ and $a \times 10^d + b = ab$.

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Sure. $p|m$ means there is a $k$ so that $m=pk$.

And $p|n$ means there is a $j$ so that $n =pj$.

And so If $n$ has $c$ digits then $m.n = 10^c*m + n$

And $10^c*m + n = 10^c(pk) + (pj) = p*[10^ck + j]$.

That's it.

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or another way of looking at it. If $p|m$ and $p|n$ then $p| Am + Bn$ for any integers $A,B$. And $m.n = 10^cm + n$ for some $c$.

Note: $p$ being prime has nothing to do with it. This is true of all integers.

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Since $23\mid69$, $23\mid69\,000$. So, $23\mid(69\,000+460)$; in other words, $23\mid69\,460$.

This argument always works.

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  • $\begingroup$ So, this divisibility after concatention is always true for all the prime numbers right? $\endgroup$
    – entropy
    Commented Nov 26, 2019 at 21:50
  • $\begingroup$ At no point I used the fact that $23$ is prime. It works for all natural numbers. $\endgroup$ Commented Nov 26, 2019 at 21:51
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The set $M$ of multiples of $23$ (or any integer) are closed under addition and integer scalings, $ $ i.e.

$$ a,b\in M\,\Rightarrow\, a+b\in M,\,\ na\in M,\ {\rm for\ all\ } n\in\Bbb Z$$

In particular $\,a,b\in M\,\Rightarrow\, 10^k a\in M\,\Rightarrow\, 10^k a + b\in M,\ $ which is said radix $10$ concatenation when we take $k$ to be digit length of $b$.

Remark $ $ The same closure properties hold for the set of all common multiples of any set of integers. This innate algebraic structure of common multiples is a prototypical example of an ideal - an algebraic structure that is fundamental in ring theory and number theory.

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