Two numbers divisible by the same prime. If concatenated still divisible by the same prime number? 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. 
 A: 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 $a \times 10^d + b = ab$.
A: 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.  
A: Since $23\mid69$, $23\mid69\,000$. So, $23\mid(69\,000+460)$; in other words, $23\mid69\,460$.
This argument always works.
A: 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.
