# 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

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 $$a \times 10^d + b = ab$$.

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.

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

This argument always works.

• So, this divisibility after concatention is always true for all the prime numbers right? Commented Nov 26, 2019 at 21:50
• At no point I used the fact that $23$ is prime. It works for all natural numbers. Commented Nov 26, 2019 at 21:51

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.