divisibilityprime numbers
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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What you discovered is that
For $p$ a prime number, and $n \in \{1, \dots, p-1 \}$, the binomial coefficient $$\binom{p}{n} = \frac{p!}{n!(p-n)!}$$ is divisible by $p$.
The proof is very simple. The fraction $$\frac{p!}{n!(p-n)!}$$ has a numerator divisible by $p$ ($p!= p \cdot (p-1) \cdots$) and has a denominator not divisible by $p$ (all prime factors of $n!(p-n!)$ are at most $p-1$).
Thus the fraction is divisible by $p$.
This is a very important fact in math, since it is used to prove that $$(a+b)^p \equiv a^p+b^p \pmod{p}$$ for all primes $p$.
Best Answer
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 $a \times 10^d + b = ab$.