ABABAB Numbers

 This tweet caught my interest:

It's not too hard to prove this is true. Let's begin by writing $ababab$ in terms of place rather than face values. We get:$$100000a+10000b+1000a+100b+10a+b=101010a+10101b$$Now we just need to show that 37 divides both $101010$ and $10101$. We find that:$$\frac{101010}{37}a+\frac{10101}{37}b=2730a+273b$$Thus indeed does 37 divide every number of the form $ababab$.

What's interesting is to extend this idea to other patterns, let's say $abcabc$. Investigating this we find that:$$100000a+10000b+1000c+100a+10b+c=100100a+10010b+1001c$$It's apparent that every number of the form $abcabc$ is thus divisible by 1001. However, 1001 is factorisable: $$1001=7\times 11\times 13$$Thus every number of the form $abcabc$ is divisible by 7, 11 and 13.

Let's try another one: $abba$. This gives:$$\begin{array}{rl}1000a+100b+10b+a& =1001a+110b\\ & =11\times 91a+11\times 10b\end{array}$$Clearly numbers of this form are divisible by 11. The same is true (pretty obviously) of the pattern aabb where we have:$$\begin{array}{rl}1000a+100a+10b+b& =1100a+11b\\ & =11\times 100a+11\times b\end{array}$$How does this tie in then with the divisibility rule for 11? Well one rule is:

If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.

Let's apply this rule to $abcabc$, $abba$ and $aabb$ which patterns we know are divisible by 11:

• $abcabc$ --> $(a+c+b)-(b+a+c)=0$
• $abba$ --> $(a+b)-(b+a)=0$
• $aabb$ --> $(a+b)-(a+b)=0$

Thus the divisibility rules are in conformity with what we already knew.

There  is an inherent ambiguity in presenting numbers in the form of $ababab$, $abcabc$, $abba$, $aabb$ etc. because algebraically these numbers would be rendered $a^3{b}^3$, $a^2{b}^2{c}^2$, $a^2{b}^2$ and $a^2{b}^2$ respectively. This is because, when writing $ababab$ for example, the common interpretation is $a\times b\times a\times b\times a\times b$. 

It might be better to use a tuple, $(a,b,a,b,a,b)$ and say that it is equivalent to $a\times {10}^5+b\times {10}^4+a\times {10}^3+b\times {10}^2+a\times 10+b$. However, I guess things are clear enough if you explicitly state that each pronumeral represents a digit between 0 and 9 and that place as well as face value needs to be taken into consideration..

The patterns of the pronumerals are reminiscent of rhyming schemes in literature and many have specific names e.g. Balliol rhyme AABB. This topic could obviously be investigated more extensively but I'll leave off there for the moment.