Analysis for Realsies (2015): 1

Hey guys! Sooo, last week in Tute A y'all saw this cool fact that one way to check the divisibility of a (natural) number by $3$ is to add up all of its digits and check to see if the resulting sum is divisible by $3$.

E.g.: take $832745812349$, now $8+3+2+7+4+5+8+1+2+3+4+9=56$ and $56$ isn't divisible by $3$. Therefore $832745812349$ isn't divisible by $3$.

And in fact, this is also true for $9$, you can add up the digits of a number and if this sum is divisible by $9$, then the original number must also be divisible by $9$ (and vice versa).

I then asked you to try to figure out a similar trick for $11$, and you can find the rule (and a derivation/proof) here!

Well, I then issued like a super-ninja-hax-challenge version of this where I asked you to derive a similar rule for divisibility by $101,1001,10001,100001,\ldots$ How did you go with this? I'll write the answer below, but I kinda want you to take a look at the above blog post first.

Here we go:

So, the trick for divisibility with respect to $11$ of a $k$-digit whole number (with digit representation) $\overline{n_0 n_1 n_2\ldots n_k}$ is to take the following "alternating sum" $$\begin{align} n_k-n_{k-1}+n_{k-2}-\ldots+(-1)^kn_0,\end{align}$$ and if this sum is divisible by $11$, then the original number is divisible by $11$. Note in particular, the order in which we've go - we've taken the last digit, subtracted the second last...etc. And the reason that we've done this is because this presentation generalises more easily to $101,1001,\ldots$

So, what about $101$? Well, the rule is to take the last two digits, subtract the next pair, then add the next pair, and so forth. So, let's see an example:

the number $74914071682$ satisfies that $$\begin{align} 82-16+7-14+49-7=101,\end{align}$$ which is clearly divisible by $101$. Thus, $74914071682$ should be divisible by $101$ and indeed:  $74914071682=101\times 741723482$.

And the generalisation to $1001$ involves us taking the last three digits, then subtracting the next triple, and adding the next triple...etc. And I the pattern goes on...

What about the proof!? I hear you ask. Well, it's pretty similar to the proof for our rule regarding divisibility by $11$. So, I'll let this to you guys as a healthy challenge exercise. =)