## Monday, 23 March 2015

### 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. =)