Nerd to Jock

I'm making the transition from nerd to jock. It doesn't surprise me that you don't believe me, which is why I'm presenting evidence!

A couple of years ago, I went to Microsoft Tech-Ed and got 14 t-shirts! A couple were medium-sized and so I gave them away, but I'm pretty sure I kept the rest. And added them to my collection of IBM t-shirts. That is to say, virtually all my t-shirts were related to technology companies. My mom tried her best to counteract this by buying me the occasional t-shirt that she found hilarious, but I don't think I'd go so far as to say that they made me less nerdy.

Recently, however, I've been running [I didn't end up running in this 5K due to food poisoning] some 5Ks and other benefit things that give out t-shirts. Thus, my t-shirt wardrobe is gradually transitioning from technology t-shirts to running t-shirts. Plus, I wrote recently about football, though admittedly I was on the outside of the conversation. Still, incontrovertible evidence that I am changing from being a nerd to being a jock!

To be fair, there is some evidence that this is a very slow process in its earliest stages. But only a little bit of evidence, really. And I suppose that this post kind of contradicts itself. And all but one of the fitness-related tags below are new....

Divisibility by Nine

When I was a GTA at The University of Kansas, I once taught a proof for a number being divisible by nine if and only if the sum of its digits is divisible by nine. I stumbled across the proof the other day and thought I'd go ahead and share it. It's one of my favorites, though I've since learned that it is by no means the simplest. The pdf linked to at the bottom includes multiple proofs related to divisibility rules. It uses modular arithmetic, though, and I've never had an intuitive grasp of that. This proof makes more sense to me, even if it is hard to read. Let me tell you, it was even harder to write in HTML, as I recall - so many subscripts and superscripts!

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If a number is divisible by 9, then the sum of its digits is divisible by 9

A number n with digits [most significant] d_k, d_k-1, ... , d₁, d₀ [least significant] can be rewitten as d_k*10^k + d_k-1*10^k-1 + ... + d₁*10 + d₀

We start by observing that 10^x = 10^x - 1 + 1

Note: 10^x - 1     =     9*10^x-1 + 9*10^x-2 + ... + 9*10² + 9*10 + 9     =     9*(10^x-1 + 10^x-2 + ... + 10² + 10 + 1)

Thus, our number in convoluted form is:
d_k*(9*(10^k-1 + ... + 1) + 1) + d_k-1*(9*(10^k-2 + ... + 1) + 1) + ... + d₁*(9*(1) + 1) + d₀*(1)

If we factor out everything with a 9 in it, we are left with:
9*(d_k(10^k-1 + ... + 1) + d_k-1(10^k-2 + ... + 1) + ... + d₁(1)) + [d_k + d_k-1 + ... + d₂ + d₁ + d₀]

We can see from here that if a number is divisible by 9, then a 9 must be able to be factored out of the part in []'s (a 9 is already factored from the left part. Note that if the digits are divisible by 9, then the number is divisible by 9, too. This follows since we did direct algebraic manipulation of the number without any implications (which are only one way). Thus, a number is divisible by 9 IF AND ONLY IF the sum of its digits is divisible by 9

The same proof except for the last paragraph is used to prove the same about divisibility by 3, since 9 is divisible by 3 on the left side.

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http://www.ms.uky.edu/~lee/as100fa02/divide.pdf [pdf]