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## A fun identity

So I’ve been working on one of my papers where we need to compute some numbers. They end up being determined by a cubic equation.

However, one often finds surprising identities when Mathematica spits out a bunch of numbers expressed in algebraic form. Here is one of them:

$\frac{1}{42} \sqrt[3]{\frac{1}{2} \left(90720 \sqrt{7446}-2859138\right)}-\frac{7893}{7\ 2^{2/3} \sqrt[3]{2859138+90720 \sqrt{7446}}}=0$

PS: I use italics in the word surprising above only because if one does not know the origin of these identities they might seem surprising.

## Random mathematics factoid.

Well, sometimes I get five minutes off to goof around and I find silly facts, which as far as I can tell serve little to no purpose. Hence the name factoid is sometimes applied to them. The random novelty of the day is that the number

$1111111111111111111$

is prime. Moreover, it is a factor of $1010101010101010101010101010101010101$. The only other prime factor is almost as pretty.

$909090909090909091$

Beware  of $909090909090909090909090909091$ which surprise, surprise, it  also happens to be prime. And so seems to be

$11111111111111111111111$

If someone knows anything more about these primes, please let it be known. I found them by accident, but they seem to have been discovered before my time, so I can not name them after myself, nor get a patent for them.

Today’s puzzle is really simple. It is a single number, and don’t worry about the formating: it is not essential.

Your job, if you decide to take it, is to figure out how this number was chosen.

3983166922118810678205990336564718434224120605664143701183608

7681419052507877828777197836786790614849623650815370930359235

0129704516432578314074470098769495753238915840085811644293357

6455898753759925348456032608650278150327110180834816760596303

09728652360528042842943610644525298135991037793081877485511245

2474332249873412259136132368731917415053983296042291396691439150

43327960030646869030670768248956738775708735485960504256334453

736488021205358726905569574806264667486581193984051009097136483

506525271749484080035733863870881763567000457650973784539927240

2043442156827044195038802613229625783142492845706197046207004041

55134209534719408986385965928333597344225594042950352896

In case you ever wondered what the most prestigious mathematical competition for high school students looks like, you get to solve 6 problems, in two sets of three problems. For each set, you get 4.5 hours of uninterrupted work, with all the scratch paper you might need, you bring your standard geometry drawing tools (ruler and compass) and you go.

For the list of problems, the IMO organization has a list of the official problems. Some are quite entertaining to do. This year, after being piqued a bit, I worked problem one in about half an hour to 45 minutes (I still got it). I used to compete in these types of competitions when I was in high school.

In any case, the problems are elementary: they just use algebra, trigonometry, basic plane geometry and elementary number theory. Here is problem one:

Let n be a positive integer and let

$a_1 , . . . , a_k (k \geq 2)$

be distinct integers in the set {1, . . . , n} such that n divides  $a_i(a_{i+1}-1)$ for i = 1, . . . , k − 1. Prove that n does not divide $a_k(a_1-1)$.

If you ever have tried to design these problems, you will notice that the conditions are somewhat contrived. n appears in two places: both in the size of the original set, and in the division condition. Could one do without the first?

The answer is no. If all of the $a_k$ are multiples of n, then the problem does not follow. But having all of them small must be important. The next thing one needs to do is understand the relation of divisibility better so that one can solve the problem. I’ll let you figure the rest out…

## Probability of seeing a stage, part II

In the previous post in this series, I described a problem dealing with a lattice with an origin, and asked what is the probability that one can see the origin from a random place in the lattice. In this post I’ll give you my *new* proof of this result, with the understanding that someone else might have done this same proof before me (I’m not aware of such person so I will claim discovery).

As Carl Brannen pointed out in the comments, this is the same as the probability that when one picks two random integers they are relatively prime. I’ll leave that to you as a proof. This is a famous result in mathematics: it is given by

$\zeta(2)^{-1} = \frac{6}{\pi^2}$

It is also sometimes stated as “Most fractions are reduced”, seeing as fractions involve the ratio of two integers.

The first time I heard about this result I was about 15 years old. It was told to me without proof and I thought it must be really hard to proof, because how does one pull all those factors of $\pi$ by counting?

Here below is an illustration of the situation.

A region near the origin.

I’ve graphed a region of the plane near the origin. If you don’t pay too much attention to the details, you will notice that the figure is rather uniform in color. We have to show that this uniformity of color persists to a large enough (infinite) size.