## June 10, 2009

### Final Exams Again

#### Posted by John Baez

I’m busy grading final exams for my undergraduate number theory class. The class went quite well — perhaps because instead of proving quadratic reciprocity, I spent time teaching them about arithmetic functions, Dirichlet convolution, Möbius inversion and the like… topics which lead to lots of fun puzzles and computations.

Nonetheless, grading finals is always mind-numbing and dispiriting. I’m sure you’ve seen it — perfectly intelligent people grading finals, trading the most mean-spirited and witless of witticisms just to keep from going insane.

In that spirit, let me report three mildly amusing things I’ve seen so far. Don’t get your hopes up — they’re not nearly as funny as the proof of the infinitude of primes that I described last time I taught this class.

Indeed, I’m sure some of you have seen funnier final exams this year. If so, tell us about ‘em!

One student said that the Fundamental Theorem of Arithmetic stated that any counting number could be uniquely factored into a product of powers of ‘distant’ primes.

Okay, not that funny.

Another question said: “Exactly one of these number is prime:

$77577$ $77777$ $77977$

Which one is it?” And one student answered: “$77977$, because the sum of the digits is prime”.

That’s a bit more funny after you solve the problem yourself.

Finally, a quite good student got tripped up on this question: what is $1000!$ mod $1000$?

Of course the answer is $0$.

And of course the dumb way to get tripped up was to attempt to use Wilson’s Theorem, which says that $(p-1)! = -1$mod $p$ when $p$ is prime. That’s for people who try to solve any problem by grasping for the nearest available theorem, whether it applies or not.

But here’s a more interesting way to get tripped up. $1000 = 0$ mod $1000$, so $1000! = 0! = 1$ mod $1000$.

Okay, back to grading. Sorry for the interruption.

Posted at June 10, 2009 7:47 PM UTC

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### Re: Final Exams Again

Not a funny final exam answer, but related to factorials and mod: if we could compute $(\sqrt{n})! \mod n$ quickly, then we could factor large numbers quickly.

Posted by: Mike Stay on June 10, 2009 8:19 PM | Permalink | Reply to this

### Re: Final Exams Again

What do you mean by that, Mike? To calculate $n!$ mod $n^2$, to calculate $\sqrt{n}!$ mod $n$ when $n$ happens to be a perfect square but we don't know that (but that's quick to discover, isn't it?), or to extend factorial to fractional numbers with the Gamma function and calculate $\sqrt{n}!$ in $(\mathbb{R},+)/n$, or what?

Posted by: Toby Bartels on June 10, 2009 8:30 PM | Permalink | Reply to this

### Re: Final Exams Again

I think he means (floor of square root) factorial, with mod as a precursor to gcd, which *can* be computed quickly.

Posted by: some guy on the street on June 10, 2009 9:40 PM | Permalink | Reply to this

### Re: Final Exams Again

Yes, that’s right. RSA numbers, in particular, are a product of two large primes, so floor(sqrt(n))! will have a factor in common with only one of the two. Then GCD picks out that common factor.

Of course, there’s no reason to expect floor(sqrt(n))! to be easy to compute.

Posted by: Mike Stay on June 10, 2009 10:48 PM | Permalink | Reply to this

### Re: Final Exams Again

Solidarity certain helps, so thanks for helping to stop me from going insane. Alas I fear it is too late, as I think I went yesterday.

Posted by: Eugenia Cheng on June 11, 2009 3:40 PM | Permalink | Reply to this

### Re: Final Exams Again

Luckily this form of insanity is almost always temporary. You sound perfectly normal now.

If anyone has ever gone permanently nuts from grading, I’d be very interested to hear about it.

Posted by: John Baez on June 12, 2009 2:32 PM | Permalink | Reply to this

### Re: Final Exams Again

I assure you, she’s not normal at all.

Posted by: Tom Leinster on June 12, 2009 3:47 PM | Permalink | Reply to this

### Re: Final Exams Again

Now, or normally?

Posted by: John Baez on June 12, 2009 3:51 PM | Permalink | Reply to this

### Re: Final Exams Again

I’m pretty sure I’m correct in saying that even among mathematicians, n-category theorists are rarely considered normal.

Posted by: Dan Piponi on June 13, 2009 1:38 AM | Permalink | Reply to this

### Re: Final Exams Again

John wrote:

If anyone has ever gone permanently nuts from grading, I’d be very interested to hear about it.

Here’s the closest reference I’ve found so far:

“University faculty are not immune from mental illness. In fact, Kilburg (1986) argued that the stresses of an academic position might actually increase the risk of mental illness among academics. No matter what, we might assume that roughly one-quarter or one-fifth of university faculty suffer from a mental illness at any one time, and that perhaps 5% or 10% suffer from an illness severe enough that it impacts their ability to complete basic occupational duties.”

Posted by: John Baez on June 12, 2009 3:58 PM | Permalink | Reply to this

### Re: Final Exams Again

Posted by: Tom Leinster on June 12, 2009 4:57 PM | Permalink | Reply to this

### Re: Final Exams Again

By the time my birthday had arrived, yesterday, life was looking up. I’d finished grading those exams, and Lisa and I had hopped aboard a plane to Paris. At 11 pm the following evening we were met at the apartment we’re renting by a friend of the owner, who kindly served us wine, bread, cheese, strawberries and melons. The place is much larger than advertised. It’s a penthouse at the top of a 7-story building, with a 360-degree view of the city. It’s ringed by a gardened patio, with a bamboo screen that provides privacy. We can eat outdoors, and even almost bathe outdoors, because the bathroom has large glass doors to the patio, and a skylight. In short, it’s so ridiculously luxurious that I keep thinking it must be the first scene of a movie where black-masked ninja terrorists rappel up the walls of the building, break in, kidnap us, and torture me for the rest of my summer vacation. But, it hasn’t happened yet. So, my birthday was very nice.

Posted by: John Baez on June 13, 2009 8:35 AM | Permalink | Reply to this

### glorious hypercrystalline shadows; Re: Final Exams Again

Happy Birthday!

This is a pretty present for you!

arXiv:0906.2109 [ps, pdf, other]
Title: Quaternionic Representation of Snub 24-Cell and its Dual Polytope Derived From E_8 Root System
Authors: Mehmet Koca, Mudhahir Al-Ajmi, Nazife Ozdes Koca

I am doing no more grading until the Fall. Today is my last day teaching at Lincoln High School (Algebra, Geometry, AP Statistics, AP Calculus).

California is de facto bankrupt. There are 250,000 students who expected summer school but won’t have one. I revert to finishing coauthored papers while spending the savings that I don’t have in an economy that doesn’t work. Chaos within chaos within chaos.

And yet we glimpse, projected onto the walls of our finite-dimensional cave, the glorious hypercrystalline shadows of a transfinite-dimensional meta-meta-cosmos…

Posted by: Jonathan Vos Post on June 12, 2009 7:46 PM | Permalink | Reply to this

### Could I see that in 3-D, color, rotating? Re: glorious hypercrystalline shadows; Re: Final Exams Again

arXiv:0906.2117 [ps, pdf, other]
Title: Grand Antiprism and Quaternions
Authors: Mehmet Koca, Mudhahir Al-Ajmi, Nazife Ozdes Koca
Subjects: Mathematical Physics (math-ph)

Vertices of the 4-dimensional semi-regular polytope, the grand antiprism and its symmetry group of order 400 are represented in terms of quaternions with unit norm. It follows from the icosian representation of the E_{8} root system which decomposes into two copies of the root system of H_{4}. The symmetry of the grand antiprism is a maximal subgroup of the Coxeter group W(H_{4}). It is the group Aut(H_{2} plus H’_{2}) which is constructed in terms of 20 quaternionic roots of the Coxeter diagram H_{2} plus H’_{2}. The root system of H_{4} represented by the binary icosahedral group \textit{I}of order 120, constitutes the regular 4D polytope 600-cell. When its 20 quaternionic vertices corresponding to the roots of the diagram H_{2} plus H’_{2} are removed from the vertices of the 600-cell the remaining 100 quaternions constitute the vertices of the grand antiprism. We give a detailed analysis of the construction of the cells of thegrand antiprism in terms of quaternions. The dual polytope of the grand antiprism has been also constructed.

Posted by: Jonathan Vos Post on June 12, 2009 8:20 PM | Permalink | Reply to this

### Re: glorious hypercrystalline shadows; Re: Final Exams Again

Thanks for the birthday presents, Jonathan. Yes, it’s strange and sad how we manage to make such a misery of our short stay in this wonderful universe — even California, such a wealthy place, has let itself go broke. I’m doing embarrassingly well myself right now, but that too shall pass… so the more permanent patterns of nature are somehow reassuring to contemplate.

Posted by: John Baez on June 13, 2009 8:50 AM | Permalink | Reply to this

### A Tribute to Isaac Newton

Totally unrelated to anything you are talking about EXCEPT maybe after finals, you might want to enjoy something silly.

#163!! “You’re All Right, Sir Isaac!”

I love this guy :)

Posted by: Eric on June 13, 2009 5:11 PM | Permalink | Reply to this

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