## COMC problem of the week

A nice problem from the 2015 archive filed under week 9, dated 27th October 2015.

Define the function $t(n)$ on the nonnegative integers by $t(0)=t(1)=0, t(2)=1,$ and for $n>2$ let $t(n)$ be the smallest positive integer which does not divide $n$. Let $T(n)=t(t(t(n))).$ Find the value of $S$ if $S=T(1)+T(2)+T(3)+⋯+T(2006)$.

My answer is 1171. Here’s my solution.
It is easy to see that $t(n)=2$ for odd $n \ge 3$. Correspondingly, $T(n)=0$. Also $T(2)=0$ so we just need to sum over all even numbers from 4 to 2006.
Claim: $t(n) =p^k$, for some prime.
Pf of claim: Suppose $t(n)= ab , \gcd(a,b)=1, a, b > 1$. Since $a < ab \Rightarrow a \mid n$. Likewise we have $b\mid n \Rightarrow ab \mid n$ which is a contradiction.
Now if $t(n)$ is odd, $t(t(n))=2 \Rightarrow T(n)=1$. On the other hand if $t(n) = 2^k, k > 1$, $t(t(n))=3 \Rightarrow T(n)=2$.

So each term in the sum contributes either 1 or 2. Now $t(n)= 16$ is impossible as we would have $5, 7, 11, 13 \mid n \Rightarrow n \ge 5005$.
In the case where $t(n)= 8$ we have $4, 3, 5, 7 \mid n \Rightarrow n = 420(2k+1)$. Within the range of 4 to 2006, the only possible candidates are 420 and 1260.
Next for $t(n)= 4$ we have $2, 3 \mid n \Rightarrow n = 6(2k+1)$ and there are 167 choices in the range.
Finally, there are 1002 even numbers with a positive contribution to the sum, so
$S= 1002 + 167 + 2 = 1171$.

A verification using mathematica

## A Sunday Morning Geometry Problem

It’s 6.15 am on a Sunday morning, 18th October to be exact. The newspapers have not been delivered and the kids are still sleeping. Some peace and quiet for me to do some reading while sipping my morning coffee … hmmm … here’s a nice problem*.

In the diagram, the three sides of a triangle have been split into thirds, then the lines have been added. What is the area of the smaller shaded triangle as a fraction of the area of the large triangle?

This looks familiar. Now, where have I seen it before? I don’t recall an immediate solution or approach to this. Let’s try to understand the problem. I need to draw. A cursory search on the table and floor revealed a piece of paper and a purple coloured magic marker. From the way the question is asked and previous experience with problems of this sort, the proportion is an invariant. Can I draw a special triangle that helps to solve the problem immediately? An equilateral triangle? A right-angled triangle? I cannot seem to control the shape of the inner triangle. It looks like another approach is needed.

Clearly triangle BCD** has 1/3 of the area, as do the other pieces ACE and BFA. Ah … inclusion and exclusion. If I take away these three pieces I would get the shaded portion but I’m double counting. This would mean that the shaded area = sum of the areas of the 3 small triangles at the vertices. Is that right? Let me check that again. … Yes, that’s right. Now let’s label the areas.

Hmm… how to proceed? … Can’t seem to get anywhere? … I wonder what if I draw another three lines joining the vertices with the other dots G, H and I. Would I get something nice here? I should. It looks like I would get a hexagon. Ok… must focus on solving the simpler triangle problem first.

Uh-oh, I hear the door opening, the boy must be awake. I remember reading that Euler could write mathematics with kids on his lap and/or running circles around him. But he is Euler. He could even write mathematics half-blind. Then again, Euler didn’t have my kids.

Ok, time to put this problem on hold.
—————————————————
It’s half an hour to noon. The kids are happy painting away in the living room and I steal some time to return a couple of emails. That ratio has been on my mind the whole morning even as I was playing and watching cartoons with the monkeys. What is that number? Since the computer is switched on, time to call in the heavy artillery — geogebra. I plotted a general triangle. Dividing the sides into three equal parts turns out to be not trivial. An easy way out is to use the distance function and move the points around until the ratio is 1:3. Geogebra automatically calculates the areas of the two triangles and I can compute the ratio directly in the input box to get the number 0.14. Hmm … looks weird. In rational terms that would be 7/50. Can this be right? How to split the triangles into 50 parts? Anyway I distorted the triangle and as expected the ratio remains the same 0.14. Hmm … . Oh time for lunch. I’ll leave the computer on and maybe return to it later.
—————————————————
It’s 3.30 in the afternoon. The girl is napping but the boy, as usual, cannot sleep and is playing with lego. Time to work on my geometry problem again. The purple marker and the sketch from the morning is still on the table. Let me make a neater sketch which may help me spot something that I missed. … Yes! I can draw an auxiliary line and the area will be 2X. Doing this for Y and Z, I now have 4 (X+Y+Z) with 3 more regions to compute. The 7/50 puzzles me, I cannot see how the fractions will sum to 7/50.

But it looks like I have enough the solve the problem. The unknown regions at the lower left and right have area A/3 -3Z-Y and A/3 – 3X-Z respectively. Now by focusing on the correct inner triangle (i.e. AKB in Diagram 1) we have
$(X+Y+Z)+ (\frac{A}{3}-3Z-Y) +2Z = 2 ( \frac{A}{3} -3X-Z +Z)$
$\Rightarrow 7X = \frac{A}{3}.$

By symmetry (in this case algebraic symmetry in the sense of the same computations), we must have X=Y=Z. So in fact
$\frac{(X+Y+Z)}{A}=\frac{3X}{A} = \frac{1}{7}$.
So I was deceived by geogebra. The ratio was not 0.14 but 0.142857 … which was rounded to two decimal place. There is a lesson here about using computers/software and not paying attention to the assumptions and possible pitfalls.

Ok problem solved. Very nice. There is the natural generalization of whether the problem has an invariant solution for other ratios besides 1:2 but that is for another day. The girl has woken up from her nap and it’s playtime again.
—————————————————
Monday, 19th Oct, 10.30 am. Lunch with colleagues. Between the few of us, we rummaged up a pen and some useless piece of paper and I explained the problem. D quickly recognized it as a problem he has previously solved, took over the paper and furiously calculated away. E quickly worked out the relation depicted in diagram 2. While D computed on, we mused about generalisations and whether adaptations to arbitrary quadrilaterals can be solved and by and by the conversation drifted to another topic.
—————————————————
12 noon. D came into my office and presented the following generalization. Given a triangle of size 1, with one side divided into the ratio 1:t and another into the ratio 1:s, where we only require t, s to be positive. By drawing the auxiliary line from F, the area A (and hence B) can always be computed.

We have the following pair of equations:
$(t+1)A+B=\frac{1}{1+s}$
$tA + (s+1)B = \frac{t}{1+t}$
The system can be solved if and only if
$\frac{t+1}{t} \ne \frac{1}{1+s}$.
The above is clear since the left side is strictly greater than 1 while the right is strictly less than 1. Let me check by substituting t=s=2, … Yes, A= 1/21.
—————————————————
20th Oct, Tuesday morning at the office. Perhaps it would be helpful to record down the thoughts and processes that went through my head over the past two days while thinking about the problem on my own and with colleagues. Writing is an integral part of discovery because it allows one to rethink the problem, consolidate the understanding, check the solutions and possibly further generalize — what Polya calls Look Back. … In both of the solutions, the drawing of the auxiliary line seems to be critical. As an experiment, let me try to label the three unknown regions in Diagram 2 as L, M, N. Assuming the total area is 1, I now have six variables and I can form 7 linear equations, for example of the form
$X +L + Z =\frac{1}{3}$
$M + Y+ (X+Y+Z)+ N =\frac{2}{3}$
$X+Y+Z+L+M + N + (X+Y+Z) =1$.
Let’s see if this system can be solved. Oops, that last one there is just the sum of the previous two. Well I still have six equations with six variables. Instead of doing it by hand, I use matlab. … Ok, it does not work. The matrix has rank 3. … Back to writing things down. … Hey in D’s general solution, although the system is consistent would it be possible to end up with non-positive solutions? Ok, might as well try to get a closed form for A and B. … Uh-oh, time to go for a meeting. Have to put this aside again.
—————————————————
Tuesday 6.10 pm. At the classroom with the part-time Masters students. I just passed out quiz 4 and the students are working on it. Good. I have some time to continue with my calculations … The algebra is slightly messy but the end result is nice.
$A = \frac{1}{(1+t)(1+s+st)}, B= \frac{st}{(1+s)(1+s+st)}$
Quite neat. Better check again for s=t=2. Yes, it agrees. How about s=t=1? A=B=1/6. That’s right, since 3 medians of a triangle are concurrent and results in six triangles of equal area. … 20 minutes have passed, time to put this aside again and get ready to collect the quiz and continue with the lesson.
—————————————————
Wednesday 21 Oct, 7.40 am at the office. No classes today, maybe I can finish up my writing before embarking on the other boring admin duties. I’m pretty sure the closed form I obtained was right, but let us substitute it into the equation to be sure. … Yes, the solutions check out. … Oh, I still have my hexagon generalization unresolved. Let’s try. … Hey, the general formula still works!

Using the ratios t=1, s=1/2, area of the three red triangles are all equal to A=2/15.
Similarly, using the ratios t=1/2, s=2, the area of the three green triangles are all equal to sB=1/6. Hence the hexagon has area = 1-1/2-2/5 = 1/10. Just to be sure, let me use geogebra to check on a general triangle.

I’m almost ready to move on to other things, but maybe it is a good time to do a search on google. The search terms “area triangle 1/7″ look promising. … As expected the problem is well known. The problem is sometimes called Feynman’s Triangle after the physicist but it certainly dates back further. In fact, there is a general theorem called Routh’s theorem. In diagram 4, assuming the third side is partitioned into the ratio 1:r, then the inclusion-exclusion argument would give the area of the middle triangle as
$1- \frac{1}{1+t} -\frac{1}{1+s} – \frac{1}{1+r} + \frac{1}{(1+t)(1+s+st)} + \frac{1}{(1+s)(1+r+rs)}+ \frac{1}{(1+r)(1+t+tr)}$
$= \frac{(rst-1)^2}{(1+s+st)(1+r+rs)(1+t+tr)}$.
In the above, I used Maple to simplify the algebra. Wikipedia has a very good write-up on the theorem which appears in Routh’s text from 1891, together with links to the corresponding pages of the text. The Feynman connection is also mentioned and many other references including several ingenious proofs of the 1/7 result can be found by following the various links.

At this point, 21st Oct 12:19pm, the narrative caught up with the writing in real time and I think I should stop here.

*I saw the problem in Issue 1 of chalkdustmagazine.
** There were no labellings in the original diagram and they were included for clarity.

## Byss and Abyss

Saw this quote attributed to Winston Churchill. A search on google books revealed that it came from chapter III of his autobiography, My Early Life: 1874-1904.

I had a feeling once about Mathematics, that I saw it all — Depth beyond depth was revealed to me — the Byss and the Abyss. I saw, as one might see the transit of Venus — or even the Lord Mayor’s Show, a quantity passing through infinity and changing its sign from plus to minus. I saw exactly how it happened and why the tergiversation was inevitable: and how the one step involved all the others. . . . But it was after dinner and I let it go!

A little digging into the context and the book seems to be in public domain. So here goes the entire bits leading to the above paragraph.

CHAPTER III

EXAMINATIONS

IT took me three tries to pass into Sandhurst. There were five subjects, of which Mathematics, Latin and English were obligatory, and I chose in addition French and Chemistry. In this hand I held only a pair of Kings — English and Chemistry. Nothing less than three would open the jackpot. I had to find another useful card. Latin I could not learn. I had a rooted prejudice which seemed to close my mind against it. Two thousand marks were given for Latin. I might perhaps get 400! French was interesting but rather tricky, and difficult to learn in England. So there remained only Mathematics. After the first Examination was over, when one surveyed the battlefield, it was evident that the war could not be won without another army being brought into the line. Mathematics was the only resource available. I turned to them — I turned on them — in desperation. All my life from time to time I have had to get up disagreeable subjects at short notice, but I consider my triumph, moral and technical, was in learning Mathematics in six months. At the first of these three ordeals I got no more than 500 marks out of 2,500 for Mathematics. At the second I got nearly 2,000. I owe this achievement not only to my own back-to-the-wall’ resolution — for which no credit is too great; but to the very kindly interest taken in my case by a much respected Harrow master, Mr. C. H. P. Mayo. He convinced me that Mathematics was not a hopeless bog of nonsense, and that there were meanings and rhythms behind the comical hieroglyphics; and that I was not incapable of catching glimpses of some of these.

Of course what I call Mathematics is only what the Civil Service Commissioners expected you to know to pass a very rudimentary examination. I suppose that to those who enjoy this peculiar gift, Senior Wranglers and the like, the waters in which I swam must seem only a duck-puddle compared to the Atlantic Ocean. Nevertheless, when I plunged in, I was soon out of my depth. When I look back upon those care-laden months, their prominent features rise from the abyss of memory. Of course I had progressed far beyond Vulgar Fractions and the Decimal System. We were arrived in an Alice-in-Wonderland’ world, at the portals of which stood A Quadratic Equation.’ This with a strange grimace pointed the way to the Theory of Indices, which again handed on the intruder to the full rigours of the Binomial Theorem. Further dim chambers lighted by sullen, sulphurous fires were reputed to contain a dragon called the Differential Calculus’. But this monster was beyond the bounds appointed by the Civil Service Commissioners who regulated this stage of Pilgrim’s heavy journey. We turned aside, not indeed to the uplands of the Delectable Mountains, but into a strange corridor of things like anagrams and acrostics called Sines, Cosines and Tangents. Apparently they were very important, especially when multiplied by each other, or by themselves! They had also this merit — you could learn many of their evolutions off by heart. There was a question in my third and last Examination about these Cosines and Tangents in a highly square-rooted condition which must have been decisive upon the whole of my after life. It was a problem. But luckily I had seen its ugly face only a few days before and recognised it at first sight.

I have never met any of these creatures since. With my third and successful examination they passed away like the phantasmagoria of a fevered dream. I am assured that they are most helpful in engineering, astronomy and things like that. It is very important to build bridges and canals and to comprehend all the stresses and potentialities of matter, to say nothing of counting all the stars and even universes and measuring how far off they are, and foretelling eclipses, the arrival of comets and such like. I am very glad there are quite a number of people born with a gift and a liking for all of this ; like great chess-players who play sixteen games at once blindfold and die quite soon of epilepsy. Serve them right! I hope the Mathematicians, however, are well rewarded. I promise never to blackleg their profession nor take the bread out of their mouths.

I had a feeling once about Mathematics, that I saw it all — Depth beyond depth was revealed to me — the Byss and the Abyss. I saw, as one might see the transit of Venus or even the Lord Mayor’s Show, a quantity passing through infinity and changing its sign from plus to minus. I saw exactly how it happened and why the tergiversation was inevitable: and how the one step involved all the others. It was like politics. But it was after dinner and. I let it go!

The practical point is that if this aged, weary-souled Civil Service Commissioner had not asked this particular question about these Cosines or Tangents in their squared or even cubed condition, which I happened to have learned scarcely a week before, not one of the subsequent chapters of this book would ever have been written. I might have gone into the Church and preached orthodox sermons in a spirit of audacious contradiction to the age. I might have gone into the City and made a fortune. I might have resorted to the Colonies, or ‘Dominions’ as they are now called, in the hopes of pleasing, or at least placating them; and thus had, a la Lindsay Gordon or Cecil Rhodes, a lurid career. I might even have gravitated to the Bar, and persons might have been hanged through my defence who now
nurse their guilty secrets with complacency. Anyhow the whole of my life would have been altered, and that I suppose would have altered a great many other lives, which in their turn, and so on. …

But here we seem to be getting back to mathematics, which I quitted for ever in the year 1894. …

## SEND SOME MORE MONEY

It’s been 9 years since I blogged on SEND MORE MONEY. I got back into this because I found out that the author of SEND MORE MONEY is Henry Dudeney and the puzzle apparently appeared in Strand Magazine vol. 68 (July 1924), pp. 97 and 214. Dudeney calls this Verbal Arithmetic. The Wikipedia entry on Dudeney casts doubts that he was the inventor and cites this 1864 example. However, the cited example does not have the feature that the letters make up sensible words. In fact, the words often make up sensible sentences.

The SEND+SOME+MORE=MONEY is my own invention but I don’t claim it to be original since it is not much of a stretch from Dudeney’s problem. Unfortunately, teaching duties beckon and I cannot spend time cracking it. Mathematica comes to the rescue and I already found four distinct solutions by simply assuming M=2.

## Prime (Car) Number

It has always been a regret of epsilon proportion that my car license plate number doesn’t have particular significance. For example it is is divisible by 7 and hence not a prime, unlike the fact that my apartment number and my level number are both primes. In fact, both are one of half some twin prime pairs. My block number is perfect.

Well, I will be getting a new car soon and when asked whether I wished to retain the current license plate number, the voice within me shouted no! (It also saves me a hundred bucks from the processing fees.) Mathematica tells me there are 1229 primes less than 9999, which means I have a 12% chance — fingers crossed.

Incidentally, 1229 is itself a prime. Furthermore, the possibility of getting a prime (car) number should be higher than 12%. The local population being predominantly chinese likes special numbers like 8888, or 9999. So some favourite numbers are reserved and available if one is willing to pay more. My guess is that there are more composites than primes in this reserve list.

Posted in Fun Stuff, Number Theory | 1 Comment

## Fermat

was born on 17th August 1601. Incidentally
$1601 = 1^2 + 40^2$
is a prime.

## baby rudin

Define a real valued function on the reals by $f(x):=\begin{cases} 0, & x \not\in \mathbb{Q} \\ \frac{1}{q}, & x = \frac{p}{q}, \gcd(p,q)=1, q >0 \end{cases}$.
Where is this function continuous?

The problem appeared in Hyman Bass’s article in the June 2015 issue of the notices of AMS. I casually dismissed it as another standard problem, having “seen it somewhere before” but it caught the eye of one colleague and another colleague, an analyst, said it was non-routine. Given the interest, I thought about where I could have seen it and my first instinct was to check baby Rudin. There it was in chapter 4, problem 18. Although I have to admit, if I had seen it before, it was probably not from Rudin.

Fact check: Bass mentioned he learnt about real numbers in 1951 at Princeton while Rudin wrote his textbook when he teaching at MIT in the early 1950s. The first edition appeared in 1953. So the source might have pre-dated Rudin.

## two thousand years of combinatorics

by Don Knuth is the opening chapter to Combinatorics:Ancient & Modern, which according to its preface is perhaps the first book-length survey of the history of combinatorics. Knuth’s chapter is actually taken from his Art of Computer Programming Volume 4 and contains the following two snippets.

John Wallis was credited for explicitly considering the null case of combinations. This quote is from his Discourse of Combinations (1685):

It is manifest, that if we would take none, that is, if we would leave all; there can be but one case thereof, what ever be the number of things exposed.

On integer partitions, Knuth wrote:

Mersenne listed the partitions of 9 into any number of parts on page 130 of his 1636 Traitez de la Voix et des Chants (Treatise on the Voice and Singing). For each partition $9= a_1 +a_2 + \cdots a_k$ he also computed the multinomial coefficient $9!/(a_1!a_2! \cdots a_k!)$; as we have seen earlier, he was interested in couting various melodies, and he knew (for example) that there are $9!/3!3!3! = 1680$ melodies on the nine notes {a, a, a, b, b, b, c, c, c}. But he failed to mention the cases 8+1 and 3+2+1+1+1+1, probably because he had not listed the possibilities in any systematic way.

## An identity

The following problem is apparently a bonus question for 13 year olds at a local girls school: Evaluate the sum
$\displaystyle \frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} + \cdots + \frac{1}{99} – \frac{1}{100}}{\frac{1}{1+101}+ \frac{1}{2+102} + \cdots + \frac{1}{50+150}}.$
It’s not that easy if you ask me. I had to work out the following identity first before I managed to solve it.
$\displaystyle \sum_{k=1}^N \frac{1}{2k-1} – \frac{1}{2k} = \sum_{k=1}^N \frac{1}{k+N}.$
The identity can be proved via induction.

## 3.14.15 and Einstein

14 March 2015, was supposed to be the Pi day of the century — for obvious reasons. 14 March is also Einstein’s birthday, and 2015 interestingly marked 100 years of the theory of relativity. I liked this article written by Jeff Edelstein that described how Einstein tutored a 12 year old girl in maths. As Edelstein wrote, this could be a hoax, but personally for me, some stories (or myths) are worth retelling. There are two lovely quotes in the article, both recollections of the girl being tutored.

He’d say we’re not going to bother with the homework problem. First he’d give me his own math problem, help me work it out, and then we’d go to the homework. He’d simply tell me the things we were doing, showed me the best ways to do it. And he would always tell me it’s not the answer that counts, but how you solve the problem.

But he was just so nice. Even when I told him I hated math. He said, ‘you shouldn’t hate math, math is the center of the universe, and anyone who knows math knows everything.’

Two more titbits about the great man. My paper with Hirschhorn appeared this year, and thanks to Mike — M.V. Subbarao — Ernst Straus, I now have an Einstein number of 4. And an Erdös number of 3. Hurray!
Finally my favourite Einstein story. Einstein and wife (apparently in 1931) toured Hubble’s lab and viewed his telescope. When told that the impressive instruments were what Hubble used to study the shape of the universe, she replied:

Well, my husband does that on the back of an envelope.

As Sheldon Cooper would say:”Bazinga!”

## Under Promise and Over Deliver

Mr Lee Kuan Yew, the man responsible for making modern Singapore what it is today, passed away on 23 March 2015, aged 91. Incidentally it was also Emmy Noether’s birthday. I have wanted to start off my class by showing the google doodle but the lesson plan had to change in view of the more sombre and relevant news. The mathematical exploits and Emmy and Sophie will have to wait for another day.

Throughout this whole week of national mourning, much have been wrote about his sagely advice. In particular, the current minister of defence said that Mr Lee often reminded the younger ministers to “under promise and over deliver.” Something worth remembering in this blow-your-own-trumpet kind of world. There is also much mentioned from down under when he harshly commented in 1980 that if Australia did not shape up, it would become the “poor white trash of Asia.” Talk about not mincing his words. Well, the positive side is Australia as a nation — and the former PM acknowledged it on record — took the advice seriously, reformed and emerged after a quarter of a century to be a strong economic force.

## Kobon Triangles

Students occasionally have great ideas. Was discussing a problem that originated from some students but was not very well posed. We managed to reformulate it as the maximum number of triangles that can be formed with n lines. This turns out to be well known and already discussed by Gardner who stated that the problem came from Kobon Fujimura. A link to a MAA column by Ed Pegg Jr as well as the OEIS entry. The problem is incidentally still not completely solved.

## Translate

The new book on my desk eta products and theta series identities has the following quote in the preface.

In der Theorie der Thetafunctionen ist es leicht, eine beliebig grosse
Menge von Relationen aufzustellen, aber die Schwierigkeit beginnt da,
wo es sich darum handelt, aus diesem Labyrinth von Formeln einen
Ausweg zu finden. Die Besch¨aftigung mit jenen Formelmassen scheint
auf die mathematische Phantasie eine verdorrende Wirkung auszu¨uben
– G. Frobenius, 1893

The quote was rendered as:
In the theory of Thetafunctionen it is easy to an arbitrarily large Establish set of relations, but the difficulty starts here where it is a question of this labyrinth of formulas a
To find a way out. The preoccupation with those formula masses seems auszuuben to the mathematical imagination a searing effect.

Perhaps, I can feed the same passage in a years’ time to see whether the translation has improved.

I was trying to solve an olympiad type problem involving a nested radical of the form
$\sqrt{a+b\sqrt{r}}.$
I had managed to discover that $\sqrt{a^2- b^2r}$ is an integer but it turned out the trick is to rewrite $\sqrt{a+b\sqrt{r}} = c + d\sqrt{r}.$

Of course, one naturally asks if this is a specific incident or is there a general theory. This lead to digging up an article that I painstakingly photocopied from the library from back when photocopying was the norm. The article in question is by Susan Landau from 1994 in the Math. Intelligencer titled “How to tangle with a nested radical.”

A simplified version of Theorem 1 is this:
Let $k$ be a field extension of the rational numbers and $a, b, r \in k$ but $\sqrt{r} \notin k$. Then
$\sqrt{a^2- b^2r} \in k$ is equivalent to $\sqrt{a+b\sqrt{r}} \in k(\sqrt{s}, \sqrt{r})$ for some $0\ne s \in k$.

For example, one may check that
$\sqrt{5+2\sqrt{6}} =\sqrt{2}+\sqrt{3}$.

## Millenium Bookball

is a pretty cool mathematical sculpture by George Hart. Gone with the Wind, Charlie and the Chocolate Factory, The Cat in the Hat, Green Eggs and Ham. What is not to like? Speaking of which, I still cannot believe that it was only because of my son who is now four that I read Dr Seuss for the first time in my life. Growing up with parents who do not speak English, I only started reading English books in Primary School. I still remember the joy when I did well in school examinations and the prize was to go to the school hall and select one free book for keeps.

## On practice

I’m sure I am interpreting this in a context different from Vygotsky who was quoting Lenin

Man’s practice, repeated a billion times anchors the figure of logic in his consciousness.

## Math videos

“Speaker, I’d like to talk about twin prime numbers …” goes McNerney in the US Congress. This took place on 11 Feb 2014. More details may be found here. There is still hope in politics afterall. I would love to see the day when mathematics is discussed in our parliament.

I’ve longed known about the discussion on the Monty Hall problem in the movie 21. But it was preceded by a short take on Newton’s or Newton-Raphson method. See the clip here at Mathematics in Movies site.

## Prime Mystery

A nice puzzle from Aziz Inan on plus.maths.org

The number N represents the first 6 digits of a special number. N consists of three prime numbers put side by side. These three prime numbers come xth, yth and zth on the list of primes, where x, y and z are themselves three consecutive primes (for example, x, y and z could be 3, 5 and 7, in which case we’d be looking at the third, fifth and seventh prime numbers). In addition, if N is split in the middle into two separate numbers, the prime factors of the left part of N add up to its right part. What is N? And what is the special numbers whose first 6 digits it forms?

## 10 surprising things about our brains

an article from huffington. Worth the read.