Archive for the ‘Mathematics’ Category

I think of myself as being very fortunate for having had the opportunity to see Stephen Hawking deliver a lecture on Cosmology, many years ago in a lecture theatre at Oxford University.

I am thinking of a pair of scales. One pan holds his brilliant career in Physics, the other his physical condition. I think that he achieved a perfect balance and that this must be deemed as a part of his legacy.

It so happens that I do not agree with his negation of an afterlife, so I will wish him the very best in that.



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Do you remember that famous carol “The Twelve Days of Christmas”? You can find some historical information about it at Wikipedia. There is also a section on its financial implications.

What I find interesting about this carol, from a mathematical point of view, is the structure of the underlying collection of gifts as well as the calculation of their total number.

The carol starts with a seemingly innocent gift of a partridge in a pear tree sent on the first day of Christmas, which is 25 December. Then, however, as the days proceed through the Christmas period, the gifts received on any one day are:

  • a completely new set of gifts equal in number to the number of the day within Christmas (e.g. on the third day I get three French hens for the first time, on the fourth I get four calling birds for the first time and so on)
  • the whole lot of gifts that were sent on the preceding day

This process, where the set of gifts on any one day is linked directly to the gifts of the preceding day, is an example of a concept known as recursion. Recursion is widely used in both Mathematics and Computer Programming. Many interesting books have been written, which address this concept. If you have ever come across those beautiful images known as fractals, then I should tell you that recursion plays a great part in their production.

Let me turn to the second reason for my interest in the carol: How do I calculate the total number of gifts received from my true love?

Well, if you examine its verses, you will see that I get:

  • 1 partridge in a pear tree on each of the 12 days of Christmas, hence 1 × 12 partridges in their pear trees
  • 2 turtle-doves on each of the last 11 days, hence 2 × 11 turtle-doves
  • 3 french hens on each of the last 10 days, hence 3 × 10 french hens
  • 4 calling birds on each of the last 9 days, hence 4 × 9 calling birds
  • 5 golden rings on each of the last 8 days, hence 5 × 8 golden rings
  • 6 geese a-laying on each of the last 7 days, hence 6 × 7 geese
  • 7 swans a-swimming on each of the last 6 days, hence 7 × 6 swans
  • 8 maids a-milking on each of the last 5 days, hence 8 × 5 maids
  • 9 ladies dancing on each of the last 4 days, hence 9 × 4 ladies
  • 10 lords a-leaping on each of the last 3 days, hence 10 × 3 lords
  • 11 pipers piping on each of the last 2 days, hence 11 × 2 pipers
  • 12 drummers drumming on the final day, hence 12 × 1 drummers

Phew! That’s the whole lot… So, the total number of gifts is the sum of all the above, namely:

(1 × 12) + (2 × 11) + (3 × 10) + (4 × 9) + (5 × 8) + (6 × 7) +

(7 × 6) + (8 × 5) + (9 × 4) + (10 × 3) + (11 × 2) + (12 × 1)

Now, the individual products of the first line are all repeated once on the second line, so this sum is actually twice the value of the first line i.e.

2 × ((1 × 12) + (2 × 11) + (3 × 10) + (4 × 9) + (5 × 8) + (6 × 7))


2 × (12 + 22 + 30 + 36 + 40 + 42)

or by doing the sum in brackets

2 × 182

and finally 364.

Now, isn’t that strange? Because the total we have arrived at is precisely one less than the number of days in a year (forgetting leap-years). So, by sending me all these gifts and with the proviso that I resort to one gift per day, my true love has taken care of me for all but one of the days making up one whole year beginning on Christmas Day. These gifts will take me up to the day before next Christmas Eve. That Christmas Eve there will be no gift to use as it will be the 365th day. I should then spend that day thanking my true love and anticipating the flurry of gifts, which will follow on the next day, if she still has the means!

I cannot help wondering whether the creators of this carol designed it on purpose so that the total would come out to 364 or whether this fact completely escaped them. In any case, it is an interesting calculation.

Thank you for bearing with me. May you have a Merry Christmas and a Happy New Year!

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I have every good reason to recommend watching “The Man Who Knew Infinity”, an account of the life of the Indian mathematician Srinivasa Ramanujan.

Ramanujan was an amazing figure who had practically no formal training in Mathematics, yet made extraordinary contributions to various fields of the subject. The film centres on his years at Trinity College, Cambridge University, where he worked closely with the English mathematicians G. H. Hardy and J. E. Littlewood.

The film can easily be followed by non-mathematicians. I particularly liked the scene where Ramanujan tries to explain to his wife (and to the general audience) that his love for Mathematics comes from the tendency of mathematical patterns to appear in ways that cause surprise. Hardy also contributes to this by explaining to his butler what Ramanujan was trying to do when tackling the problem of partitions.

The film presents us with some lovely poetic images of India and I cannot forget that scene where Ramanujan’s wife looks on, as her husband sails away in the boat that will take him to a ship bound for England.

Jeremy Irons gives a brilliant performance as the cantankerous G. H. Hardy and at the end of the film quotes from Hardy’s famous and haunting book A Mathematician’s Apology. Dev Patel is also very convincing as the youthful and enthusiastic Ramanujan. The culture shock that he experienced at Cambridge is illustrated well and one can only feel sorry for him, as well for the fact that his life ended so soon.

The only departure from historical fact that I managed to pinpoint concerns the exchange between Ramanujan and Hardy concerning the number 1729. The film shows this as taking place when Hardy bids farewell to Ramanujan, as the latter sets off on his return journey to India. In fact, the exchange took place when Ramanujan was in hospital.

In closing, I must again say ‘bravo’ to the filming world for yet another good film about Mathematics and mathematicians. It has already given us ‘Agora’, ‘A Beautiful Mind’ and ‘The Imitation Game’. I should also include ‘The Theory of Everything’, as Stephen Hawking has used Mathematics so much in his explorations as a cosmologist.

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Recently, the following question was set in a GCSE maths exam:

Liz buys packets of coloured buttons.

There are 8 red buttons in each packet of red buttons.
There are 6 silver buttons in each packet of silver buttons.
There are 5 gold buttons in each packet of gold buttons.

Liz buys equal numbers of red buttons, silver buttons and gold buttons.

How many packets of each colour of buttons did Liz buy?

A certain student answered by calculating 8 × 6 × 5 = 240, concluding that this was the number of beads of each colour and then deducing that Liz bought 30 red, 40 silver and 48 gold buttons.

The student lost marks, apparently because the instructions given to the examiners stated that the answer, that had to be arrived at for the number of buttons, of each colour was 120.

Yet it is the question that is at fault, because although 120 is the least common multiple of 8, 6 and 5, the wording does not include any information pointing to the fact that it is this, which is being sought. In fact any common multiple will do, hence the easy choice of 240.

This is quite unfair on the student, but the damage most probably cannot be undone.

There are better ways of phrasing the question and I have thought of one, which in my view also serves to instruct and teach, besides stating the question for the purposes of the exam:

Liz buys packets of coloured buttons.

There are 8 red buttons in each packet of red buttons.
There are 6 silver buttons in each packet of silver buttons.
There are 5 gold buttons in each packet of gold buttons.

Liz wishes to buy the same number of buttons of each of these three colours and, in such a way, so as to obtain the least possible number of buttons.

Calculate how many packets of each colour of buttons Liz should buy, the number of buttons of each colour thus bought and the total number of buttons.

I believe this way of phrasing the question has some benefits:

  1. It illustrates a more natural way in which the problem could occur in real life (Why miss an opportunity to bring Maths a little closer to our daily lives for the sake of students?)
  2. The last sentence helps the student make a distinction between the number of packets and the number of buttons. This is important, because one quite easily could mistake packets for buttons.

And here is a solution to the problem:

Whatever the number of packets of red buttons Liz chooses, she will always end up with a multiple of 8. Similarly, she will end up with a multiple of 6 for the silver buttons and a multiple of 5 for the gold buttons.

Therefore, if Liz wants the same number of buttons of each colour, this common number must be a common multiple of 8, 6 and 5. Since she wants this number to be the least possible, it follows that it must the least common multiple (LCM) of these numbers.

To calculate it, let’s decompose these numbers into prime factors:

8 = 2³

6 = 2 × 3

5 is prime

Hence the LCM of these three will be 2³ × 3 × 5 = 24 × 5 = 120.

[Digression: If one is not happy with this calculation, then one might consider that LCM(8, 6, 5) = LCM(LCM(8, 6), 5).

For the LCM of 8 and 6, we list the multiples of these two:

for 8: 8, 16, 24, …

for 6: 6, 12, 18, 24, …

from whence it follows that LCM(8, 6) = 24. The multiples of 24 and 5, now, are:

24: 24, 48, 72, 96, 120, …

5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125…

and the result follows.]

So, Liz must buy 120 buttons of each colour. To achieve this, she will have to buy:

120 ÷ 8 = 15 packets of red buttons

120 ÷ 6 = 20 packets of silver buttons

120 ÷ 5 = 24 packets of gold buttons

The total number of buttons is 3 × 120 = 360.



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Thumbs up to the film “The Imitation Game” and to Benedict Cumberbatch for giving such a convincing portrayal of Alan Turing.

The film is interesting, well-done and easy to understand by people who have little or no knowledge of Mathematics.

One of Turing’s achievements was the concept of a test, which could test a machine’s ability to exhibit intelligent behaviour equivalent to, or indistinguishable from, that of a human. This has come to be known as the Turing Test. I was pleased to notice how subtly the film handled and paid homage to this concept in one of the last scenes. There, Turing is shown concluding his conversation with the detective and saying: “So, am I a man, a machine, or a war hero?” It was as though Turing was turning the test onto himself…

The detective replies “I cannot judge you, sir”. It goes without saying that Alan Turing’s final years and treatment by the British judicial system were tragic and disproportionate to his contributions to victory in World War II, Mathematics and modern Computing. He died at the age of 41 on 8 June 1954. Had he lived for longer, he might have borne witness to the grand advances in Computing during the 60s and later: Unix, the Mac, maybe even Windows. Surely, he would have made his mark there…

PS As a Mathematician I could not help noticing a small slip by Keira Knightley: In the scene when Turing and Joan Clarke are on the lawn solving a mathematical problem, Keira Knightley alludes to a theorem of Euler, pronouncing the “Eu” as in “yew”. It is actually pronounced “oy” as in “boy”.

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A Chance Discovery

One evening, I was on the telephone and the conversation had got rather boring. Not being able to hang up, I started aimlessly keying numbers into my desktop calculator…

At some point, I typed in 47 and repeatedly pressed the square root key. After three times, my calculator showed me a result, which made me raise my brows:


I was very surprised to see the first four digits of the golden ratio making up the result. Since I pressed the square root key three times, I must have taken the 8th root of 47, so that one can write:

\sqrt [8]{47} \approx \phi

where \phi is the usual symbol for the golden ratio.

Throughout history, the golden ratio has attracted the interest of artists and architects. The proportion has been used in many of their works and a lot has been written about this particular number.

For more information, see the Wikipedia entry.

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Hello, readers.

About a year ago, I wrote a post explaining why 2010 did not usher in a new decade.

However, according to this, 2011 is indeed the first year in a new decade. So, I wish you happiness for both!

Take care,


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