Greenwich Maths Time

Greenwich Maths Challenge 6

2011-05-29T19:00:00.000+01:00

Here is the sixth Greenwich Maths Challenge. As usual there will be a small prize for the first correct solution emailed to A.Mann@gre.ac.uk by a Greenwich undergraduate.

If you specify the value of y at n different values of x there is a unique polynomial y=p(x) of degree n-1 which takes these values. But suppose I have a polynomial p(x) of unspecified degree, whose coefficients are all non-negative integers. If you give me a value of x I will tell you the value of p(x). What is the minimum number of evaluations you need to make, to be able to identify all the coefficients of my polynomial? To win the prize you must justify your answer.

Greenwich Maths Talent 2011

2011-05-20T15:34:00.002+01:00

a graduation showcase supported by the Institute of Mathematics and its Applications (IMA).

King William Court 315, University of Greenwich, Friday 27 May, 6 - 7:30pm

Cheese, wine and soft drinks will be served.

This is an opportuty to celebrate the success of graduating maths students and to find out about the exciting projects they have worked on during the final year.

The showcase will be opened by Professor Tom Barnes, Deputy Vice-Chancellor.

The twitter hashtag is #GMT2011. There is also an event website.

All welcome - if you are coming, please tell Noel-Ann Bradshaw (N.Bradshaw@gre.ac.uk) for catering purposes.

Two new books (and three old ones)

2011-04-27T15:18:00.005+01:00

Biology is a big application area of mathematics at the moment, and here are two new books which I am looking forward to reading this summer:

Stewart says that

"Mathematical theory and practice have always gone hand in hand, from the time primitive humans scratched marks on bones to record the phases of the Moon to the current search for the Higgs boson using the Large Hadron Collider. Isaac Newton's calculus informed us about the heavens, and over the past three centuries its successors have opened up the whole of mathematical physics: heat, light, sound, fluid mechanics, and later relativity and quantum theory. Mathematical thinking has become the central paradigm of the physical sciences.

"Until very recently, the life sciences were different. There, mathematics was at best a servant. It was used to perform routine calculations and to test the significance of statistical patterns in data. It didn't contribute much conceptual insight or understanding. Most of the time, it might as well not have existed.

"Today, this picture is changing. Modern discoveries in biology have opened up a host of important questions, and many of them are unlikely to be answered without significant mathematical input, The variety of matheamtical ideas now being used in the life sciences is enormous, and the demands of biology are stimulating the creation of entirely new mathematics, specifically aimed at living processes. Today's mathematicians and biologists are working together on some of the most difficult scientific problems that the human race has ever tackled - including the nature and origin of life itself.

"Biology will be the great mathematical frontier of the twenty-first century."

Curiously enough, three of the most exciting maths books I have come across, and which have influenced and inspired my teaching, as students will have noticed, also relate to the mathematics of life.

Big Noise

2011-04-27T08:12:00.002+01:00

No, not one of our first year lectures, but the Bexley Big Noise STEM Careers Fair. This was an event for year 9 and 10 students in Bexley to promote careers in science, engineering, technology and maths. Kevin and I went with two of our students (thanks Ameli and James) , and we had a wonderful time, demonstrating code-breaking, mathematical modelling and actuarial mathematics, to lively and enthusiastic mathematicians of the future.

Young Operational Research Society Conference

2011-04-09T17:45:00.003+01:00

Last Monday I travelled to Nottingham to the Young Operational Research Society Conference.

My main reason for attending was to present a paper on my research concerning a new Multi-Objective Evolutionary Algorithm for Portfolio Optimisation. This took place on the Tuesday Morning and was well received.

However of even greater benefit was the information I picked up from those working in OR about the skills they needed graduates to have. Nowadays it is important that, as well as having good technical skills, graduates also need good interpersonal skills and business awareness. Much of this can be acquired through the group projects we set at Greenwich and our focus, in the second year, on employability.

Many of the people I spoke to were involved in building simulation models. Something I will emphasise when I teach this next year. I took part in a demonstration of a software called SIMUL8 (we will be using this in OR next year). We had to simulate a nightclub and despite the fact that I have never been to one and do not know what Indie music is our team won!

More information about the OR Society can be obtained from: http://www.orsoc.org.uk/orshop/(f3aerozzof4l3o45irqjymax)/orhomepage2.aspx

MathSoc's Pi Day Party

2011-03-18T15:25:00.007Z

On 14th March (3.14) the maths staff and students celebrated pi day. We had a short talk on the history of pi which included learning some fun mnemonics like 'How I Need A Drink Alcoholic Of Course' (3.1415926) and having a competition to see who could memorise the most digits of pi. This was won by Nic Mortimer (3rd year - BSc Decision Science) who memorised 101. Karen Richardson from the library came a close second.

We also listened to a pi song - available here: http://bit.ly/49CFj

This was followed by some pi pie and pi cake made by Noel-Ann Bradshaw, Ameli Gottstein (3rd year - BSc Maths) and James Howe (3rd year - BSc Maths). Pictures by Michael Dullaway (3rd year - BSc Maths).

Greenwich Christmas Quiz

2011-03-18T10:34:00.005Z

Congratulations to Ameli Gottstein who won this year's Greenwich Christmas Maths Challenge. This completes a clean sweep for Ameli who has won in each of her three years as a Greenwich undergraduate!

Here are solutions to the questions.

Q1. Identify the mathematicians whose names are given below as anagrams. Accents and punctuation marks such as hyphens are omitted, and spellings are taken from the MacTutor History of Mathematics website http://www-groups.dcs.st-and.ac.uk/~history/

a) ENRW
b) EIKLNV
c) EEKLPR
d) AAIIJNX
e) AHILMNOT
f) BEILLNORU
g) ACDEEHIMRS
h) AEMNNNNOUV
i) GKLMOOOORV
j) AAAEKKLOSVVYQ

Answers: (a) Wren, (b) Kelvin, (c) Kepler, (d) Jia Xian, (e) Hamilton, (f) Bernoulli, (g) Archimedes, (h) Von Neumann, (i) Kolmogorov, (j) Kovalevskaya

For information on these mathematicians see the MacTutor website.

2. A ship is at anchor in a harbour. A spectator sees a ladder dangling from her deck. The bottom four rungs of the ladder are submerged, each rung is two centimetres wide and the rungs are eleven centimetres apart. The tide is rising by eighteen centimetres per hour. After two hours, how many rungs will be submerged?

Answer: still only four rungs. The ship and ladder rise with the tide!

Q3. For Spanish, Russian or Hebrew, it’s 1. For German, it’s 7. For French, it’s 14. What is it for English?

Answer: The answer is 7. It’s the first integer whose name in the language has more than one syllable.

Q4. How many people is "Twice two pairs of twins"?

Answer: Eight

Q5. Each integer from 1 to 10^10 (ten billion) is written out in full (for example 211 would be "two hundred eleven" and 1042 would be "one thousand forty two" - the word "and" is not used), and the numbers are then listed in alphabetical order (ignoring spaces and hyphens). What is the first odd number to appear in the list?

The book answer is 8,018, 018,885 (eight billion, eighteen million, eighteen thousand, eight hundred eighty five).

Q6. The Ruritanian National Library contains more books than any single book on its shelves contains words. No two of its books contain the same number of words. Can you say how many words are contained in one of its books?

Answer: At least one book contains zero words. Use the Pigeon-hole Principle!

Q7. A maths lecturer has a collection of eighteenth-century mathematical pornography in two bookcases in a room 9 by 12 feet (a foot is an archaic unit of measure about 30cm long). Bookcase AB is 8.5 feet long and bookcase CD is 4.5 feet long. The bookcases are positioned centrally on each wall and are one inch from the wall, as shown in the diagram.

Some students are going to visit the lecturer and she wishes to protect the students from the books and vice versa. The lecturer decides that the best way to do this is to turn around the two bookcases so that each is in its starting position but with the ends reversed so that the books are all facing the wall. The bookcases are so heavy that the only way to move them is to keep one end on the floor as a pivot while the other end is swung in a circular arc. The bookcases are so narrow that for the purpose of this problem we can consider them as straight line segments. What is the minimum number of swings required to reverse the two bookcases?

(For more about eighteenth-century mathematical pornography, see Patricia Fara's recent Gresham College lecture.)

Answer: Eight swings are enough. For example, (1) Swing end B clockwise 90 degrees; (2) swing A clockwise 30 degrees; (3) swing B anti-clockwise 60 degrees; (4) swing A clockwise 30 degrees; (5) swing B clockwise 90 degrees; (6) swing C clockwise 60 degrees; (7) Swing D anti-ckockwise 300 degrees; (8) swing C clockwise 60 degrees.

Q8. The number 2 to the power 29 has nine digits, all different: which digit is missing? (Calculator not required.)

The answer is 4.

The way to do this without a calculator is to use the fact that, if you divide a number by 9, the remainder you get is the same as the remainder when you divide the sum of its digits by 9. (This is the basis of lots of maths tricks you can impress your friends with.) For example, 123456 divided by 9 has remainder 3; so does 1+2+3+4+5+6 = 21.

Consider the powers of 2 divided by 9 (draw your own table). It’s easy to see that the pattern repeats every six powers. (You may know Euler’s Theorem relatign to this.) So 229 will have the same remainder divided by 9 as 25 does: this is, 5, which must be the sum of the nine digits. Now the sum of all the ten possible digits 1+2+3+…+9+0 is 45, so the sum of eight of them is between 36 and 45, and the only number in that range which leaves remainder 5 when divided by 9 is 41, which means 4 must be left out.

Q9. What is the 99th digit to the right of the decimal point in the decimal expansion of (1+√2) to the power 500? (Calculator not required.) (In case this isn't displayed correctly by your browser, the expression is (1+sqrt(2))^500, where sqrt(2) means the positive square root of 2.

The answer is 9.

At first sight this seems an impossible difficult question without doing detailed calculations to a huge number of decimal places. But consider the following.
Let x = (1+√2) and x’ = (1-√2). What happens when we add powers of x and x’?
Let y = (1+√2)^500 + (1-√2)^500

We can use the binomial theorem to expand the two powers. Now, for odd powers of √2, the coefficients will have opposite signs in the expansions of x^500 and x’^500 and they will cancel, so we have an expression for y in terms of even powers of √2. But an even power of √2 is a power of 2, so it is an integer. So y is an integer.

Now, what is x’^500? Well, x’ = 1-√2 is about -.414: it’s a negative number with absolute value less than ½. So the 500th power of x’ is positive and is less than (1/2)^500, which is very small – about 4x10^-192. So the decimal expansion of x’ begins with more than 100 zeroes.

So x plus a positive number beginning with 100 zeroes gives an integer: so the first 99 digits after the decimal point in x must all be 9s.

When I first saw this problem I thought it was impossible so I looked at the answer straightaway. But having read the answer it doesn’t seem so impossible – because when one has a term like (1+√2) in an expression it’s a common trick to think about (1-√2). It’s hard to see any other way one could tackle this problem! So having looked at the answer, I’m now sure I could have done it myself if I’d bothered. (I may be being over-optimistic!)

Q10. Here are two messages enciphered using substitution ciphers. What do they say?

a) PREEZ LAEDHKPFH FSO AFYYZ SRT ZRFE!
b) RIFFY SRPWUZQIU! OWURWVM YAE I MAAX RALWXIY!

Answers: (a) MERRY CHRISTMAS AND A HAPPY NEW YEAR!
(b) HAPPY CHRISTMAS! WISHING YOU A GOOD HOLIDAY!

(b) was designed to avoid the letter E to make frequency analysis a little harder.

Questions 3, 4, 8 and 9 came from Mathematical Mind-Benders by Peter Winkler and Question 5 from the same author’s Mathematical Puzzles: A Connoisseur’s Collection. Questions 2, 6 and 7 are from David Wells's The Penguin Book of Curious and Interesting Puzzles.

Greenwich Christmas Maths Challenge 2010

2010-12-21T11:38:00.007Z

A small prize will be offered for the best set of solutions emailed to A.Mann@gre.ac.uk by a Greenwich student before 5pm on Monday 10 January 2011. In the event of a tie, a winner will be chosen randomly. The judges' decision is final. For obvious reasons, the source of these questions won’t be revealed until afterwards. The quiz has ten questions of varying degrees of difficulty: some I think are pretty hard! On past experience, you may well not need to answer them all in order to win.

Q1. Identify the mathematicians whose names are given below as anagrams. Accents and punctuation marks such as hyphens are omitted, and spellings are taken from the MacTutor History of Mathematics website http://www-groups.dcs.st-and.ac.uk/~history/
a) ENRW
b) EIKLNV
c) EEKLPR
d) AAIIJNX
e) AHILMNOT
f) BEILLNORU
g) ACDEEHIMRS
h) AEMNNNNOUV
i) GKLMOOOORV
j) AAAEKKLOSVVY

Q2. A ship is at anchor in a harbour. A spectator sees a ladder dangling from her deck. The bottom four rungs of the ladder are submerged, each rung is two centimetres wide and the rungs are eleven centimetres apart. The tide is rising by eighteen centimetres per hour. After two hours, how many rungs will be submerged?

Q3. For Spanish, Russian or Hebrew, it’s 1. For German, it’s 7. For French, it’s 14. What is it for English?

Q4. How many people is "Twice two pairs of twins"?

Q5. Each integer from 1 to 10^10 (ten billion) is written out in full (for example 211 would be "two hundred eleven" and 1042 would be "one thousand forty two" - the word "and" is not used), and the numbers are then listed in alphabetical order (ignoring spaces and hyphens). What is the first odd number to appear in the list?

Q6. The Ruritanian National Library contains more books than any single book on its shelves contains words. No two of its books contain the same number of words. Can you say how many words are contained in one of its books?

Q7. A maths lecturer has a collection of eighteenth-century mathematical pornography in two bookcases in a room 9 by 12 feet (a foot is an archaic unit of measure about 30cm long). Bookcase AB is 8.5 feet long and bookcase CD is 4.5 feet long. The bookcases are positioned centrally on each wall and are one inch from the wall, as shown in the diagram.

Some students are going to visit the lecturer and she wishes to protect the students from the books and vice versa. The lecturer decides that the best way to do this is to turn around the two bookcases so that each is in its starting position but with the ends reversed so that the books are all facing the wall. The bookcases are so heavy that the only way to move them is to keep one end on the floor as a pivot while the other end is swung in a circular arc. The bookcases are so narrow that for the purpose of this problem we can consider them as straight line segments. What is the minimum number of swings required to reverse the two bookcases?

(For more about eighteenth-century mathematical pornography, see Patricia Fara's recent Gresham College lecture.)

Q8. The number 2 to the power 29 has nine digits, all different: which digit is missing? (Calculator not required.)

Q9. What is the 99th digit to the right of the decimal point in the decimal expansion of (1+√2) to the power 500? (Calculator not required.) (In case this isn't displayed correctly by your browser, the expression is (1+sqrt(2))^500, where sqrt(2) means the positive square root of 2.

Q10. Here are two messages enciphered using substitution ciphers. What do they say?

a) PREEZ LAEDHKPFH FSO AFYYZ SRT ZRFE!
b) RIFFY SRPWUZQIU! OWURWVM YAE I MAAX RALWXIY!

Greenwich Maths Challenge 5

2010-12-06T21:09:00.003Z

The problem posed was as follows:

"In the small country of Mathsland, the citizens are obsessed with politics. Each one passionately supports one of the three political parties, which are the Coffee, Milk and Water parties.

Whenever two citizens meet, they discuss politics. If they support the same party, they don’t change their allegiance, but if two citizens who support different parties meet, they are both so persuasive that each of them abandons their previous allegiance and supports the third party. Thus if Milk and Water supporters meet, they both change to support Coffee.

The repressive laws of Mathsland forbid any gathering of more than two people so all political discussions are limited to the above. If at any time all citizens support a single party, that party will declare a dictatorship and the other two parties will be abolished.

Initially there are 13 supporters of Coffee, 15 of Milk and 17 of Water. Find the shortest possible sequence of meetings which results in a dictatorship, or prove that under these conditions no party will ever command the support of every citizen."

This was the most popular Greenwich Maths Challenge yet, with entries almost equally divided between those who claimed to have found such a sequence and those who claimed to have proved it was impossible. The latter were correct, with several excellent answers submitted. In the opinion of the judges, the first completely valid proof came from Aaron Lang, who wins the prize.

The puzzle is an old one: the traditional scenario involves chameleons but we didn't want it to be too easy to google so we changed the setting. It can be found in several books of puzzles. One solution, given in Terence Tao's Solving Mathematical Problems, is to assign values of 0 to supporters of Coffee, 1 to Milk and 2 to Water. The initial total is 49, which gives a remainder of 1 when divided by 3. One can verify that each of the three possible moves changes the total - for example Coffee meeting Milk removes one of each, subtracting one from the total, but adds two Waters, adding 4 - but that each change does not alter the remainder after division by 3. So at any stage the remainder must remain at 1, but any of the desired solutions needs remainder zero, so they are impossible.

Greenwich Maths Challenge 5

2010-11-07T08:55:00.005Z

A small prize will be awarded for the first correct solution sent to Tony Mann (A.Mann@gre.ac.uk) by a Greenwich undergraduate.

In the small country of Mathsland, the citizens are obsessed with politics. Each one passionately supports one of the three political parties, which are the Coffee, Milk and Water parties.

Whenever two citizens meet, they discuss politics. If they support the same party, they don’t change their allegiance, but if two citizens who support different parties meet, they are both so persuasive that each of them abandons their previous allegiance and supports the third party. Thus if Milk and Water supporters meet, they both change to support Coffee.

The repressive laws of Mathsland forbid any gathering of more than two people so all political discussions are limited to the above.

If at any time all citizens support a single party, that party will declare a dictatorship and the other two parties will be abolished.

Initially there are 13 supporters of Coffee, 15 of Milk and 17 of Water. Find the shortest possible sequence of meetings which results in a dictatorship, or prove that under these conditions no party will ever command the support of every citizen.

COMING SOON

2010-11-04T08:14:00.002Z

Coming soon - Greenwich Maths Challenge 5! A tough maths puzzle with a small prize and a lot of glory for the first correct solution from a Greenwich maths undergraduate. GMC5 will be posted at or soon after 10am on on Sunday morning, 7 November 2010. To practice, why not have a look at GMCs 1-4 posted earlier on this blog.

Maths Puzzle Competition

2010-09-26T15:00:00.002+01:00

The entries for the freshers' maths puzzle competition have now been marked. There were 4 very close groups with marks between 60 and 55 - Roxy, Lena, Shanka and Tammy but way out in front with 71 points was Emma M's group.

Well done to all those who took part. Answers are not being posted but some questions will be gone through in MaTT at some point.

Pi Day at the British Science Festival

2010-09-16T19:43:00.004+01:00

Pi Day (Twitter #PiHunt) yesterday was a great success, with a brilliant audience whose success with Buffon's Needle surpassed our expectations and forced us to depart from our script. And it gave rise to Peter Rowlett's wonderful post-event pi mnemonic

Now I away,
I leave enlivened by having great and happy
historic, enjoyable, happily unanimous joy
at the exciting BSHM #pihunt

So no excuse for anyone to use 22/7 ever again.

(find more at http://travelsinamathematicalworld.blogspot.com/2010/09/how-i-wish-i-could-calculate-pihunt.html)

Pi-Hunting

2010-09-14T15:02:00.004+01:00

This is Ludolph van Ceulen, who along with Noel-Ann Bradshaw and Tony Mann of Greenwich and Mark McCartney of Ulster will be presenting an event about the digits of the mathematical constant pi at the British Science Festival on Wednesday 15th September. Robin Wilson will be there too but is unlikely to be around at the same time as Ludolph. A highlight wil be the "Pi Moment" at 3:14pm (why?) To find out more, come to the event!

Greenwich Maths Challenge 4 result

2010-06-24T09:09:00.003+01:00

Congratulations to Mike Wakeling who provided the first correct answer to GMC4. This one attracted a lot of interest with many attempted solutions. Many people thought the problem was insoluble (as I did when I first saw it!) This problem came from the late Martin Gardner, and I think it's a gem. If you haven't tried it yet, have a go at the problem before reading the solution below.

Here's the solution. You have the given quantities of red, yellow, green and blue paint and have to colour the diagram in four colours so that any two regions with a common boundary are different colours. This is impossible if the colours are red, yellow, green and blue. So you have to mix the blue with an equal quantity of red, so that you now have enough yellow for 24 square metres and enough purple, green and red for 16 square metres each: the problem can now be solved!

Greenwich Maths Challenge 4

2010-05-29T11:48:00.005+01:00

Here, to give people something to do now that the exams have finished, is the final Greenwich Maths Challenge for the 2009/2010 academic year. A prize will be awarded for the first correct solution emailed to A.Mann@gre.ac.uk by a Greenwich maths undergraduate.

You have to paint the marked regions in the diagram using four colours so that each region is a single colour and no regions with a common boundary are the same colour. Each region is 8 square metres in area except for the top area which is 16 square metres. You have only the following paint available: enough red for 24 square metres, enough yellow for 24 square metres, enough green for 16 square metres and enough blue for 8 square metres. Can you find a way to do this?

Greenwich Maths Challenge 3

2010-05-24T17:33:00.006+01:00

The winner of the third Greenwich Maths Challenge was Nic Mortimer, who was the first (and only) person to break the Vigenere cipher. The text was Hardy's famous reminiscence of Ramanujan,

I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number **** and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."

Here the missing number is of course 1729. The keyword to the cipher was "Hardy".

Greenwich Maths Challenge 4 will be posted here on or around Friday May 28, to mark the end of the exams at Greenwich!

Greenwich Maths Challenge 3

2010-02-07T09:30:00.003Z

This was posted at 9:30am on Sunday 7 February but because I had prepared it in advance it doesn't appear as the newest item in the blog. Here is a direct link to it.

Tomorrow's Mathematicians Today

2010-02-07T09:00:00.004Z

This was a wonderful conference with excellent presentations by undergraduates from all round the country, crowned by an exhilarating keynote address by Ian Stewart. The friendly atmosphere and evivent enjoyment of all the participants made it an extremely rewarding day.

Greenwich Maths Challenge 3

2010-02-06T20:38:00.004Z

The following text has been encrypted with a Vigenere cipher. (Simon Singh's Code Book CD-rom, which can be downloaded free, has a useful tool for deciphering such ciphers.) The text which has been enciphered contains one number which has been omitted. The prize for this challenge will be awarded to the first Greenwich maths undergraduate to email to A.Mann@gre.ac.uk two pieces of information: 1) the missing number and 2) the Vigenere keyword.

PRVPCTBVUMUCVJMPNXWMZEVKGTWYHLOENDQPLCDRWUKQCFIYD
BYIUGCUIEWYEITDZUUDECYAEGPLMRUILDKKYATYHLBMSHPZEV
PCKTFPCYAKKCYAUXJSOEHYUDKKYAIYRNLDZWUHSERRHNLQDHV
FUYILVRKLNERFLRVSJPEULRPSRYCYYZQRLRVVRPNXQSTBVUGA
IJWFLSDDJSEJWLBMSHPLXGUCZSZEJLAJWFLSLPMMTNRABBVVG
UTNRBPFWHPLNKZYFS

Greenwich Maths Challenge 3 - COMING 7 FEBRUARY

2010-01-23T20:30:00.003Z

The third Greenwich Maths Challenge will be posted here on Sunday 7 February at 9:30am (approximately). We have postponed it from the previously announced date so that those preparing presentations for the Tomorrow's Mathematicians Today conference on 6 February will not be distracted! It will be a cryptography challenge so speed will be of the essence. If you don't already have it you might like to download Simon Singh's free "Code Book" CD-rom and practice cracking Vigenere ciphers!

Christmas Quiz results

2010-01-23T16:56:00.004Z

Congratulations to Ameli Gottstein, who has (for the second year in a row) won our Christmas quiz. Here are the answers:

Q1. Identify the mathematicians whose names are given below as anagrams. Accents and punctuation marks such as hyphens are omitted, and spellings are taken from the MacTutor History of Mathematics website

a) ENRW - Wren

b) EIKLNV - Kelvin

c) EEKLPR - Kepler

d) AAIIJNX - Jia Xian

e) AHILMNOT - Hamilton

f) BEILLNORU - Bernoulli

g) ACDEEHIMRS - Archimedes

h) AEMNNNNOUV - von Neumann

i) GKLMOOOORV - Kolmogorov

j) AAAEKKLOSVVY - Kovalevskaya

Q2. What is the smallest two-digit integer?

-99. Silly, but it caught me out when I was asked.

Q3. Is 10^2010 + 1 prime? Justify your answer.

The trick here is that for any integers x and y, x^3 - y^3 has a factor x-y. So if n is a multiple of 3, then 10^n+1 is the difference between two cubes because it is 10^(n/3)^3 - (-1)^3. So 10^2010+1 is not prime - in fact it has a factor 10^670+1.

Q4. On average, how many times do you need to toss a fair coin before you have seen a run of an odd number of heads followed by a tail?

This was from Peter Winkler, Mathematical Puzzles: A Connoisseur's Collection (A.K. Peters, 2004). Suppose that the answer is x. Consider the first couple of tosses. If the first coin comes up tails (which it will with probability 1/2), then we have made no progress and the expected number of tosses is now x+1. If the first coin comes up heads and the second toss is also heads (which happens one time in four), then we again expect to take a further x tosses so the expected number of tosses is now x+2. And if the first toss is heads and the second is tails (again one time in four), we have finished and it took two tosses. So we have
x = (1/2).(x+1) + (1/4).(x+2) + (1/4).2 so x = 6.

Q5. Consider a chessboard – an 8x8 grid of squares in which each row and column is alternately black and white. You have 31 2x1 rectangles each the size of two adjacent squares of the chessboard. Remove the top left and bottom right corner squares. Is it possible to cover the 62 remaining squares with your 31 rectangles? Provide a solution or show that it can’t be done.

This is a lovely old favourite. Colour the chessboard in the normal way, so that each row and each column contains alternating balck and white squares. Each of your rectangles, wherever it is placed, must cover one black and one white square. But the 62-square board we have to cover has 32 black and 30 white squares, so it can't be covered by our 31 pieces.

Q6. A, B and C are candidates in an election. There is an odd number of voters. The votes are counted and there is a three-way tie. As a tie-breaker the voters’ are asked for their second choices, and again there is a three-way tie. A suggests that, to break the tie, there be a two-way election between B and C, with the winner then facing A in another two-way election. Is this fair? And what is the probability that A would win the election if it were held in this way, assuming no voter changes their mind?

I found this one in the same book by Peter Winkler as Q4: it was originally devised by Ehud Friedgut. The answer is that A's proposal is very far from fair: the probability that A would win under that system is 1. Suppose B beats C in the two-way election: that means that the majority of A's voters prefer B to C. But B has exactly one-third of the second preferences so, to balance, a minority of C's supporters must prefer B to A. Since C's supporters will decide the two-way election between A and B, A will win that election. A symmetrical argument shows that if C beats B in the two-way election then A will beat C in the run-off. So A is guaranteed to win.

Q7. Here are two hidden messages. What do they say?

a) MERRY CHRISTMAS AND HAPPY NEW YEAR (this was originally written in invisible ink, ie in white text on a white background. Moving your mouse over it would reveal it!)

b) AAAAA VHCTUMMLD

No-one submitted a correct solution to this one. It uses a Vigenere cipher, with the "keyword" chosen so that the first word comes out as AAAAA. My intention was that likely guesses, from the word lengths with part (a) as a clue, would be "MERRY CHRISTMAS" and "HAPPY CHRISTMAS", and testing them with a Vigenere solver would reveal which one it was. In fact the keyword was "TALLC" and the message "Merry Christmas".

Christmas Quiz deadline extended

2010-01-14T06:49:00.001Z

Because of the bad weather the deadline for the Christmas Quiz has been extended to 5pm on Monday 18 January.

CHRISTMAS QUIZ

2009-12-17T08:24:00.007Z

A small prize will be offered for the best solution emailed to A.Mann@gre.ac.uk by a Greenwich student before 5pm on Monday 18 January 2010 (deadline extended because of the bad weather). In the event of a tie, a winner will be chosen randomly. The judges' decision is final. For obvious reasons, the source of these questions won’t be revealed until afterwards. The quiz has seven questions.

Q1. Identify the mathematicians whose names are given below as anagrams. Accents and punctuation marks such as hyphens are omitted, and spellings are taken from the MacTutor History of Mathematics website
http://www-groups.dcs.st-and.ac.uk/~history/

a) ENRW
b) EIKLNV
c) EEKLPR
d) AAIIJNX
e) AHILMNOT
f) BEILLNORU
g) ACDEEHIMRS
h) AEMNNNNOUV
i) GKLMOOOORV
j) AAAEKKLOSVVY

Q2. What is the smallest two-digit integer?

Q3. Is 10^2010 + 1 prime? Justify your answer.

Q4. On average, how many times do you need to toss a fair coin before you have seen a run of an odd number of heads followed by a tail?

Q5.
Consider a chessboard – an 8x8 grid of squares in which each row and column is alternately black and white. You have 31 2x1 rectangles each the size of two adjacent squares of the chessboard. Remove the top left and bottom right corner squares. Is it possible to cover the 62 remaining squares with your 31 rectangles? Provide a solution or show that it can’t be done.

Q6. A, B and C are candidates in an election. There is an odd number of voters. The votes are counted and there is a three-way tie. As a tie-breaker the voters’ are asked for their second choices, and again there is a three-way tie. A suggests that, to break the tie, there be a two-way election between B and C, with the winner then facing A in another two-way election. Is this fair? And what is the probability that A would win the election if it were held in this way, assuming no voter changes their mind?

Q7. Here are two hidden messages. What do they say?

a) MERRY CHRISTMAS AND HAPPY NEW YEAR

b) AAAAA VHCTUMMLD

GMC2 Solution

2009-12-16T13:29:00.002Z

Congratulations again to Nic Mortimer, who wins the prize for the first corrrect solution to the second Greenwihc Maths Challenge. He deciphered the encrypted passage from G.H. Hardy's A Mathematician's Apology (which is given below). The substitution cipher was based on the word CHRISTMAS with A mapping to C, B to H, C to R and so on. But to make it harder, half the Es in the plaintext were treated as Ys, so the mapping was not one-to-one.

The next Greenwich Maths Challenge will be launched on or after Monday 11 January.

Meanwhile, you can do the Christmas Quiz, which will be posted here shortly!

Here is the passage from Hardy:

Greek mathematics is ‘permanent’, more permanent even than Greek literature. Archimedes will be remembered when Aeschylus is forgotten, because languages die and mathematical ideas do not. “Immortality” may be a silly word, but probably a mathematician has the best chance of whatever it may mean... Nor need he fear very seriously that the future will be unjust to him.
No other subject has such clear-cut or unanimously accepted standards, and the men who are remembered are almost always the men who merit it. Mathematical fame, if you have the cash to pay for it, is one of the soundest and steadiest of investments.

One in a million

2009-12-10T13:57:00.004Z

Yesterday's radio play "One in a million" presented some interestingn examples in applied probability (although the image on the BBC website, reproduced here, seems to show some quantum mechanics!) If you want to hear a play in which Bayes' Theorem plays a key role, you have another few days while its available on BBC iPlayer.

(Thanks to Steve Baker for this)

Greenwich Maths Challenge 2 - Winner

2009-12-08T17:46:00.002Z

Greenwich Maths Challenge 2 has been won by Nic Mortimer! The solution won't be published for a couple of weeks to allow the rest of you the fun of solving.

GREENWICH MATHS CHALLENGE 2!!!

2009-12-06T14:41:00.006Z

Here is the second Greenwich Maths Challenge, posted on Monday 7 December at 8pm. There will be a prize of a £10 token for the first correct solution received from a Greenwich student or a group of Greenwich students. Collaboration is encouraged. Your answers should be sent to A.Mann@gre.ac.uk. The judges' decision is final.

We have taken a quotation from a book by a famous mathematician (the views expressed are those of the mathematician in question and not ours!) and encrypted it using a substitution cipher. Your task is to decrypt the ciphertext below. (Line breaks are inserted uniformly after every 40 characters and have no other significance.) The cipher tools on Simon Singh's Code Book CD-rom may (or may not) be useful - this can be downloaded free from http://www.simonsingh.net/Code_Book_Download.html

MODSEGCQADGCQBRPBPLSOGCJDJQGKOSLDOGCJSJQ
DVSJQACJMODSEFBQDOCQUOSCORABGDISPWBFFHDO
SGDGHSODIWASJCDPRAYFUPBPTKOMKQQSJHDRCUPS
FCJMUCMDPIBSCJIGCQADGCQBRCFBISCPIKJKQBGG
KOQCFBQYGCYHDCPBFFYWKOIHUQLOKHCHFYCGCQAS
GCQBRBCJACPQADHSPQRACJRDKTWACQSVDOBQGCYG
SCJJKOJDSIADTSCOVDOYPSOBKUPFYQACQQADTUQU
OSWBFFHDUJDUPQQKABG …

JKKQASOPUHDDRQACPPURARFSCORUQKOUJCJBGKUP
FYCRRDLQSIPQCJICOIPCJIQADGSJWAKCODOSGDGH
SODICOSCFGKPQCFWCYPQAKPDWAKGSOBQBQGCQADG
CQBRCFTCGSBTYKUACVDQASRCPAQKLCYTKOBQBPKJ
DKTQASPKUJIDPQCJIPQSCIBDPQKTBJVSPQGDJQP

(And yes, the quotation in some ways reflects the period in which it was written!)

Greenwich Maths Challenge 1

2009-12-06T14:35:00.002Z

A number of answers were submitted to this problem (see below). The winner is Alex Cole, with commendations to Khadija Khairoun and Steve Baker. All of the competitors identified that the issue lies in the combination of probabilities that are not independent. Unfortunately Alex's answer, which used Bayes's Theorem, has too much mathematics to be easily turned into HTML so I'm not posting it here. The problem was taken from Raymond Smullyan's book The Riddle of Sheherazade and Other Amazing Puzzles.

Greenwich Maths Challenge 2 will be published here on Monday 7 December at 8pm (or as soon after as I remember).

GREENWICH MATHS CHALLENGE 1 !!!

2009-11-08T20:12:00.003Z

Here is what you've all been waiting for - the inaugural Greenwich Maths Challenge. The intention is that a new question will be posted monthly during the University session. There will be a prize of a £10 token for the best answer received from a Greenwich student or a group of Greenwich students. Collaboration is encouraged. Your answers should be sent to A.Mann@gre.ac.uk. The judges' decision is final.

The first challenge is this. What is wrong with Scheherazade's argument in the following story by Raymond Smullyan? The clearest explanation submitted before midnight on Sunday November 29 by a Greenwich student or students will win the prize.

"And now" said Scheherazade, "I have a paradox for you. There are three boxes labeled A, B and C. One and only one of the three boxes contains a gold coin; the other two are empty. I will prove to you that regardless of which of the three boxes you pick, the probability that it contains the gold coin is one in two."
"That's ridiculous!" said the king. "Since there are three boxes the probability is clearly one in three."
"Of course it's ridiculous," said Scheherazade, "and that's what makes it a paradox. 1 will give you proof that the probability is one in two, and your problem is to find the error in the proof, since the proof must obviously contain an error."
"All right," said the king.
"Let's suppose you pick Box A. Now, the coin is with equal probability in any of the three boxes, so if Box B should be empty, then the chances are fifty-fifty that the coin is in Box A."
"Right," said the king.
"Also, if Box C is empty, then again the chances are fifty-fifty that the coin is in Box A."
"That's right," said the king.
"But at least one of the boxes, B or C, must be empty, and whichever one is empty, the chances are fifty-fifty that the coin is in Box A. Therefore the chances are fifty-fifty, period!"
"Oh, my!" said the king.

Coming Soon - Greenwich Maths Challenge!

2009-11-06T16:00:00.004Z

We're about to introduce the Greenwich Maths Challenge! Each month, a tough maths question will be posed on this site. There will be a prize for the best (or, sometimes, first) solution received from a Greenwich student or group of students before the deadline specified. Others are welcome to submit solutions but won't be eligible for the prize. The problems will be varied but they will all be demanding. The first problem will be posted on this blog on the evening of Sunday 8 November. Are you ready for the Greenwich Maths Challenge?

More on the THE Award

2009-11-06T09:08:00.003Z

Professor Chris Bailey, leader of the Greenwich team which has just won the Times Higher Education Award for the Outstanding Engineering Research Team of 2009, writes about the work of his team:

The field of Computational Mechanics and Reliability uses mathematical techniques and software to predict how products (i.e. laptops, aircraft engines, automobiles, ships, etc) will behave when subjected to different environmental conditions and most importantly how long they will survive. This area of research within the school has aided many companies design and maintain reliable products from aerospace electronics to heritage structures.

Award triumph for Greenwich mathematicians!

2009-10-24T10:43:00.003+01:00

Chris Bailey and his team at Greenwich have won the Times Higher Award for 'Outstanding Engineering Research Team of the Year" for 2009. This is wonderful news for the University and the Department of Mathematical Sciences!

The mathematics of Discworld

2009-10-21T15:35:00.001+01:00

I have only just discovered, and can't resist posting the link to, Ian Stewart's pages on the mathematics of Discworld.

Greenwich podcasts!

2009-10-14T13:02:00.002+01:00

A number of Greenwich mathematicians have made podcasts about their work which you can download from Travels in a Mathematical World. (This site, maintained by Peter Rowlett of the Institute of Mathematics and its Applications, has many fascinating podcasts about mathematics and mathematicians.)

A new podcast is added every week: this week's one (Number 41) is by Professor Ed Galea of the Fire Safety Engineering Group at Greenwich. Earlier podcasts by Greenwich people about their work at Greenwich are numbers 16 (Professor Chris Bailey), 20 (Professor Choi-Hong Lai), and 29 (Noel-Ann Bradshaw). In addition Noel-Ann Bradshaw has recorded a series of podcasts on the history of mathematics - these are numbers 2 (Newton and Leibniz), 6 (Galois), 11 (Euler), 13 (Nightingale), 17 (Al-Khwarizmi), 21 (Turing), 25 (Fibonacci) and 30 (Ramanujan).

The great Tom Lehrer

2009-10-11T09:41:00.003+01:00

Mathematicians of my generation were brought up on the records of Tom Lehrer, who sang satirical songs as a sideline to his maths career. I'm grateful to Sandy Galbraith for sending this link to Lehrer singing about maths: some of these I didn't know!

Keith Moffatt awarded 2009 David Crighton Medal

2009-10-02T20:28:00.002+01:00

Professor Keith Moffatt FRS of Cambridge University has been awarded the 2009 David Crighton Medal by the LMS and the IMA. You can read about Professor Moffatt's work on the LMS website (follow the link "Prizes").

Two colleagues at Greenwich have expressed particular pleasure at the announcement.

Professor Ed Galea of our Fire Safety Engineering Group says "This is wonderful news. Much of my PhD was based around the work of Keith Moffatt. I can remember spending months trying to understand some of the work in his papers on magnetohydrodynamics and star formation. And when I finally did understand it I thought how elegant and simple it was, which made me feel both immensely pleased and immensely stupid at the same time!"

Valdis Bojarevics, Reader in our Computational Science and Engineering Group, says, "Good news indeed! This jiggles my memory to one of the first encounters with this remarkable man back in 1981 when I was demonstrating to the prominent visitor, Prof. Keith Moffatt, a rotating vortex generation in a mercury experiment passing a thousand amperes current. As usual the so called 'generals' effect happened, and the thing did not work initially as expected: the 'pinch' effect with a loud bang and splashing of the weak acid solution on top of the mercury. We ended both hiding at the bottom of the experimental table... Keith was so impressed, that we were invited to publish this in the Journal of Fluid Mechanics ( he was the editor at that time), [V.Bojarevics, E.Shcherbinin. Azimuthal rotation in the axisymmetric meridional flow due to an electric current source. - Journal of Fluid Mechanics, (1983) vol. 126, 413.] I met him on several occasions, his talks are always the top attraction at the conferences. His publications are available online at http://www.igf.fuw.edu.pl/KB/HKM/"

Turtles all the way down

2009-09-30T21:19:00.003+01:00

From Logicomix: an epic search for truth by Apostolos Doxiadis and Christos H. Papadimitriou, art by Alecos Papadatos and Annie Di Donna. More about this when I find time to read it!

The three jars problem

2009-09-27T20:05:00.004+01:00

Here is a relatively well known puzzle (it appears, for example, in the film Fermat's Room.

"You are given three closed tins of sweets that are labelled "Lemon Sherbet", "Toffees" and "Mixture". One contains lemon sherbets, one contains toffees and the third contains a mixture of the two, and they are all wrongly labelled. What is the minimum number of sweets that you have to remove in order to ascertain which jar contains which variety?"

If you don't know the problem, work out the answer now before reading any more. So that you can't immediately see the answer I give below, I'm interposing a totally irrelevant image:

(Bottles of sherbet powder in Goodies Sweet Shop, Steep Hill, Lincoln: photograph by Andy Dingley from Wikipedia Commons.)

The book answer is one. If you take one from the tin labelled "mixture" it tells you what is in that tin, since it isn't the mixture. You can then easily identify the other two, knowing that neither is correctly labelled.

I think this answer is wrong. The true answer is zero. (And this is not because of any issue with the wording, that you might be able to feel or smell what is in the mixture tin without actually removing anything.)

You don't actually have to remove a whole sweet. You could open the mixture tin and break one of the sweets into two, remove half a sweet and work out the answer. You could take a smaller piece than a half - it could be a third, or a quarter, or a fifth, or even smaller.

In fact, for any epsilon greater than zero, you could remove a piece smaller than that size and it would give you the solution. So the number of pieces you have to examine is smaller than any positive number. The only such number is zero, so zero pieces suffice.

Maths puzzle competition result

2009-09-27T19:56:00.002+01:00

The entries for the freshers' maths puzzle competition on Friday have been marked and we can now announce the winner. It was very close, with all teams scoring well, but Adam's group scored 78/100 to beat Andrew and Daniel's group by a single point.

Congratulations to the winners. And, while some of the expositions could have been clearer, remarkably, every script submitted was easily legible. Keep that up, please!

Greenwich Scavenger Hunt Result

2009-09-26T11:26:00.002+01:00

Winners: Jodie, Chris, Kieron, Corin etc.
(Note: I'll act as middleman and get you all a discounted membership of the OR Society for only £150 a month each)
Highly commended: Roxy
Finding out the hard way that rebels don't win anything: Nic and his group

Games in Greenwich Park with new students

2009-09-25T20:43:00.002+01:00

A very successful afternoon of mathematical games in Greenwich Park with new first year students and their mentors. Naturally the staff team, with superior knowledge of the mathematics of statics and dynamics, scored a decisive victory at Jenga.

iSquared special issue

2009-09-24T22:18:00.002+01:00

The latest issue of the excellent maths magazine iSquared is a special issue about "Women in maths". It contains a number of thought-provoking articles, interviews, reviews and quotes, from Emilie du Chatelet on the prejudice which excluded women from the sciences in the eighteenth century to Julia Robinson wishing to be remembered as a mathematician regardless of her sex.

iSquared has been nominated for the Maggies awards for the best magazine covers of the past year. A tip - if you are thinking of subscribing to iSquared (which I recommend), if you vote for the awards you get a voucher which gives you a discount on a subscription.

Science Museum

2009-09-23T21:39:00.005+01:00

A visit to the Science Museum is always fun. Highlights today were seeing the original book with Kepler's view of the planets fitting into the Platonic solids, Henry Perigal's geometric pen, ome Indian weights and measures (the same object being both), a model of Kelvin's tide-predicting machine, Alan Bennett's blown glass Klein Bottles, and the wonderful range of polyhedra models.

Any visit to the Science Museum reminds me of previous visits, from my first visit to London when I was 11, a later visit with my father when I finished my sixth form, visits during my undergraduate days and ever since. And of course my favourite object of all, Bill Phillips's water-powered economic computer Moniac, which I remember admiring with my father on that first visit forty years ago. It's a false memory - the Museum didn't have the object then - but it's still part of my personal story of how I became interested in mathematics, mathematical modelling and computing: the areas in which I have spent my entire working life.

Coincidence?

2009-09-22T23:37:00.003+01:00

One of the freakest random events ever seems to have occurred in the Bulgarian national lottery, when the same six winning numbers were selected on two consecutive draws on 6 and 10 September. An investigation has apparently found no evidence of wrong-doing: it was just a remarkable coincidence. I find that hard to believe. What do you think?

http://news.bbc.co.uk/1/hi/8259801.stm

Ian Stewart coming to Greenwich!

2009-09-17T20:15:00.003+01:00

I'm thrilled to hear that Ian Stewart has accepted an invitation to give the keynote speech at the Undergraduate Mathematics Conference which the maths department at Greenwich is organising, with support from the IMA, next February. (For more details of the conference contact the organiser Noel-Ann Bradshaw.)

Ian Stewart is the author of many wonderful books including Why beauty is truth: a history of symmetry, Letters to a young mathematician, Professor Stewart's Cabinet of Mathematical Curiosities and Taming the Infinite: The Story of Mathematics. This summer he was awarded the Christopher Zeeman Medal, awarded jointly by the LMS and the IMA for his work on promoting mathematics. He is a brilliant speaker, and it will be wonderful to hear him at Greenwich again.

(Photo by Avril Stewart from Wikipedia Commons.)

A fascinating novel

2009-09-10T20:16:00.002+01:00

I've only just got around to reading this fascinating novel, published last year. It fictionalises the relationship between the mathematicians Hardy, Littlewood and Ramanujan almost a hundred years ago. Historians may object to some of the liberties taken by the novelist, but it is a loving examination of a mathematical world, not ar removed in time, but now almost unthinkable. It deals with the characters' dilemmas sympathetically. As one who was brought up on the story of Hardy and Ramanujan as it has become part of mathematical folklore, I was particularly interested to see a non-mathematician's take. A thought-provoking novel which I strongly recommend.

Simon Singh at the British Science Festival

2009-09-10T08:28:00.003+01:00

A wonderfully entertaining and thought-provoking lecture by Simon Singh at the British Science Festival on "Why journalists love stupid equations and other problems in the media". The issue of how to promote maths is not straightforward and Simon brought out the complexities through a range of examples ranging from the blatant commercial creation of an equation designed solely to gather press coverage for a PR company's client. to serious research which appealed to the press: Simon showed that it's not always easy to distinguish these. In my opinion (for what it's worth) we have to live with the media we have and it's not worth getting worried about the appeal of stupid equations. Much more serious is the second issue Simon raised: that of the way the libel laws are constraining scientific debate and suppressing the expression of serious scientific comment. For more about this see the Sense about Science campaign.

The London Mathematical Society

2009-04-22T10:06:00.002+01:00

Readers of this blog may be aware of the controversial proposal to form a New Unified Mathematical Society in the UK and consequently to disband the London Mathematical Society and the Institute of Mathematics and its Applications. Yesterday I attended the first of the necessary two Special General Meetings of the LMS which passed by a narow majority the motion to proceed with the winding up of the Society.

For such a historic event it was all rather low-key. There was an hour of procedural wrangles before the vote, but no discussion of the substantive issue.

I suppose the vote of the Scottish Parliament in January 1707 for the Treaty of Union must have been rather similar.

A good time-waster

2009-03-30T20:40:00.001+01:00

Thanks to Claudia for introducing this game which is wasting a lot of my time:

Maths and making fun of Arsenal

2009-03-05T19:51:00.002Z

http://www.guardian.co.uk/science/2009/feb/27/tottenham-hotspur-david-bentley-symmetry

I rather regret doing this now: Arsenal supporters are having such a tough time that it seems a bit unfair to have mocked poor Marcus in this way. In my defence, at the time I named the Bentley group, it wasn't obvious just how poor Arsenal were going to be this season.

Wii-maths

2009-02-23T12:36:00.002Z

An exciting evening for MathSoc with Peter Rowlett leading a session on exploring Newtonian dynamics with a Wii. Peter explained the mechanics behind pool and then had us playing the game to see how it works in practice. We then moved onto doubles matches in tennis, with perhaps slightly less maths in evidence, but I'm proud of saving three match points against Justin.

Royal Institution Masterclasses

2009-02-08T10:47:00.002Z

We're running our annual series of sixth-form Royal Institution masterclasses once again. We've had two so far - Chris Walshaw on how abstract algebra made the founders of Google very rich, and Graham Hoare on the joys of number theory. I am amazed yet again at the enthusiasm of the students who participate. Having people arrive an hour early for a Saturday morning maths lecture is rather remarkable (and rather different from my own early-morning lectures where students tend to stumble in sleepily after I've started). But each year I'm also impressed by our undergraduate helpers who always manage to be enthusiastic, welcoming, willing to do whatever may be required, and totally reliable. All of the speakers really enjoy doing these classes, and that's a wonderful reflection on the Greenwich students who help and the sixth-formers from all over London who attend.

A new symmetry object

2009-01-08T13:59:00.002Z

This is to report that Ameli Gottstein, who won the Christmas Maths Quiz, has chosen as her prize to have a new mathematical structure, a group of symmetries in multi-dimensional space, named after her. (Others similarly honoured include the footballers Jermain Defoe and David Bentley.)

A description of Ameli's group will shortly appear. If you want to name one of these amazing mathematical structures after someone, all you have to do is make a donation to Marcus du Sautoy's favourite charity and Marcus will name a group for you. This makes an excellent present (I think) and is in a very good cause.

Christmas maths quiz - solution

2009-01-06T18:21:00.004Z

Here are the answers to the Christmas Maths Quiz posted last month. The best entry received from a Greenwich undergraduate came from Ameli Gottstein, who wins the prize. Well done Ameli!

Many of the questions (2, 3, 4, 5 and 6) were taken from Peter Winkler's Mathematical Puzzles: A Connoisseur's Collection (A.K. Peters).

The answers were:

Q1. Identify the mathematicians whose names have been shuffled so that the letters are in alphabetical order.
a) AEHLST - Thales
b) AAHIPTY - Hypatia
c) AAHKMYY - Khayyam
d) DEE - Dee
e) AEFMRT - Fermat
f) AHILLOPT - L'Hopital
g) EELRU - Euler
h) ADEGMNOR - De Morgan
i) AAAJMNNRU - Ramanujan
j) ACDIR - Dirac
As a tie-break, can you suggest a candidate for the mathematician with the longest name in which the letters are in alphabetical order, so that (like one of the above) they would appear unchanged if their name were included in the above list?
Peter Rowlett wrote a computer programme to address this and provided a list of all mathematicians in the St Andrews website whose names have this property. The longest such name waa that of Edwin Abbott (author of the classic mathematical novel Flatland). Peter also found mathematicians whose names are in reverse alphabetical order - the longest this time was Rolle.

Q2. Can you find English words containing the following letters consecutively? a) WKW b) HIPE c) ZV d) HQ e) NSW
The answers I was expecting were AWKWARD, ARCHIPELAGO, RENDEZVOUS, EARTHQUAKE and ANSWER. Alternatives proposed included (c) JAZZVOCALIST and (d) MATHQUIZ, which I liked.

Q3. Which of the United States is closest to Africa?
Maine. (If you don't believe this, have a look at a globe! Map projections distort distances!)

Q4. There are several different time zones in the United States. New York on the east coast is normally three hours ahead of Los Angeles on the west coast. A phone call is made from an East Coast state to a West Coast state and it is the same time at both ends. How can this be?
One for Americans, I think. Florida is an east coast state and part of Florida (for example Pensacola) is in the Central Time Zone, while parts of Oregon (including Ontario) observe Mountain Time, so Ontario, Oregon and Pensacola are only an hour apart. The trick now is to call from Pensacola to Ontario between 2 and 3am on the morning in October when Daylight Saving Time ends, so that the clock has gone back an hour in Pensacola but has not yet done so in Ontario, and the time in each is the same.

Q5. There are 100 lightbulbs and switches, numbered 1 – 100, and 100 students. Initially each bulb is off. The first student switches on every light. The second then switches off every second light. The third student now changes the state of every third bulb, so 3, 9, 15,… are switched off and 6, 12, 18, … are switched back on. The fourth student changes the state of every fourth bulb, and so on. After the 100th student has done this, which bulbs are left on?
The ones which are left are exactly the perfect squares: 1, 4, 9, 16, ...

Q6. On average, how many times would you have to roll a fair die before all six numbers have appeared?
If we have already thrown n of the six numbers, then the expected number of rolls before a new number is thrown is 6/(6-n). So the answer to the question is 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 which is about 14.7.

Q7. Sasha and Siobhan are in the same class. They look alike, have the same birthday, were born in the same year, and have the same parents. Yet they are not twins. Can you explain this?
They are triplets.

Q8. Let n be a natural number. Show that there is an integer (non-zero) multiple of n which (in base 10) contains only 1s and 0s.
We use the famous Pigeon-Hole Principle (PHP). Consider the integers 1, 11, 111, 1111, 11111 ... up to the integer with (n+1) ones. There are n+1 such integers and only n possible remainders when you divide by n, so two of these integers have the same remainder. If they were 11111111 and 1111, for example, then the difference would be 11110000 which is a multiple of n of the required form.

Q9. Can you decipher this message? SZKKBSLORWZBGLBLFZOO
HAPPY HOLODAY TO YOU ALL - this used the Atbash cipher.

Q10. And can you decipher this one? JUTZ JU ZNOY WAOF ATZOR EUA NGBK JUTK GRR EUAX IUAXYKCUXQ!
DON'T DO THIS QUIZ UNTIL YOU HAVE DONE ALL YOUR COURSEWORK. This was a Caesar shift cipher.

A Christmas maths quiz

2008-12-14T11:57:00.002Z

A small prize will be awarded for the best answers submitted to A.Mann@gre.ac.uk by a University of Greenwich maths undergraduate by midnight on Sunday 4 January 2009. The rest of you are doing it for fun only!

Q1. Identify the mathematicians whose names have been shuffled so that the letters are in alphabetical order. (Some are easier than others!) As extra help, they are given in chronological order. Accents and punctuation marks such as hyphens are omitted, and spellings are taken from the MacTutor History of Mathematics website
http://www-groups.dcs.st-and.ac.uk/~history/

a) AEHLST
b) AAHIPTY
c) AAHKMYY
d) DEE
e) AEFMRT
f) AHILLOPT
g) EELRU
h) ADEGMNOR
i) AAAJMNNRU
j) ACDIR

As a tie-break, can you suggest a candidate for the mathematician with the longest name in which the letters are in alphabetical order, so that (like one of the above) they would appear unchanged if their name were included in the above list? (My best so far is five letters long.)

Q2. Can you find English words containing the following letters consecutively?
a) WKW
b) HIPE
c) ZV
d) HQ
e) NSW

Q3. Which of the United States is closest to Africa?

Q4. There are several different time zones in the United States. New York on the east coast is normally three hours ahead of Los Angeles on the west coast. A phone call is made from an East Coast state to a West Coast state and it is the same time at both ends. How can this be?

Q5. There are 100 lightbulbs and switches, numbered 1 – 100, and 100 students. Initially each bulb is off. The first student switches on every light. The second then switches off every second light. The third student now changes the state of every third bulb, so 3, 9, 15,… are switched off and 6, 12, 18, … are switched back on. The fourth student changes the state of every fourth bulb, and so on. After the 100th student has done this, which bulbs are left on?

Q6. On average, how many times would you have to roll a fair die before all six numbers have appeared?

Q7. Sasha and Siobhan are in the same class. They look alike, have the same birthday, were born in the same year, and have the same parents. Yet they are not twins. Can you explain this?

Q8. Let n be a natural number. Show that there is an integer (non-zero) multiple of n which (in base 10) contains only 1s and 0s.

Q9. Can you decipher this message?

SZKKBSLORWZBGLBLFZOO

Q10. And can you decipher this one?

JUTZ JU ZNOY WAOF ATZOR EUA NGBK JUTK GRR EUAX IUAXYKCUXQ!

Ingenious

2008-12-07T14:02:00.002Z

Noel-Ann and I have been playing a new board game (or at least, one new to us) called "Ingenious". Tiles, in the shape of two coloured hexagons joined domino-style are placed on a hexagonal grid and one scores points for each hexagon of the same colour in an unbroken line from the ones one has placed. The catch is that the loser is the one whose lowest-scoring colour has the lower total, regardless of how many points one has for one's other five colours. The game in the photo is typical - I have won because Noel-Ann's score for yellow is less than mine for green and red, my joint worst-scoring colours.

I have found this an unusually difficult game in which to plan strategically. If anyone has any tips please email me privately.

9th IMA Younger Mathematicians Conference

2008-11-18T19:41:00.010Z

What a truly inspirational day.......

So many eye-opening talks including, the importance of eigenvalues and eigenvectors within Google's framework for ranking the search results, the future plans for preserving the Cutty Sark as a national treasure and the use of the Monte Carlo Simulation methods for calculating the most likely ruin event within the financial industry.
Noel-Ann Bradshaw (University of Greenwich) gave a brilliant talk on how multi-objective evolutionary algorithms can be used within portfolio optimisation in the financial markets.
Gareth Howell (Cardiff University) gave us a sneak preview into the life of a PhD student and talked about research he's conducting on Theoretical Statistics, including the Farey Series for finding all rational fractions between 0 and 1.

David Youdan also told us about the possible merge between the IMA and LMS into a more unified mathematical society.
All in all, it was a stimulating day and one I certainly will never forget.

My favourite Quote of the day:

"The openness to keep learning is as important as the subject itself" - Andrew Smith (Deloitte)

Maths Inspiration

2008-11-15T10:21:00.004Z

Lovely to see Rob Eastaway's Maths Inspiration going so well, as shown on the BBC morning news yesterday. When Rob set this up a couple of years ago, one of the first venues was the Greenwich Theatre and we organised an outing for students - we even cancelled a Discrete Maths lecture for the day! And I suspect Rob's event, which featured Claire Ellis, Hugh Hunt and Colin Wright as well as Rob himself, was almost as entertaining as my class would have been.

Mark Haddon - Proof-reading at the Royal Society

2008-11-15T08:45:00.003Z

A fascinating discussion at the Royal Society this week between the novelist Mark Haddon, author of The Curious Incident of the Dog in the Night-time, and the mathematician Marcus du Sautoy. It was full of fascinating insights into the nature of creativity, artistic and mathematical: it was a remarkably stimulating evening (and there is a webcast available (click on "Physics and Mathematics" and look for "Proof-reading: Telling stories with numbers, telling stories with words"). I was left excited by illuminating ideas which emerged in the interview. It also inspired me to investigate Haddon's website.

"A webcomic of romance, sarcasm, math and language"

2008-11-04T07:45:00.002Z

One of many from http://xkcd.com/. Thanks to Ian for telling me about this!

Blanc Mange, bags and stick goblins – Math Lecture with John Mason

2008-11-01T21:35:00.003Z

It’s 1 o’clock on a windy Wednesday. We wait for the lecture, enjoying a free lunch, socializing and solving problems for each other, or just causing more. Then the time comes, we enter the room, crowding in the middle, waiting again for a speech to start.

We are asked to fill the front rows so we move. We are asked to participate in conversation so we try. Short introduction: Who? Why? What? (John Mason to educate us holds a lecture on thinking mathematically.) Then he walks to the middle fires up his Mac and soon we stare at the blue screen filled with circles, numbers and colour. The lecture is called Thinking Mathematically. So the circles – at least for me – bring up memories of Maths and stick figures (And stick figures make me - like a dog of Pavlov - to associate with a giant in the playground. Hence the goblins in the header.) Of course they should mean numbers and addition and powers and laws but that’s why we have this lecture. Not to train our strictly meant mathematical knowledge but to make us better understand what we already should now.

We are sorting colourful objects, putting them in bags and then generating functions like wings of angels

All this to learn something most people would think impossible to learn. But we do, and we like it.

After all this is what maths about. Understanding what you already know to make something entirely new and astonishing. Or help someone else make it.

A MathSoc event

2008-10-14T21:04:00.002+01:00

Greenwich maths staff and students took part in a MathSoc field trip to Thorpe Park today:

One from the Independent

2008-09-19T09:51:00.002+01:00

"The art of mathematics"

2008-09-18T20:06:00.000+01:00

Thanks to Tony Ackroyd for this one

http://news.bbc.co.uk/1/hi/sci/tech/7617191.stm

Newton discovers Surrealism

2008-09-09T20:25:00.001+01:00

from Private Eye no. 1218, 5 September 2008

Darth Vader explains the Pythagorean Theorem

2008-09-06T12:40:00.002+01:00

Thanks to Don Cook and the Philomathes list for this one.

Mathematical Chat-up Lines

2008-09-01T16:45:00.003+01:00

Try these at your own risk (and report the results here):

"If I were sin^2 and you were cos^2 together we would be 1."
"I wish I were your derivative sp that I could lie tangent to your curves."
"That dress would look even better at 32 feet per second per second."

(My thanks to Daniel for sending me these, from a letter to the Times from Dom Rowland and Ed Smith of Trinity College, Cambridge.)

Maths at the Edinburgh Fringe

2008-08-14T20:41:00.002+01:00

In a rather warm studio at the top of several flights of stairs sat a girl, Rachel, reading Ian Stewart’s From Here to Infinity and a man, Colwyn, strumming a rather out-of-tune guitar. This was the opening scene of the Edinburgh Fringe production of the play ‘The root of minus 1’ performed by Hartshorn-Hook Productions in association with Angel and Virgins Theatre Company.

The play unfolded into a poignant insight into this couple’s struggle to come to terms with the death of their sister / sister-in-law. The sister, Michelle, had been a budding mathematician at university but had met with a fatal accident before completing her degree. She had developed a very close and possibly intimate relationship with her lecturer, Karen, who helped Rachel and Colwyn find out more about the maths in Michelle’s life. Discussions with Karen covered a wide variety of mathematical issues: certain historical mathematics topics such as counting, infinity and the Pythagoreans and other mathematical areas such as topology, calculus and as the title suggests – imaginary numbers.
The actress playing the part of Karen also cleverly played the character Emily the counsellor who, without giving her own opinion, helped the couple to understand their own fears and feelings.

The mathematical content came over with a passion that I hope would inspire others to take more than a passing interest in the subject as well as a number of amusing insights about mathematicians. If the theatre company ever staged a production in London it would make a great trip for maths students – I would enjoy seeing it again.

What is 7 times 13?

2008-08-04T09:59:00.001+01:00

http://www.youtube.com/watch?v=7WMi5TUJDso

(Thanks to Don Cooke and Udai Venedem of the Philomathes email group for this.)

How many socks make a pair?

2008-08-01T11:56:00.003+01:00

Last night I was at the launch party for Rob Eastaway's new book, How many socks make a pair? Like Rob's previous books, including Why do buses come in threes? and How long is a piece of string?, this presents interesting mathematics from everyday life. It contains a number of gems. I was pleased to learn the name of Penney Ante, which I came across as a scholboy wihtout ever knowing its origins, and to discover the Saddam Puzzle.

"The Football Stadium" puzzle asks whether, if 101 metres of bunting is laid along a 100-metre touchline, pinned to each corner spot, is there enough play to allow someone to pass underneath the bunting at the centre line? The answer is surprisingly counter-intuitive (well, counter-my-intuition, anyway).

Strongly recommended!

The statistics of sex in your 70s

2008-07-29T15:47:00.001+01:00

A Swedish study of attitudes to sex among the older generation provoked some interesting correspondence in the Independent over the last couple of weeks. The survey showed that 68% of men and 54% of women were continuing to have sex into their old age. Some readers jumped to the conclusion that 14% of men must either be lying or being unfaithful to their partners. But of course that superficial conclusion is wrong. Since women live longer than men, there were many more women than men in the age group, so it turns out that it's the women who are lying or being unfaithful (or having younger partners).

A neat illustration of how easy it is to misinterpret statistics?