Greenwich Christmas Maths Challenge 2010

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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