CHRISTMAS QUIZ

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

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

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

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!!!

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

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).