Tuesday, 21 December 2010

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/

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.

Diagram for bookcase problem 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?


Monday, 6 December 2010

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.