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/
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"?
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.