The RSS Christmas Quiz 2017: Solutions

Last month, we published the Royal Statistical Society’s dastardly Christmas Quiz – a formidable test of general knowledge, logic, and lateral thinking.

If you would like to take a look at the quiz questions before seeing the solutions, they are still available here.

We were delighted to receive a large number of high-quality entries, several of which scored in excess of 90%. However, both markers were in agreement that the top two prizes should be awarded as follows:

First place – Andrew Garratt

Second place – The team comprising Maurice Cox, Alistair Forbes, Andy Knott and Sam Knott.


An honourable mention to the team of Toril Opsahl and Ole Martin Halck from Oslo, who are not members of the RSS, but submitted an excellent set of answers that placed them just a whisker behind our prize-winners.

And now, the solutions…


1 …GO! [10 points]

(a) All six are named “Aloysius”: Aloysius Snuffleupagus from Sesame Street, who is friends with Big Bird; Police Chief Aloysius from “It’s a Mad, Mad, Mad, Mad World”; Aloysius, the teddy-bear from “Brideshead Revisited”, who was modelled on John Betjeman’s teddy-bear, Archibald Ormsby-Gore; Aloysius O’Hare from “The Lorax”, whose name evokes Chicago O’Hare; Aloysius the janitor from “The Hudsucker Proxy”; Aloysius Pig from “Garfield and Friends” / “U.S. Acres”.

(b) All six have the surname “Parker”: Eleanor Parker, known as “Woman of a Thousand Faces”; Ray Parker Jr, writer of the “Ghostbusters” theme; Dorothy Parker, a founding member of the Algonquin Round Table (also known as the “Vicious Circle”); Bonnie Parker, the female half of Bonnie and Clyde; Trey Parker, co-writer of “The Book of Mormon” and the voice of Balthazar in “Despicable Me 3”; Sarah Jessica Parker, regularly seen with Kim Cattrall, Kristin Davis, and Cynthia Nixon in “Sex and the City”.

(c) In “Thunderbirds”, FAB 1 is a pink, six-wheeled car, FAB 2 is a yacht, and FAB 3 is a racehorse – all owned by the character Lady Penelope Creighton-Ward. The clue obtained by combining parts (a) and (b) is “Aloysius Parker”, who appears in the show as Lady Penelope’s butler and the driver of FAB 1. The question title (“…GO!”) alludes to the famous catchphrase “Thunderbirds Are Go!”


2. POLYMERISATION [4 points]

The polymer £5 note introduced in 2016 by the Bank of England features the quote “I have nothing to offer but blood, toil, tears and sweat" (initial letters IHNTOBBTTAS), while the new polymer £10 note features the quote “I declare after all there is no enjoyment like reading!” (initial letters IDAATINELR). The polymer £20, due for release in 2020, will feature the quote “Light is therefore colour” – so the answer is “LITC”.


3. ROCK LEGENDS [9 points]

As hinted at by the question title, and the reference to “outlandish” arithmetic, this question concerns asteroids or minor planets (“rocks”) named after famous people (“legends”). The bizarre arithmetic in the question is correct when applied to the official numbers of these asteroids, according to the International Astronomical Union’s Minor Planet Center. (A searchable list can be found online at the Wikipedia page “List of minor planets named after people”.)

In order, the sets of initials in the question refer to the following individuals (with the corresponding numbered asteroids indicated in brackets):

William Shakespeare (2985 Shakespeare) + Johann Wolfgang von Goethe (3047 Goethe) = Alfred Bernhard Nobel (6032 Nobel)

Albert Einstein (2001 Einstein) + Charles Robert Darwin (1991 Darwin) = Wilhelm Richard Wagner (3992 Wagner)

Edmund Percival Hillary (3130 Hillary) + Hendrik Johannes Cruijff (14282 Cruijff) = Thomas Sean Connery (13070 Seanconnery) + Sigismund Schlomo Freud (4342 Freud)

Rembrandt Harmenszoon van Rijn (4511 Rembrandt) + Johannes Chrysostomus Wolfgangus Theophilus Mozart (1034 Mozartia) = Galileo Galilei (697 Galilea) + Tutankhamun (4848 Tutenchamun)

Bilbo Baggins (2991 Bilbo) + Johannes Gensfleisch zur Laden zum Gutenberg (777 Gutemberga) = Norma Jeane Mortenson (3768 Monroe)

Carl Edward Sagan (2709 Sagan) + Edwin Eugene Aldrin Jr (6470 Aldrin) = Louis Daniel Armstrong (9179 Satchmo)

Thomas Alva Edison (742 Edison) + Christopher Columbus (327 Columbia) = Max Karl Ernst Ludwig Planck (1069 Planckia)

Dmitri Ivanovich Mendeleev (2769 Mendeleev) + Christian Andreas Doppler (3905 Doppler) = Paul Cézanne (6674 Cézanne)

The person represented by the question mark is Professor James Moriarty, the arch-enemy of Sherlock Holmes, since (3047 Goethe) + (2001 Einstein) = (5048 Moriarty).

As several entrants highlighted, Professor Moriarty is a particularly appropriate choice here, as he authored the fictional book “The Dynamics of an Asteroid”.


4. CAN YOU DIG IT? [11 points]

Each answer contains the name of a digit (“zero” through to “nine”, in order), split across two words: “Mansize Rooster” (by Supergrass); “Don’t Go Near The Water”; “The Lost World” (the Arthur Conan Doyle novel where Professor Challenger first appeared); Gareth Rees; “World Of Our Own” (an anagram of “NOW FOR LOUD ROW”); Earl of Iveagh; Pope Pius IX; “Everything’s Eventual” (a collection of short stories by Stephen King, who ordered them using playing cards); Totleigh Towers (with the “Plum” in the clue being P. G. Wodehouse); and Hanin Elias (co-founder of Atari Teenage Riot, name-checked in the song “Hot Topic” by Le Tigre).

A clue to the connection is provided by the question title, which contains the word “digit” split in the same fashion.


5. … JOE [5 points]

(a) The three film titles are “The Moderns” (featuring Geneviève Bujold, Geraldine Chaplin, John Lone, Keith Carradine, Kevin J. O'Connor, Linda Fiorentino, and Wallace Shawn), “The Meaning of Life” (Eric Idle, Graham Chapman, John Cleese, Michael Palin, Terry Gilliam, and Terry Jones), and “The Comedians” (Alec Guinness, Elizabeth Taylor, Peter Ustinov, and Richard Burton). The first title, “The Moderns”, appears within the last three words of question 5b, if case and spacing are ignored.

The connection is that each of the three titles contains the name of a type of average (mode, mean, median) – which ties back to the question title via the phrase “Average Joe”.

(b) Between 2008 and 2011, the character Mainframe from “G.I. Joe” (a second connection to the question title “… Joe”) was renamed “Dataframe” – a term that any modern statistician who uses statistical software will recognise. He would certainly feel at home in question 1, as his real name is Blaine L. Parker.


6. INK-COGNITO [6 points]

The left ink blotch obscures the number 395, while the right ink blotch obscures the numbers 86637 and 15913.

Here is one approach to reach the answer. Consider firstly the top line, where all the numbers are visible. For each number N on the left-hand side, if we multiply together the four other numbers on the left-hand side, add one, and then divide by N, we obtain the corresponding number appearing on the right-hand side. To illustrate, for the first number (N=2) we find that (3 x 11 x 23 x 31 + 1)/2 = 11765; similarly, for the second number (N=3), we find that (2 x 11 x 23 x 31 + 1)/3 = 5229; and so on. For this pattern to hold for the second line as well, the left blotch must obscure the number 395, since (3 x 7 x 47 x 395 + 1)/2 = 194933. The numbers obscured by the right blotch can then be calculated as (2 x 7 x 47 x 395 + 1)/3 = 86637 and (2 x 3 x 47 x 395 + 1)/7 = 15913.

The two sets of whole numbers given on the left – (2, 3, 11, 23, 31) and (2, 3, 7, 47, 395) – are very unusual, because the five right-hand side numbers generated by the above process are also whole numbers. These are, in fact, the only two sets of five whole numbers with this property (unless some of the numbers on the left or right are allowed to be equal to 1). Larger sets that follow the same pattern are also possible – for instance, (2, 3, 7, 47, 583, 1223) is such a six-number set. As several entrants noted, the reference to “Stefan” in the question text is a nod to Stefan Znám, who studied sets of numbers with this property. (For further information, search online for “Znám’s problem”.)


7. OD & DS [8 points]

A question of “odds & ends” – in the first two quotes, only the odd-numbered letters of each word are visible, while in the final two quotes, only the two letters at the end of each word are visible. (The two words in the question title have been transformed similarly, with “odds” becoming “od”, and “ends” becoming “ds”.)

All four of the quotes have a statistical flavour:

“The best thing about being a statistician is that you get to play in everyone’s backyard” [John Tukey]

“Politicians use statistics in the same way that a drunk uses lamp-posts – for support rather than illumination” [Andrew Lang]

“Statistical thinking will one day be as necessary for efficient citizenship as the ability to read and write” [Herbert George Wells]

“Statisticians, like artists, have the bad habit of falling in love with their models” [George Box]


8. OLIVE (_____GREEN) [8 points]

The eight clues all refer to Santa’s reindeer – in order: Dasher (the nickname of Kevin Wheatley, an Australian recipient of the Victoria Cross); Dancer (a film by Steven Cantor about the Ukrainian dancer Sergei Polunin); Prancer (a reference to the Tartan Prancer in “Vacation”); Vixen (Rorie’s sweetheart in Mary Elizabeth Braddon’s 1879 novel “Vixen”); Comet (the role played by dog actor Air Buddy in “Full House” – the term “boat” is poker slang for a full house); Cupid (Johnny Nash’s cover of the famous Sam Cooke song – the second line being “And let your arrow flow”); Donner (the character who falls downstairs in Tom Stoppard’s “Artist Descending a Staircase”, and is found by Martello and Beauchamp); and Blitzen (a miniaturised computer system designed for NASA in the 1980s).

Of course, the missing reindeer name is “Rudolph”, who shares his name with Alan Rudolph, the director of “The Moderns” from question 5a. Alan Rudolph is linked to question 1b via the film “Mrs Parker and the Vicious Circle”, which he directed. This film is centred on Dorothy Parker, and the cast includes Sarah Jessica Parker’s spouse, Matthew Broderick.

The list of eight reindeer-based clues in the first part of this question omitted Rudolph, and could therefore be said to feature “all of the other reindeer” – a lyric from “Rudolph the Red-Nosed Reindeer” that has famously been misheard by some listeners as “Olive, the other reindeer”. A misheard lyric such as this is known as a “mondegreen” – hence the cryptic question title, in which the letters “MONDE” have been omitted from “OLIVE (MONDEGREEN)”.


9. STRENGTH IN NUMBERS [12 points]

(a) The table identifies, by their initials, musical artists or groups that have released UK-charting singles (S) or albums (A) whose titles are years in the 20th or 21st century – with the year being indicated by the first column. To qualify for inclusion, the title must be a four-digit number in the appropriate range (with no additional words or symbols) and must appear in the Official UK Chart archive ( In cases where the same title is shared by multiple artists or groups, all options are recorded in the table, separated by slashes.

As many entrants noted, the question title cryptically hints at this theme – we are interested in musical “numbers” (both literally, and figuratively) with enough “strength” to register on the Official UK Chart. (Moreover, the title “Strength in Numbers” itself appears within the archive as a single and album by the band “Music” – appropriately enough.)

The completed table contains the following items (which, strictly speaking, entrants should have represented as initials – e.g. NO for New Order):

SINGLES (column S) – “1962” by Grass-Show; “1963” by New Order; “1973” by James Blunt; “1979” by Smashing Pumpkins; “1980” by Estelle; “1985” by Bowling For Soup; “1991” by Azealia Banks; “1998” by Binary Finary; “1999” by Binary Finary / Prince / Paradox Featuring Devorah; “2000” by Binary Finary.

ALBUMS (column A) – “1916” by Motörhead; “1965” by Afghan Whigs; “1977” by Ash / Jam; “1982” by Status Quo; “1984” by Eurythmics / Rick Wakeman / Van Halen; “1987” by Whitesnake; “1989” by Ryan Adams / Taylor Swift; “1992” by Game; “1999” by Cassius / Prince; “2001” by Dr Dre.

The Jam album “1977” and the Whitesnake album “1987” in the above list were both released very recently (November 2017) – which served as the original inspiration for the question.

Finally, the inclusion of the item marked with an asterisk (“1982” by Status Quo) is debatable, as the title is rendered as “1+9+8+2” by some sources – but it is included in the table as it appears in the archive as “1982”.

(b) These decimal numbers respectively represent “7/11” by Beyoncé, “7/27” by Fifth Harmony, and “2√231” by Anticappella. (The symbol appearing before “231” is a square root sign.) All of these releases appear as singles or albums within the Official UK Chart archive.

(c) The answer is “4x4=12” by deadmau5, a Canadian who wears a large mouse head as a mask during performances. This calculation is clearly wrong in our standard base 10 counting system, but it is correct in base 14 (where the number sixteen is written as “12”). Once again, this release appears in the albums section of the Official UK Chart archive.


10. BIZARRE TALES [7 points]

Dropping one letter from each word (as indicated by “almost all”) and then anagramming gives the names of nine prominent scientists / engineers: EDISON, ARCHIMEDES, HOPPER, DESCARTES, CURIE, BOHR, HERTZ, DIRAC, and THOMSON.

The word that is left over (i.e. formed by the nine omitted letters) is INTESTINE; applying the same transformation (omitting a letter, and anagramming) produces EINSTEIN.

The remaining letter, T, symbolises TESLA – another scientist. This final answer is hinted at by the question title, since TESLA is an anagram of TALES.


11. JUMPIN’ JIVES [6 points]

Question 11 concerns authors who have book titles containing multiples of 11 – specifically, the first column of the table provides an encrypted version of the author’s first name, while the second column indicates which multiple of 11 is present.

The encryption scheme being applied to the first column is the Atbash cipher, which simply turns the alphabet “upside-down” (i.e. A maps to Z, B maps to Y, and so on). The unencrypted names are therefore “Joseph”, “Tom”, “Dean”, “Frederic”, “Bill”, “Stephen”, and “Laurie” respectively.

Moving to the second column, if we regard the Greek letter “Xi” on each row as representing the number 11 (by analogy with the Roman numeral XI), and treat the number beforehand as a multiplier, we find that the second column is a cryptic representation of the values 22, 44, 77, 99, 451, 1001, 1408, and 1793 respectively.

Combining these two columns of information, solvers should then have deduced that the seven first names refer respectively to Joseph Heller (“Catch 22”), Tom Rob Smith (“Child 44”), Dean Koontz (“77 Shadow Street”), Frédéric Beigbeder (“99 Francs”), Bill Willingham (“1001 Nights of Snowfall”), Stephen King (“1408”), and Laurie Halse Anderson (“Fever 1793”). The remaining number, 451, immediately suggests “Fahrenheit 451” by Ray Bradbury, which means that the empty cell should read “Izb” – the Atbash encoding of “Ray”.

The majority of entrants correctly noted that the popular Netflix series “Stranger Things” (which was highlighted in italics within the question text) features an alternate dimension called the “Upside-Down” – making the use of an “upside-down” cipher particularly appropriate. However, only a handful of entrants also spotted that the main character in “Stranger Things” is called “Eleven” – the number that links all the titles in the question. There is also one further connection hidden in the bizarre question title, “JUMPIN’ JIVES” – Eleven’s real name is Jane Ives (i.e. “J. Ives”), and so the title is a hint that the question involves numbers that jump up in steps of “J. Ives” (i.e. multiples of 11).

For completeness, it is worth noting that Eleven also has an alternative name of “Jane Hopper” – which provides another connection to “jumping” – and that Stephen King’s short story “1408” appears in the collection “Everything’s Eventual”, which featured in question 4. However, neither of these observations were required to score full marks.


12. GUIDING AMERICA [7 points]

(a) If we combine the standard US state / territory abbreviations corresponding to the capital cities named in the question, we obtain MA + LA + WI = MALAWI, a country in south-east Africa, and VI + CT + OR + IA = VICTORIA, which points to the Australian state, the Canadian city, or Victoria Falls and Lake Victoria (which are partially located, respectively, in Zambia and Tanzania – both adjacent to Malawi). For the final part, someone about to head down a CO + AL + MI + NE = COALMINE might obtain added peace of mind by bringing along a “Canary” (perhaps even Big Bird from question 1?)

Even the most eagle-eyed of our entrants did not spot that the question title, “GUIDING AMERICA”, also follows the same pattern as MALAWI, VICTORIA and COALMINE, being composed of seven further US state / territory abbreviations (ignoring the space between the words): GU + ID + IN + GA + ME + RI + CA. This was the only available point in the quiz that went unclaimed by any entrant – so many congratulations if you spotted this link.

(b) The “halfway” relationships stated in the question are satisfied by the three-letter abbreviations for the respective countries (ISO ALPHA-3), when considered letter by letter. For instance, GNQ (Equatorial Guinea) is halfway between AGO (Angola) and MUS (Mauritius), since the letter G is halfway between A and M, the letter N is halfway between G and U, and the letter Q is halfway between O and S. In the same way, NOR is halfway between SLV and IRN, FJI is halfway between ESP and GAB, and IRL is halfway between KOR and GUF. The required answer is therefore the country whose three-letter code lies halfway between SAU and USA, which is Tajikistan (TJK).


13. PRESENT PERFECT [7 points]

The smallest possible number of moves required for Rudolph to deliver the presents and return to the North Pole is 30, and this can be achieved by the following route (and numerous others):

D4 - E2 - G3 - F1 - E3 - G2 - G4 - F2 - E4 - D2 - B1 - A3 - C2 - C4 - B2 - D1 - D3 - B4 - A2 - A4 - A6 - A7 - C6 - C7 - D5 - E7 - G6 - G7 - F5 - D6 - D7

(This particularly elegant solution, which methodically visits the four quarters of the board in turn, was submitted by Maurice Cox, Alistair Forbes, Andy Knott, and Sam Knott.)

A route with fewer than 30 moves is impossible, since Rudolph must visit an empty (non-present) square immediately before visiting A2 or G2, and also immediately after visiting A7 or G7. It is easy to see that these empty squares cannot overlap (i.e. the exit square from A7 or G7 cannot serve as the entry square for A2 or G2), which means that Rudolph must visit at least four empty squares along the route. Since he can only make at most one present delivery per move, and at least one additional move is required to return to the North Pole (D7), the minimum total number of moves is 25 + 1 + 4 = 30. This reasoning demonstrates that a 30-move solution, such as the one given above, must be “perfect” – i.e. no shorter route can possibly exist.

In the 30-move route given earlier, the four empty squares that Rudolph visits are E3, B4, C6, and F5, and it’s easy to check that he successfully delivers a present on every other move – except, of course, for his final move back to the North Pole.


The 2017 Royal Statistical Society Christmas Quiz was set by Dr Tim Paulden. Many thanks to David Edelman and Jon Nelson for their helpful contributions.

Christmas Quiz

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