Christmas Quizzes 2017–2018

Thirteenth Question, solved

Well done! You've identified a correct question to suit each of twelve answers. A prize is within your grasp.

All you have to do now is convert those choices into the tiny small miniature little one-letter answer that will (possibly) win you your choice of fab prizes.

There's a simple method ... no surprises there, we're sure. It goes like this ...

Each question contains an encoded letter of the alphabet, which you can easily work out like this ...

Find the number-value of the first letter of the question, and the number-value of the last letter of the question (disregard punctuation).

So if the question were 'Ships [meaning all kinds of marine craft] come home to die?', you'd find the initial S and the final E
The examples on this page are all written in green, as you'll notice.

Convert those two letters – let's call them 'bookends' – into numbers, using the alphabetical table here

So that would be 19 (S) and 5 (E)

What do you do with those two bookend numbers?

To determine that, you need to count the number of words in the question in question. Please note these terms and conditions: acronyms (eg BBC, PNV, CD) count as one word each; abbreviations such as 'eg' and 'ie' count as one word each; Roman numbers (eg MDMLIII, X) count as one word each; a hyphenated expression (eg ‘word-count’) counts as only one word; words within square brackets [thus] are not included in the word-count, but words inside round brackets (thus) are included; items of punctuation including – = > ! and so on are not included in the word-count.

'Ships [meaning all kinds of marine craft] come home to die' ... that's five words (we're not counting the words inside the square brackets, see?)

So now you've got the word-count of the question, and two bookend numbers. Some of those word-counts will be odd, and some even.

If the word-count number for a question is odd, subtract the lower bookend number from the higher bookend number, to get what we'll call the 'nearly-there number'.

So in our example, the word-count number is five, which is odd. So we're going to subtract the lower bookend number (5) from the higher one (19) ... giving a value of 14 ... the 'nearly-there number'

If the word-count number for a question is even, add the two bookend numbers together, to get what we'll also call the 'nearly-there number'

So if our next question had been 'Identify the graveyard of a liner, kayak, tugboat or skiff,' the bookend numbers would be I (9) and F (6), and the sentence contains ten words, an even number, so we add together the 9 and the 6, making 15 ... the 'nearly-there number'

Start back at Christmas Day's puzzle, and work through the questions you've chosen, and you'll soon have twelve nearly-there numbers.

Good work!

Now convert the twelve nearly-there numbers back into alphabetical letters, using the alphabetical table here ...

And you've got the twelve nearly-there letters.

In our example, you'd have an N, then an O (could be the beginning of a message like 'Noel' for instance)

Do your twelve actual nearly-there letters spell a little message? If so, you're on the right lines.

If not, you may have misfired on one or more questions. There's time to go back and reconsider.

But the next step, assuming you're going forward, is easy: take the twelve nearly-there letters, and find an anagram (ie shift the letters about a bit) that expresses a period of time (eg 'for six months', or 'several weeks' or something like that) and when you've worked out what that is – there's a cumulative visual / aural clue, of a sort, running through these twelve or thirteen puzzle-pages – you have only to convert it into the Roman system (using this handy device) and send it to us at webmaster@procolharum.com (now closed)

Remember: you're sending us a single letter only (it doesn't have to be in red, incidentally). And you're also sending us your choice of prizes, in the form of a twelve-letter string (like 'MICEGOSQUAWK').

Very best of luck to all you thinkers and lateral thinkers!

For those who care about such things, we were searching for a theme for this puzzle while in Italy, having chatted to Stefano Carbone; and the answer appeared midway between Milan and Rome, both cities in which Procol Harum gave superb concerts to superb audiences.

And many thanks as ever to the vigilant Jane Clare of Perth, WA, who has worked all these puzzles through to make sure they cohere, and contributed a lot to their careful wording.

In the unlikely event of there being fewer winners than prizes this year, early claimants will get more than their fair share!
Prizes will of course be awarded at the absolute discretion of Roland and Jens, who run 'Beyond the Pale', and whose decision will be final; their families are not eligible to enter.

Back to the how-to-play page for the 2017 'Beyond the Pale' Christmas puzzles | Twelfth puzzle