[If you’re attending York S&B please see comments 32&33] - here
Not only a new compiler this week but also a new type of puzzle. Those who regularly complete the Listener will be familiar with the format and the neat logic required to compile and solve these numerical grids. So throw away your Chambers and dust down your O’ level text books (GCSE text books for our younger readers). This may also be a first for fifteen squared too!!
I’ve tried to break down the solving process as best I remember it. For the various lists of numbers I used google – the multiplication bits were pen and paper or in my head. These puzzles are a lot easier than many people think.
Where to start is always the first thing to look for in these puzzles – one clue will always leap out as being the first one to solve.
Here it is 3 across – the answer is either the year of the Battle of Trafalgar or Waterloo so it’s 1805 or 1815.
3ac The time the match started in the 24 hr clock. It’s either the Battle of Trafalgar or Waterloo but I can’t remember which.
5 The number of fireworks let off at the end of the match, a triangular number.
18-5 can be entered and 5 down too falls into place. Triangular numbers make triangles i.e. 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91. So 5 down is 55.
That’s as far as that part of the grid can go for the minute.
My next inroad was 24dn
24 The most number of runs scored in an over by either team, a Fibonacci number
The 2-digit Fibonacci numbers are 13 21 34 55 89 and combining this with the fact that 28 ac is a time
28 Witrush batted first, and their innings closed at this 24 hr prime time.
Then 24dn is 21
Next stop was the 2 clues with crossing 3 digit Fibonacci numbers
25 The number of sandwiches consumed, a Fibonacci number.
20 The number of spectators at the match, a Fibonacci number.
There are 5 3 digit Fibonacci numbers 144 233 377 610 987. The grid layout means that these can be 610 down 144 across or 233 down and 377 across. Both allow a valid time for 28ac but as 13 down has to be a prime then it can’t end in a 4 so the combination is 233 and 377.
26 The number of lemonades drunk, a prime number.
26 down is a prime beginning with 7 so can be 71, 73 or 79
28 across has to be a prime after 1805 (as this is the end of the first innings) and less than 1999 with a third number of 3. 28 ac can be 1831 or 1931 or 1933.
21 Wedlock’s innings began at this 24 hr prime time. They had less than 20 minutes to get ready
18 The total number of dot balls bowled, a square number.
The first number has to be a 1 and the second an 8 or a 9 – However 18 dn is a 2 digit square so must be a 9. 18dn is 49, 21ac is 19– and 28ac is 193-.
22 The total number of legs in the ground. Hope you’re remembering about the 12th men, both umpires, and Tango the three-legged cat, which makes it a prime number.
This can be calculated from 20dn and is 521 (233×2 + 24×2 + 2×2 + 3)
21ac is a prime number within 20 of the 2 possibilities for 28ac and is thus 1951 making 28ac 1933 and 26dn 73
easy so far – the next stop for me was a little trickier
8 Wedlock’s innings closed at this 24 hr time, a square number.
This must begin with a 2 as 1 down is a multiple of 19 ac which ends in 2. Can be 2025, 2116, 2209 or 2304 (Can’t be last 2 as 9 down would then begin with a zero). It can’t be 2025 as 2dn is a palindrome and would then begin with zero so 8ac is 2116. and 2dn is 1-1.
6ac is only 2 digits
6 Witrush’s final total less their extras. which equals 29ac minus 9dn
9 The number of extras in Witrush’s innings, a prime number and a factor of 15ac.
29ac must be 11- and 6ac must begin with a 9 as the maximum 9dn can be is 19.
As Wedlock batted second then their score must also begin 11- (1ac) as the game completes when the 2nd side outscores the first or fails in the attempt. So back to 19ac now than 1dn is known.
19 The average age of the Witrush team, a factor of 6ac.
19ac is 32 and 6ac is 96.
14 The number of extras in Wedlock’s innings, which is a prime factor of 3ac.
The factors of 1805 are 1,5,19,95,361,1805 and of 1815 are 1,3,5,11,15,33,55,121,165,363,605,1815 – combine this with 1dn = 23dn + 14ac then the second digit of 23dn is a zero 10- and 14acmust begin with a 1 so can be 11 or 19.
As 27ac is now known as 20 and 7ac is a multiple of it then 4dn is 80-. Going to 4ac and 11ac – knowing that there are 233 spectators
11 The number of children spectating at the match, a prime number.
4 The number of pints of cider drunk by the spectators. Only those of legal age drank, and each had the same whole number of pints.
This means that 11dn has to be 31 and 4dn 808
Looking at the totals and extras with 9dn being either 17 or 19 then Wedlock’s total is 114, Witrushs 113 and 14ac is 11
12ac is a palindrome and is 7-7.
Then it all gets quite easy 15ac is 51; 7dn is 2628; 7ac is 220 with 16dn 13and 13dn 7817 and 3ac 1815.
So who won – WEDLOCK starting at 18:15.
Many thanks Oyler – a welcome change from the word play and hopefully a permanent feature of the EV.
Not my cup of tea, I’m afraid. EV should be a crossWORD.
Whether or not we will see more puzzles like this will depend on how this one was received. It was an experiment worth trying, and Oyler certainly offered up the sort of puzzle that it ought to be. Nothing too involved in the number-work, and wrapped up by a pleasant theme — and indeed a bit of cryptic wordplay, exploiting cricketing terminology to provide some maths terms too.
Number puzzles in the Listener usually attract a very high number of newcomers (at least, over the course of a year). For example, last year, Oyler’s Alma Mater puzzles, 4270, had 38 entries from people who hadn’t previously solved a Listener in the year, while the other 9 puzzles in the final ten had 0, 0, 1, 2, 2, 0, 1, 3 and 1 new solvers. So number puzzles can, potentially, be fairly useful ways of bringing new EV solvers into the fold.
At any rate, I’m sure that this won’t mark the end of EV’s tradition of crosswords through the year, and there likely won’t be any more than two of these a year, although that’s obviously not my decision really. Still, isn’t it nice to have a bit of variety? And number puzzles offer the potential for a whole raft of thematic material that previously hasn’t been exploited. One or two a year doesn’t detract too much from the bulk of word-based puzzles.
Thanks, twencelas: that’s a comprehensive blog. I bet this one confused some Americans and other non-cricketing nations.
Sorry to digress a little but I was intrigued by Jaguar’s second paragraph. Do they really keep a database of all entries to the Listener crossword? I had always assumed that as soon as they found three correct entries, they threw the rest away unopened.
Personally I think it would be more reassuring if EV did not accept emailed entries. Some weeks I know I am wasting my time posting an entry because they must decide at some point whether the winner is going to come from the emails or the post.
Re query from Tony at #3, all Listener entries are checked and a record kept of whether an entrant solved the puzzle or not. This is done by the Listener statistician i.e. not the newspaper. This forms the basis of the annual stats, all of which are available on the Listener crossword website. I think the entries themselves may be kept for a while in case there were any disputes about whether a solver was marked correct, but not kept indefinitely, I think.
The annual stats for the Listener are incredibly comprehensive and it’s a credit to JE Green that he’s put in such an effort to compile personal records and full statistics for over 1,000 separate solvers submitting around 20,000 entries. Anyway, the result is that you can see patterns, one such pattern being the usual spikes in newcomer entries for the Numerical puzzles. It’s not clear how many go on to submit further entries, but they can certainly be fairly popular and form a welcome addition to the puzzle set. This might be true at the EV as well, though it remains to be seen whether or not we’ll continue to have one or two such puzzles per year at the EV. But variety is the spice of life, eh?
Thanks, nmsindy. That encourages me to enter more often knowing that every submission counts in some way. I’d had this image of my letter unopened at the bottom of a big sack that was being dumped into a skip.
Thanks, Twencelas, for the blog on the puzzle. A numbers puzzle was indeed a first for the EV, and I hope didn’t surprise solvers too much.
Tony, you’ll be pleased to know that the process for picking the prizewinner means that each submission for a week, whether electronic or hard copy, has an equal chance of being picked. If you’d like the very long full explanation, please email me and I’ll be happy to oblige.
Thanks for the blog, twencelas. I’m glad it wasn’t my week as I failed to solve it.
My problem was that 20dn is ‘the number of spectators’ and 11dn says that the children are spectating. It seems that, at least according to your blog, 20dn should refer only to adult spectators… or the children should be playing their Nintendo.
Dave.
Dave – 233 is the total number including children – take away the children (11dn) leaves 202 for 4dn to allow them all to drive home after 4 pints of cider. Of the 202 adults 13 are female (16dn). It’s good to know that the children were also well lubricated – the 31 of them downing 73 lemonades (26dn – I assume). So all teh adults had half a gallon of strong cider and the children were over sugared. No doubt the number of trees (27ac) surrounding the ground was the subject of some adult debate.
Like I said… glad it wasn’t my week 😯
twencelas says there are 202 adults. But 7ac is the number of males = 220, which means there are -8 females!
No, I’m afraid the setter has got himself in a twist and confused ‘total spectators’ and ‘total adults’. The numbers only work if you assume that at some point the 233 spectators change into 233 adults