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Archive for June, 2013

Listener 4244: At Spes Non Fracta by Chalicea

Posted by Dave Hennings on 21 June 2013

Another new face to the Listener this week, although she (I feel sure Chalicea is a she!) has an Inquisitor and an EV later this month, not to mention a Magpie, and none are [oops!] a first appearance.

Listener 4244Just 15 misprints to find in this week’s puzzle, a letter omitted from some wordplays, and then a bit of highlighting and line drawing at the end. The puzzle’s title is one of the Phrases and quotations from foreign languages at the back of Chambers, and is given as Latin for “but hope is not yet crushed” (although I’m not sure where the “yet” comes from). That didn’t help much at all. I resisted the temptation to google “anniversaries June 2013”.

1ac Japan accepted an Indian copper was JAWAN with the W omitted from the wordplay. A speedy solve seemed ahead of me. A few minutes later, after my first pass through the clues was complete, I wasn’t too sure, as some clues which looked straightforward, obviously weren’t. I put it down to the missing letters and misprints. Luckily, among others, SHIV SENA, GNARLED and AVOWABLE across, plus MAAING, ETNAEAN and VEENA down gave enough of a start for me to be reasonably optimistic.

After my third clue where the wordplay omitted a W, I wondered if the World Wide Wide was the theme. The fourth was also a W, and the fifth, and… Well, I guess you’ve been there, so you know that there were nine Ws omitted from the wordplay, and they affected fourteen clues.

The clues were solid, but I think my fellow blogger, Shirley, would have been quite suspicious of Chalicea (!) as she has no alcoholic ‘tipples’ anywhere in her clues. Indeed, she positively eschews the stuff with 6dn Stink about Alcoholics Anonymous sounding nannyish and 31dn After dinner drink, local tea, assuming not out of date, although the answer here was, indeed, a type of port not a pot of tea!

My favourite clue was 12dn Takes away the pips from close of BBC in heartless acts which conjured up the nightmare of the rights to the time signal being taken away from Radio 4.

The correct letters of the misprints spelt out Cherchez la femme, and there was EMILY DAVISON in the bottom row. Of course, I didn’t immediately spot ANMER in the top row since my knowledge of the Derby is minimal, despite being born and raised in Epsom. A bit of research and he was spotted, and a double-check showed him to be a him (a colt in racing parlance), whereas Emily Davison was definitely a woman and was the ‘character’ that needed highlighting. Well, given the theme of the puzzle, it had to be her really.

I wasn’t too sure of the thematic reason for dropping the Ws from the wordplay, even though they obviously meant Women. I guessed it was because they weren’t allowed anywhere near the ballot box.

Listener 4244 My EntryFinally, two phrases had to have lines drawn through them “as demanded by one of the characters and other 27ac”. 27ac, unclued, was BLUESTOCKINGS. The two main daigonals were the obvious places to look, and VOTES FOR WOMEN and DEEDS NOT WORDS ran the whole length of each. Two lines through gave them the vote in the form of a cross.

We finally had a grid packed with thematic entries and images that must have been tricky to put together. An excellent puzzle, so thanks to Chalicea.

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Pipes by Samuel

Posted by shirleycurran on 14 June 2013

Pipes by SamuelOh no! Not a carte blanche! The numpties read on in disbelief. There is an initial grid so there must be a different final grid and there are going to be six thematic clues with word-play only. Pipes? One of the numpties plays the bagpipes – are we on Scottish territory again? A touch of the pibroch? Music for a change from poker games? We abandon an early attempt to understand the preamble which is telling us how to rotate clues but highlight the word CAREFULLY lower down. That must be there for a reason (Don’t Listener solvers always attempt to obey instructions carefully?) and two final items to highlight. (We highlight that instruction too – what a gaffe to complete the whole thing and forget the highlighting – and it wouldn’t be the first time!)

A quick read through the clues shows us that Samuel isn’t in corny joke mode (as in so many of his hilarious crosswords like last month’s SHY NIECE WITH PURSE in the Magpie) but qualifies as usual for the Listener Setters’ tippling confrerie with is ‘Unoccupied evil journalist in New York, maybe, napped after drinks (8)’ Ah, but the drinks seem to be TEAS followed by E[vi]L + ED, as that is clearly a clue to rotate. I think ‘Unoccupied evil journalist’ has to go to the end of the clue which leaves 31 letters unmoved and that would give me an E.

I am beginning to despair of Samuel’s tippling tendencies as a series of rather negative surface readings appears. We have ‘frosty look’, ‘foolery’, ‘cruel’, ‘evil’, ‘All Saints’ missing out on ‘virility’, smacking a ‘tart’s bottom with a rubber bat’, ‘clash’, ‘weep’, ‘strive’, ‘spoil’, ‘accursed … about love’ and it is only in his last two clues that he redeems himself with ‘Dash of Rum and Orange, say, in rough bar in Glasgow (4) R[um] + ISP – well, my ISP is Orange – what a lovely clue) and Bottle the first of absinthe in village (4) A in VIL. Of course there is a different device operating for the down clues and I laboroiously work out that those two give me an S and a T.

March to May 2013 114Perhaps we are lucky that our solve gives us MUMBLERS as our first down clue (‘Bunglers getting money for fine could be unclear speakers (8) FUMBLERS with M for F) and a few other downs follow. That gives us an inkling about what is going on as our solving proceeds. The across clues don’t tally with the downs and we have a set of words that might well be thematic (GONG, PEAL, VESPER, BIG BEN) and a message emerging from the remaining across clues that is telling us to ROTATE  … We were told initially that the grid represents the curved surface of a thematic object so we begin to suspect that we are dealing with some sort of pipe or BELL? A quick visit to our Bradford confirms that those are indeed bells and she adds CROTAL and LUTINE to our list, providing the six we need.

Oh wonderful Mrs Bradford, she gives us another bell ‘TUBULAR!’ Haven’t I been playing that all week and isn’t this just about the fortieth anniversary? A quick Google check gives me May 25th 1973, so we have the theme and a good idea what is going to occupy one column and who the creator is (Mike Oldfield) who is going to occupy the symmetrically opposite column.

Oh but our solve isn’t over yet! Getting the across clues into the right order (and how clever of Samuel to make sure that there was only one order possible in each of them) was only half the battle. Counting the letters that hadn’t moved and converting them to letters was almost equally demanding. Take ‘Surrounding new style knight inhibits British noblemen (8)’ We unsnarl the clue to get Inhibits British noblemen surrounding new style knight (EARLS round NS + N = ENSNARLS), then we have to count those 23 letters that didn’t move to produce the last letter of the instruction: ROTATE EVERY ROW!

Wrong grid!

Wrong grid!

We haven’t finished yet. Our down clues give us a message too. First solve, then identify the letter that gives us the position of a key letter in the clue and hey presto! ‘THEME OF THE EXORCIST‘. Wiki told us that too so we know now what we have to do. We find an M and a T on the first line, and I and a U on the next, a K and a B on the third. Simples! We rotate the rows and what do we have? MIKE OLDFIELD/ TUBULAR BELLS.

We are patting ourselves on the back and preparing to copy this very satisfactory solution onto our sending copy when we remember that we highlighted that word ‘carefully’ in the preamble. What are Samuel and our Editors up to? Haven’t we rotated every row? Well, have we? I check carefully Tubular bells Right grid!and daylight dawns. How did they engineer that twist? We haven’t rotated row six! Back to the drawing board. (Actually I just cut my grid in half and got the desired result – every row rotated and there were TUBULAR BELLS followed by their creator MIKE OLDFIELD in the order given by Samuel in the preamble.

How beautifully this all came together with the circular nature of the thematic device echoing the tubular nature of the bells. And we got our pipes too, even if they were rather grunty and ughy Exorcist pipes. This was a massive challenge; thanks Samuel! It will certainly silence the critics who want a difficult one.

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Listener 4243: Pipes by Samuel

Posted by Dave Hennings on 14 June 2013

It was time for one of our more prolific setters this week and the man who is the new editor of the Enigmatic Variations series in the Sunday Telegraph. Hopefully that role won’t diminish his setting output too much. On top of everything else, Samuel seems to come up with novel clueing methods, and here we had a bit of shuffling required before the across clues could be solved. However, before that, I need to go back in time…

Listener 4243As I was ‘researching’ Skippers by Ifor, I googled ‘anniversaries in May 2013’. Not surprisingly, on one of the web sites that I visited was mention of the Dambusters raid by 617 Squadron in 1943 which was thte theme for that puzzle. Out of interest, I scanned other entries for May 2013 and came across 40 years ago (25 May 1973): “The album ‘Tubular Bells’ by Mike Oldfield was released. It was the first album released by Virgin Records.” Hmmm…Mike Oldfield, 12 letters; Tubular Bells, 12 letters; grid size of Pipes, 12×12. I resisted the urge to write to the editors and advise them against publishing thematic crosswords too near the date of any associated anniversary!

Anyway, back to the puzzle. Despite being on the lookout for an M and a T in the first row, I decided to tackle the down entries first. Here the clues were normal, and extracting a given letter (this seemed very generous) gave a thematic hint.

1 was MUMBLERS: well, that could be the M of MIKE, but the IKE didn’t fit. 3 ONRUSH and 5 TEXTER came next: that could be the T of TUBULAR, but again the other letters didn’t fit. 7 ORGY and 9 A-PER-SE finished my run of odd-numbered clues, 12 TRAINEE, 16 DOOMED and 20 VIAL coming next. Not a bad first trawl I thought, but my knowledge of the theme hadn’t helped at all.

What I did have, however, was the basis of the thematic hint: T•e•e•r•m••e•r•••t. That looked like it could start Theme from the, but I didn’t dash off to Google to see what our theme music had been used for.

On to the across clues. Here, apart from the six thematic entries, one or more words had to be moved from the beginning of the clue to the end before solving. My head spun! 4 King and unknown Belgian, often victorious in Tours, hired soldier (6). ‘Hired soldier’ shouted MERC, and Eddie MERKXC fitted the bill. That seemed too good to be true, and I put him in the third row and slotted MUMBLERS, ONRUSH and TEXTER through him.

The top half of the grid was taking shape as I pencilled in ORGY and A-PER-SE. I revisited 6dn Cut off control before eccentric Doctor’s first materialised (7;2) and, after a brief thought about the upcoming 50th anniversary of Doctor Who in November, I got REIFIED. This in turn led me to get 2ac Clubs finally fail following round (6) for the undefined CROTAL. Well, it didn’t need a definition as it was a bell, and all the thematic acrosses would probably be so too. I even pencilled LUTINE into the symmetrically opposite entry, but time would show that to be two rows too low.

The thematic tube that the grid represented (I didn’t submit the two edges stuck together) caused a bit of a hiccup with entries that started on the right of the puzzle and cycled round to the left: CLOSURE, TEASELED, FARRIERS and OBOISTS were the culprits here.

The grid was complete a relatively short while later, and the message spelled out from the across clue shenanigans was Rotate every row. Of course, I had the advantage of knowing the musical title and its creator, but those that were still in the dark were told by the down clues that it was the Theme from The Exorcist. If I had ever known that, I had also forgotten it.

We were told to carefully follow the across instruction — as if we don’t try and do everything in a Listener carefully! The preamble told us that MIKE OLDFIELD and TUBULAR BELLS were symmetrically placed, but this still gave two possible options for the finished grid. Rotating every row ensured that the title went on the left and the creator on the right.

Listener 4243 My EntryAnd so an excellent puzzle came to an end. I think I shall resist the temptation to google ‘anniversaries’ at the beginning of every month unless I get well and truly stuck on a puzzle. As it was, it didn’t take too much enjoyment away from Samuel’s offering, but that was more than made up for with the wonderful clue at 19dn:

Dash of Rum and Orange, say, in rough bar in Glasgow (4;3)

I felt a twinge of sympathy for overseas solvers who may not have known that Orange is the name of a European Internet Service Provider.

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Listener 4242 Killer Queen: A setter’s blog by gwizardry

Posted by gwizardry on 8 June 2013

As much as writing this is a pleasure in revisiting the subject following the thrill of the puzzle reaching publication, rather unfortunately I hadn’t at all thought about doing this at the time of creating the puzzle, and as such, much of the thought that went into it, as well as the luck (put that down to hard work, time and the fruitful results of trial and error if you will), has been forgotten.

So where to begin. Well a little about my solving history. I only came to the Listener on discovering Pentominos by Oyler, which in turn only came about from starting to read the Times on the basis of an excellent tongue-in-cheek opinion column on skiing from Giles Coren. Pentominos remains my favourite numerical Listener puzzle to date for its simple elegance. Since, I have at least completed the numericals that have followed and quite a few of the more interesting ones from the archives. On the literary side I fare rather less well, a half-cheated completion of Safe-Cracking by Kea is my best yet (stunning puzzle nonetheless). Still I’m making progress, so hopefully will be able to achieve a submission outside the 4 times per year I could currently manage.

So as many of you with a good memory will note, Pentominos shares the core mechanism with Killer Queen in that it is based on the unique factorisation of a number of products, as well as the use of the ‘0’ multiplier. However, that’s rather where the similarities end. I was browsing a couple of the excellent series on recreational mathematics by Martin Gardner which I had hoped to work into a puzzle, but had also been playing around with the idea of something based on games (games are covered a fair bit in those books). I don’t play cards much now, but I have a fair bit in the past and so somewhat accidently I came up with the idea of hands of cards which have to be deduced. It was a simple matter to work out a game which can meet the A*B<52 requirement (its nicer to use something closer to a full pack i thought) – so two players with 3 cards and we have a 6 by 8 grid. Incidently, although I myself had played ‘Brag’ (the three card variant of the poker class), I felt it might appear a little more accessible to readers of the preamble if called ‘poker’ as it is. This seems to have inevitably led to a mistake on my part, as a couple of solvers have pointed out to me – straights rank above flushes in brag, if not in 5 card variants I am more recently accustomed to – thank goodness there was no round with a straight vs a flush or it could have been catastrophic!

Anyway, back to the process, after selecting the game dynamic of 3-card Brag, I began to work around the idea of ‘a club sandwich’ either being the title or solution. incidentally that’s why the clubs appear on the same rows. After the first draft I had thought about reworking them to avoid this strange anomaly, but mostly laziness and lack of time but also an acknowledgment that the 1st draft wasn’t so bad after all kept me from doing so.

So as for the construction process. As Queen would be ranked 12 following a standard ordering, it was initially fairly clear that using it as a 12 would be heavy due to it having 2*2*3 as factors. I guessed this would make the solution either very easy or tedious (even more tedious than it is already!). So as J and K naturally fitted primes in 11 and 13, it seemed a good idea to use Q=0. Almost immediately the line She’s A Killer Queen hit me (or at least I could hear the song in my head and thought it fit well). At this point I dropped the club sandwich and thought of having a title answer which matched the Killer Queen theme. From this, the only way to work in an anagram was to use coordinates. I let this rest for a little, before working out a few of the hands roughly with the principle of using as many different types of deduction to be included in the solution. Such logical steps could use the following information:

  1. Remaining cards in the deck
  2. Remaining cards in each suit
  3. Remaining cards of certain face values
  4. Possible face value card positions based on the factors
  5. Possible suit positions based on suits listed
  6. Possible hands which fit the standard ordering rules
  7. Possible hands which fit the ‘winner’ result
  8. Possible entries which don’t invalidate what remains in other rows/columns (cross-matching)

The idea was that solvers would have to use different notions in the solution based on different combinations of information.

Early on I decided on the location of the 3 queens. This was necessary to avoid too many 0s and therefore a lack of information for solving. I didn’t want to use 4 as I wanted them to be different types of round and hands in each and also to save some of the Kings and Aces. I also had the loose idea of finding the other Queen (a hidden Queen of Hearts…), although nothing came of that.

I then proceeded to fill the face values for some of the rows (E, F and G particularly) though I forget the precise order. It was then a long process of trial and error and a strict checking of the logic to avoid missing any shortcuts. I didn’t mind the possibility of solvers being able to guess, as long as it was a precarious approach. I just had to make sure there was a single solution (this follows from the logical deduction…if you follow the logic correctly at each stage then there’s no risk in there being an alternative unless you’ve ‘guessed’ at some point).

So I’ll now review the first phase of solving (which matches well with this first phase of setting) [disclaimer….diagrams may contain errors as they were drawn up in haste for this blog and not during the setting process]

Solution 1st Phase


Keep a track of

a)      52 cards in the deck

b)      Number of Suit

c)       Number each Value

Factorise the rows and columns.

1)      GJ and GM are Q’s

2)      FJ or FM is a Q

3)      3 K’s else where so GI and GL are Aces

4)      E must contain JJJ1098 and Jacks are DHS so 10D 9D and 8D given 11s in columns I and K so  J J J 10 9 8 is hand order

5)      Row D are all clubs so must be 8 7 5 4 3 2 and so 8C in DL and so DI must be 7C

6)      3 7s remain all in rows B+C, so 2 must be in column I and as remaining factors

7)      Remaining J must be in FK and must be JC

8)      A must contain KK9553 and there must be KH, 9H, 5H, 3H

9)      AN must be 5 as no H and no K, and AK must be 3H so AJ is 5 and AL is K

10)   H contains K and no A’s as too many factors and so can’t be in middle columns, must be in HI and considering cross of suits in HI is H or S but must be S as H in and AN must be 5S


At this point the solver may reach somewhat of an impasse, in that a slight change in tack is required to continue the solving process. Although this pause does not coincide exactly with the second phase of setting there are key features which I fixed at an early stage. Whereas the the first phase used distinctive ideas for form the basis for the rounds (such as the location of the Jacks, Queens, Kings and 7s), the next phase of setting was more trial and error based. The solver must now consider in greater depth the possible combinations of each hand, particularly for rows B, C and D. The setting process involved continually taking a step back each time an element was added to ensure the placement of a new card didn’t reveal something to make things easier earlier on. I realised that if I could manage this and obtain a unique solution then the puzzle would remain challenging from start to finish. This isn’t always easy to achieve with the numericals, not due to any particular fault in the construction but simply as an artefact of the mechanism. So I was glad it seemed this puzzle would have a fairly constant or even increasing difficulty curve throughout.

The position of the remaining Ace was key here (the 4th being one of the 4 not played, in order to help maintain a viable solution) as considering it as a possible card greatly expands the combinations, this was also one reason to put 2 in front of the Queens in row (as 3 Aces in the other hands might have made it intractable).

Let’s now go through the second phase of the solution.

Solution 2nd Phase

11)   Row possibilities are 7-7-A-10-9-3, 7-7-9-A-10-3, 7-7-3-A-10-9, 7-7-9-A-6-5, 7-7-6-A-9-5 so 7 in column J

12)   Row includes either 8 8 4 2 2 or 8 4 4 4 2 so (as two 8’s placed and one 4 in row D) then this leaves one 8 and zero 4s or zero 8s and two 4s…and 7 is 7S

13)   Row must include two 10s and 6 otherwise need an 8 and a 4

14)   All remaining 2s 4s and 8s now fixed to B D H so has none, also only 10 left is 10C (others in which can’t be in C so 2 is in a 6, remaining values giving A 9 7 7 6 5 and 5 must be in CN and must be 5D (D or S but 5S taken) and AS (7s D+H taken) must be in CL (7s different suit)

15)   The 10s in H must be in J and L such that P1 wins (10H in L)

16)   FL must be 10

17)   BL BM must be 44 and then must be 8-2

18)   As in  must be H and D and so KH and 4H in column and also completes (no D for the K) and KC in FI

19)   7H JD also in I, 7D in CJ and 4D in BM

20)   Remaining and are spades


We’re now just a few elements from filling in the grid. There appeared to be some question over the uniqueness of the solution from some members of Answerbank, but as we’ll see this can be resolved using the right logic. Incidentally, this was a fault in the first draft, noted by Shane the co-editor. In order to resolve this it was necessary to move around the 6s, 3s and 2s to work in the possibility of a flush, only through avoiding this flush and therefore maintaining the ‘split’ round would the full solution be possible – as shown below.

Solution Phase 3

21)   Remaining Q must be in J

22)   GK must be S and only possibility is 3S (8S 4S 2S in B and 9S 6S in C

23)   3D in GN as 3C is in D

24)   Row  must be 742 448 for ordering

25)   8H not possible in column K so must be in and HK is D

26)   Remaining suits except Qs in and are now fixed

27)   Remaining clubs are 9 and 6 go into F

28)   No 3 factors in which fixes row and row D

29)   Row is then fixed

30)   Finally, QD must be in J to keep a split (otherwise flush)

Giving the coordinates AE GM CI AN AK OD resolving to A KIND OF MAGIC

Having completed the grid, it remains to work out the 4 word title. I was fairly sure that most could extract the ‘A _ _ _ _ of _ _ _ _ _’ or perhaps ‘A _ _ _ _ in _ _ _ _ _’ or ‘A _ _ _ _ if _ _ _ _ _’ and that seasoned solvers of the literery listeners would have no trouble at all.

In fact, working the title ‘A Kind of Magic’ into the puzzle had turned out to be quite troublesome. I had thought the song was called ‘It’s a Kind of Magic’ which had put me off it. Its of an odd number of letters and had a T and O so not fitting a simple way of defining the coordinates. This was a big disappointment as I liked how it related to my setter’s pseudonym (pronounced gee – wizardry as in ‘gwhizz’ and thus could equally be defined as ‘a kind of magic’), so I began to look at a list of Queen songs and albums but much to my delight spotted it was simply ‘A Kind of Magic’. All that remained was to arrange the rows and columns to fit the letters required. I played with quite a few names for the pack (yes pack or deck may be used depending on which side of the atlantic you’re from), in order to give the OD required.

I hope that the blog is of interest, though I fear it somewhat bursts the magic a little in that there was no ultra clever reverse logical engineering. Rather a classic exercise in the pragmatism of logical trial and error.

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Listener 4242, Killer Queen by gwizardry: Solver’s Blog by Jaguar

Posted by Listen With Others on 7 June 2013

This blog provides my solution to 4242, and may deviate from the intended route. Towards the end of the initial solve things became very messy so at times I may have changed what I did to put things in a more logical order than actually happened, but then hopefully that makes it easier to follow.

Ground Work

It’s useful to explain the main techniques first, rather than as we go along:

  1. Unless it’s vital, until all or most cards are in place I found that it can save time to neglect trying to resolve suits. The only times it matters immediately are when one hand has to be a particular flush in order to win, or if other constraints also fix a card unambiguously quickly. In general it’s enough to know that there are only at most four of each card.

  2. Notation: squares are labelled by row letter then column letter, so that the top-left is AI and bottom-right cell is HN, etc.

  3. The preamble for this puzzle uses face value to mean two different things. Firstly, the face value can mean the numerical value of the card for this number puzzle, where Queens are counting as 0 and Aces as 1. Secondly, face value means the ordering of the cards as they appear in each hand. Cards are ranked, in descending order, as A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, 2. This means that while the Queen and Ace have the lowest “number values”, they have high “ranking values”, so that a hand of a 2, Ace and Queen would appear as AQ2 not 2AQ. To avoid this confusion, this solution uses “face value” to mean the numerical values given in the puzzle; rank to mean the “ordering” of cards; and “strength” when comparing different hands such as flush and pair.

  4. We factorise all row and column numbers — which is the only real way to determine what card appears where.

  5. Finally, we will need to use the face values of all cards to factorise the entire used pack. As usual it’s best to do this in powers of prime numbers 2, 3, 5, 7, 11, 13. Note that an eight then counts as 2*2*2, so that an eight contributes three twos, and so on.


A  114,075 = 3*3*3*5*5*13*13
B  7,168 = 2*2*2*2*2*2*2*2*2*2*7 (ten twos)
C  13,230 = 2*3*3*3*5*7*7
D  6,720 = 2*2*2*2*2*2*3*5*7 (six twos)
E  958,320 = 2*2*2*2*3*3*5*11*11*11
F  unknown
G  unknown
H  124,800 = 2*2*2*2*2*2*2*3*5*5*13 (seven twos)


I  5,738,733 = 3*3*7*7*7*11*13*13
J  unknown
K  156,816 = 2*2*2*2*3*3*3*3*11*11
L  416,000 = 2*2*2*2*2*2*2*2*5*5*5*13
M  unknown
N  230,400 = 2*2*2*2*2*2*2*2*2*2*3*3*5*5 (ten twos)

The pack, meanwhile, is missing an ace, a Queen, a 6 and a 2, but otherwise has the following:

three Aces (1)
three Queens (0)
four Kings (13 each)
four Jacks (11 each)
four 7s
thirty “2s” (one each from the three 2s, three 6es and four 10s, plus twelve from the 8s and eight from the 4s)
fifteen “3s” (four 3s, eight from the 9s and three from the 6es)
and eight “5s”, one each from the 5s and 10s.

After all this laborious counting we can start to solve the puzzle!


Step 1

Listener 4242 Jaguar Blog Fig 0First, notice that in rows B, C, D all of the four 7s appear. Moreover, three of these appear in column I, so that we can immediately put three 7s in squares BI, CI and DI. In particular, cell DI is the 7 of Clubs.

Now consider row D. Because it has six cards the same suit, all face values have to be different — and none of the cards can be an Ace, because this isn’t used in the main grid. The only way to split row D’s face-value product, 6,720 = 2*2*2*2*2*2*3*5*7, into six different cards without using an Ace is to use 2, 3, 4, 5, 7 and 8. Also, Player Two wins this hand. That would mean that he either has a higher-flush (8 of Clubs beats the 7) or a straight flush, which would be 4, 3, 2, say. This option, though, is ruled out because we also know that whenever possible cards are ordered in descending order of rank. The 7 comes first in Player One’s hand, so he cannot have the 8, so Player Two has it. Therefore cell DL is the 8 of Clubs.

Now turn to row B. This was given as 7,168= 7*(ten 2s), so that now we have accounted for the 7, the remaining five cards must be 2s, 4s, 8s or (possibly) Aces. Counting powers of two if we used only 4s and 2s shows that we would fall one power of 2 short, so that there is at least one 8 in row B. Player Two must have this, for the only way Player One could have the 8 is if he had a pair of 7s (such a hand would be entered 7 7 8 in the grid). This is not possible — not enough factors of 7 in row B to use — so Player Two has the 8. How then can Player One win? He must have a flush, all Spades, so that his hand is 7S, 4S, 2S, as the remaining cards must be powers of 2 that are less than 7. Anything else would lose or would not fit — so we have Player One’s hand in row B. The 7 in CI must be the 7 of Hearts, as column I contains no Diamonds.

The next place to turn our attention to is row E. This has the factorisation:

958,320 = 2*2*2*2*3*3*5*11*11*11

Which tells us that row E has three Jacks and three other cards. The only way to absorb all of 2*2*2*2*3*3*5 into just three cards is to use 10, 9, 8. We know that columns L and N contain no Jacks so that if Player Two has any Jacks he has one in Column N. But this would result either in a pair of Jacks (imposibel) or hand that is not in descending order of rank. So Player One holds all three Jacks. How then can he lose? Note the rules that tell us that three of a kind beats a straight, but that a “straight flush” beats three of a kind. So Player Two must hold the 10, 9, 8 of the same suit — specifically, Diamonds. The remaining three Jacks are the JD, JH and JS, but we don’t yet know their order.

Step 2

Listener 4242 Jaguar Blog Fig 1The final Jack is hiding somewhere in column K, so must be in either row F or row G. What do we know about these hidden rows? Firstly, the fact that row G is a tie means that the two hands must match exactly. In particular, if there is a Jack in GK, then there would have to be a second Jack appearing in GN. This is not the case, as there is only one Jack unaccounted for. So the final Jack (which is the Jack of Clubs) is in FK.

We can also start to place the Queens. They are hiding in rows F and G, and Columns J and M. That places all Queens in three of the four cells FJ, GJ, FM, GM. In particular there must be at least one Queen in row G. And, because the hands tie, there must be two Queens in row G — specifically, in GJ and GM. What cards can go to the left of these Queens? The only possibilities, bearing in mind that hands are entered in descending order of rank or as pairs first, are:

AQx ; KQx; QQx — where x is the third (unknown) card. We discount the two Queens possibility as that would mean column I would also be “killed”. Also, there can only be one King in row G as there are two in row A and one in row H, so that only one King is missing. So the only option is AQx, so that cells GI and GL contain Aces.

There are three cards left to place in column I. These must include two Kings (the 13’s) and a 9 from the factorisation. The 9 counts as two 3s, though, and row H can only contain a single 3, so HI can’t contain the 9 and must have a King.

Returning to the hidden rows, note that in particular exactly six Clubs are hidden in these rows — only row D contains Clubs, and the Ace is not used — and these six are the K, Q, J, 10, 9, 6. That King is the only King in these two rows, so it cannot be in a pair, nor can it be in row G as it should appear to the left of a Queen. So the King of Clubs is in either cell FI, or FL (not FN as column N has no Kings). Also missing in the hidden rows is precisely one of the eight “5s” in the pack. That stops the 10 from appearing in row G, as this would force the other empty cell in row G to have a 10 so that the hands tie. Nor can it appear in FI or FJ, as that would mean a pair of 10s (the hand would be 10, 10, J). So the 10 appears in FL, FM or FN. Finally, there are also only two missing “2s” in rows F and G (you can count 28 2s in all other rows, out of 30). So the 6 cannot appear in row G either. The only place for the 6 of Clubs is then cell FN.

Recalling that cell FI is either a 9 or a King, we can try placing a 9 in FI, but find that this cannot work as then the only possible fit is 99JKQ6 with nowhere for the ten to go, or 99JK106 when the remaining Queen can’t fit anywhere either. So cell FI is the King of Clubs, and FJ must be the final Queen, so that FL and FM are the 10C and 9C. But then the only way for Player One to win in row F is if the Queen were the Queen of Clubs — “straight flush” beats “normal flush”, but “flush beats straight”. So row F is KC QC JC 10C 9C 6C. Finally, there are two missing factors of 3 in the hidden rows, so that the final two cells in row G are both 3s; and cell AI is the missing 9.

Step 3

Listener 4242 Jaguar Blog Fig 2The remaining two Kings appear in row A. With Player One holding a 9 as his first card, it must be Player Two that has the pair of Kings. That accounts for all but the factors 5*5*3 in row A; splitting these between three cards can only be done using two 5s and a 3. Again, Player One can’t hold the pair of 5s as these would appear on the left of his hand, so Player Two has one 5 and Player One the other 5 and 3.

At first glance Player One now has some junk — but we are told he wins this hand over a pair of Kings. He does not have a straight; he must have a flush, of three Hearts. (Player One’s having a lot of luck so far — two flushes to open and a straight flush in hand F!)

We’re down now to just thirteen totally empty cells. Let’s turn to row B. We know that the remaining cards contain seven factors of 2. This can be done by 448 or 882. But we can be even more specific than this, as the only 2 unaccounted for is the 2 of Diamonds — we haven’t fixed the 2 of Clubs yet, but it’s in row D somewhere, and the 2 of Hearts is unused. The two 8’s missing are the 8 of Hearts and 8 of Spades. Suppose we did finish off row B as 882. Then look at column L. On putting an 8 in BL the only missing factor in row L is 5, so that the last two cards would be a 5 and an Ace, one of which goes in HL.

Note that Player One has to win in row H. But his hand is heavily constrained already by the lack of available cards. The only hands better than “King high” are all ruled out:

i) A pair of Kings means five Kings appearing in the grid;
ii) A straight would mean KQJ using a fifth Jack (and also “killing” information about row H due to the Queen);
iii) A flush is impossible as there are only two cards in any one suit appearing in row H.

Therefore Player One has only got King high. Any hand with an Ace automatically beats this. If cell HL had a five instead, then possible hands are:

55x ; 543 ; 542 ; 532.

55x and 543 both would beat King high, while 542 and 532 both use a fourth 2 when there are only three in the grid — one appearing in BK, one in BN and the third in row D somewhere. So cell HL can have neither an Ace nor a 5, so cell BL cannot have an 8. Therefore Player Two’s hand in row B is 448, with the 8 being the 8 of Spades as column N has no Hearts. Now cell HL can be either an Ace (ruled out already as above), or a 2, or a 10. If it were a 2, then Player Two’s hand in row H could only be 22x which again beats King high. So HL is a 10, and CL is the final Ace.

Step 4

Listener 4242 Jaguar Blog Fig 3Now’s the time when it definitely helps to have a pack of cards handy. Laying aside all the cards that are accounted for, we can see that the eight cards remaining are, ignoring suits:

10, 9, 8, 7, 6, 6, 5, 2.

While we’ve not yet placed them, the 2, 3, 4, 5 of Clubs all go in row D, and there are still suits to be sorted out for a few cards yet.

Return to row H. The factorisation was:

124,800 = 2*2*2*2*2*2*2*3*5*5*13

We have so far accounted for the 13 and 5*2 = 10, leaving six 2s , a 5 and a 3. That factor of 3 has to appear as a 6 because all of the 3s are accounted for. This leaves five 2s and a 5 in three cards. The only way we can do this using the remaining card values is by using 10, 8, 2 (8, 5, 4 would use a fifth 4). The 10 cannot appear in HM, as this would give Player Two a winning pair, so Player One has that 10 in cell HJ.

Meanwhile row C contains the 9, 7, other 6 and the 5. Player One wins this hand too, and can do so with either the hand 776 or the 765 straight. In this second case, that would give Player Two a hand of A97. But row N does not contain a factor 7, so we exclude this, giving Player One 776 and Player Two A95.

To complete the number fill, we start with column K. Dividing its product 156,816 by the face values of all cards in column K leaves 156,816/(11*11*2*3*3*6) = 12. So the remaining cards are either a 6 and a 2 or a 4 and a 3. Cell HK can only contain 8, 6, or 2. It cannot be 8 (which now goes in HM) but cannot be 2 either, as DK would then be the already-used 6 of Clubs. So HK is a 6 and DK is the 2 of Clubs.

This leaves a 2 to go in HN. Now looking at column N we see that the remaining card must be 230,400/(5*8*5*8*6*3*2) = 4, or the 4 of Clubs. So DM is the 5 of Clubs, and DJ is the 3, meaning that all card ranks are known for each cell.

Step 5

Listener 4242 Jaguar Blog Fig 4

All that remains is to determine the suits of each card. For the most part this is obvious as several times only one suit is left for that number. The information given quickly resolves the remaining ambiguities. If Player Two holds all three Spades that appear in row C, for example, then he would have a winning flush, but Player One wins, so that CK is the 6 of Spades not the 6 of Diamonds; the 6 of Diamonds is then in HK, which is the only Diamond appearing in columns K or I — making EJ the Jack of Diamonds. Column K also contains two Hearts which cannot both be 3s, so that GK is the 3 of Spades, EK the Jack of Hearts and EI the Jack of Spades.

We can also use simple counting. Allowing for the unused cards — a Club, two Hearts and a Spade — then we count twelve Spades appearing in rows A to E and row H. The thirteenth Spade, we just saw, was the 3, so that no more Spades can appear in row G. This fixes CL as the Ace of Spades — making CN the 5 of Diamonds, in turn giving AN and HI as Spades and GI as the Ace of Hearts and GL as the Ace of Diamonds. Then we see that the remaining cards in column L (AL, BL, HL) are all Hearts.

Finishing off gives us:

Listener 4242 Jaguar Blog Fig 5What of the two Queens? These can be suited, because they have to be a Heart and a Diamond, but if GM held the Queen of Diamonds then that would give Player Two a winning flush. So he has the Heart and Player One the Diamond Queen.

We can complete the puzzle by reading off the coordinates of the cards:

KC, QH, 7H, 5S, 3H, 2H.

Note that the Two of Hearts is given as in the “Outstanding Deck” = OD. The other five cards are seen to be in FI, GM, CI, AN, AK respectively, giving FIGMCIANAKOD or, rearranging, “A Kind of Magic”, which, along with “Killer Queen”, is a song by the band Queen.


Final solution:

Listener 4242 Jaguar Blog Fig 6

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