Killer Queen by gwizardry
Puzzle explanation
The coordinates of the listed cards plus OD for 2H (in the outstanding deck)
gave the letters of A Kind of Magic,
a song by the band Queen, who also wrote Killer Queen.
The positions of the 48 cards could be deduced without using that information.
Puzzle solution process
This is one possible logical path to a solution.
The products of face values are first resolved into their prime factors
(none of which can be greater than 13),
which helps particularly with seeing where the Ks (13), Js (11) and 7s go.
As the QS is on the outstanding deck,
the other three queens must be in rows F and G, in columns J and M.
Row G is a split round (the face values are identical on both sides),
so cells GJ and GM must both contain Qs, the other Q being in cell FJ or FM.
The hands in row G can’t be pairs of queens,
so the first card in each must be higher, either K or A.
The prime factors for row A are 114075 = 3^{3}.5^{2}.13^{2},
so the row contains two Ks and two 5s (not 10s, as there are no factors of 2).
To account for 3^{3}, the two remaining cards must be a 3 and a 9.
Four cards in the row are hearts (therefore of different face values),
so these must be KH, 9H, 5H, 3H, and the other K and 5 are a diamond and a spade.
Column N contains no hearts, so cell AN must be D or S; it can’t be the K
(because column N’s face values are 230400 = 2^{10}.3^{2}.5^{2}, containing no 13s),
so cell AN contains 5D or 5S.
Because each hand is ordered by decreasing face value,
the only place for the 3H is then cell AK, and then the only place for the other 5 is cell AJ,
so the row’s face values are _53 __5, the remaining values being K, K, 9.
Player 2’s hand can only be KK5 (pair) with Player 1’s hand as 953,
or K95 (high card) with Player 1 having K53.
The only way Player 1’s hand can win is if it’s a flush,
so the row is either (9H 5H 3H; K_ K_ 5_) or (KH 5H 3H; K_ 9H 5_);
we can fill in (_H 5H 3H; K_ __ 5_).
Row B (7168 = 2^{10}.7) has one factor of 7,
row C (13230 = 2.3^{3}.5.7^{2}) has two
and row D (6720 = 2^{6}.3.5.7) has one, which accounts for all four 7s.
Column I (5738733 = 3^{2}.7^{3}.11.13^{2}) has three 7s,
so they must be in cells BI, CI, DI, with DI = 7C (all cards in row D are clubs)
and the other two being 7H and 7S (there are no diamonds in column I).
The other five cards in row B account for its ten factors of 2;
they can’t be five 4s (there are only four 4s in the deck),
so the row has at least one 8.
Player 1’s hand can’t have an 8 or A
(the pairs 778 and 77A aren’t possible as there’s only one 7 in the row),
so its face values must be 742 (the pairs 744 and 722 would be out of order).
Player 2’s hand has to account for seven factors of 2,
so it can only be 882 or 448.
The only way 742 can beat either of those is if it’s a flush,
and the only suit that row B has three of is S,
so Player 1’s hand is (7S 4S 2S).
As the 4S and 2S are now spoken for,
the fourth spade in row B must be the 8S (and the remaining 82 or 44 are a diamond and a heart).
As 7C and 7S are now identified, CI = 7H and the remaining 7D is elsewhere in row C.
In row D (all clubs), the five unknown cards account for 2^{6}.3.5.
There’s no A, as the AC is on the outstanding deck,
so the three cards not having a factor of 3 or 5 must be 2, 4 and 8 (to be distinct);
thus the row’s cards are {8C, 7C, 5C, 4C, 3C, 2C} in some order.
The 8 can’t be in Player 1’s hand (a flush starting with 7),
so Player 2 has a flush starting with 8 (DL = 8C).
Row E (958320 = 2^{4}.3^{2}.5.11^{3}) has three Js (JD, JH, JS),
leaving the other three cards (all diamonds) to account for 2^{4}.3^{2}.5.
The only possible combination is 10 = 2.5, 9 = 3^{2} and 8 = 2^{3}.
Column L (416000 = 2^{8}.5^{3}.13) has no J,
so Player 1 has JJJ (three of a kind) and Player 2 has 10D 9D 8D (straight flush).
The fourth J (JC) is in column K (156816 = 2^{4}.3^{4}.11^{2});
it can’t be in cell GK because the split result would require another J in cell GN,
and none of rows AD and H has a product with a factor of 11, so JC must be in cell FK.
Row H (124800 = 2^{7}.3.5^{2}.13) has a K and five other cards
to account for 2^{7}.3.5^{2}.
The 3 and the two 5s must all apply to distinct cards
(as face values of 15 and 25 aren’t possible),
and each can be combined with at most one 2
(face values of 12 and 20 aren’t possible),
so at least four factors of 2 are accounted for by the other two cards,
each of which is a power of 2 (ie, 2, 4 or 8);
no A is possible, as that would leave 16 for the other face value.
The K might be the third card in a hand (if it’s a pair),
but neither of columns K and N contains a factor of 13,
so the K is the first card in one of the hands.
It can’t be in cell HL because column L’s only K is already fixed in cell AL,
so cell HI contains a K.
It can’t be KC (no clubs in row H),
KD (no diamonds in column I) or KH (which is somewhere in row A),
so HI = KS.
Now that the KS is taken, the two Ks in row A are KD and KH (no clubs in row A),
so the spade in row A must be the 5: AN = 5S.
Row C starts with 7H and has the 7D in either CJ or CM
(none of columns K, L, N has a factor of 7).
The other four cards (one diamond and three spades) account for 2.3^{3}.5,
so their face values could be
{10, 9, 3, A}, {10, 3, 3, 3}, {9, 6, 5, A}, {9, 5, 3, 2} or {6, 5, 3, 3}.
We can rule out {10, 3, 3, 3} because
it would need three different suits but only D and S are available.
For {6, 5, 3, 3}, the 3s would have to be 3D and 3S, making the other two 6S and 5S,
but 5S is already used (in cell AN), so that’s ruled out.
Similarly, as 5S and 2S (in cell BK) are both already used,
{9, 5, 3, 2} would have to include 5D and 2D, but only one diamond is available.
For {9, 6, 5, A}, the 5 would again have to be 5D, making the rest all spades;
Player 1 would have to have a pair of 7s
(otherwise that hand’s values could only be 765,
but 5 can’t be in column K, which has no factors of 5);
column L (no factors of 3) can’t contain 6 or 9,
so the only possibilities for the row would be (7H 7D 9S; AS 6S 5D) and (7H 7D 6S; AS 9S 5D).
For {10, 9, 3, A}, the 9 and 10 would have to be spades (the 9D and 10D are used in row E);
Player 1 would again have to have a pair of 7s
(there aren’t enough values less than 7 to make a highcard hand starting with 7)
and the third card can’t be 10 (no factor of 5 in column K);
as Player 1 wins with a pair, Player 2’s hand can’t be all spades (a flush);
the possibilities are (7H 7D AS; 10S 9S 3D), (7H 7D 9S; A_ 10S 3_) and (7H 7D 3D; AS 10S 9S).
So we can enter CJ = 7D, CK = _S, CM = _S, and we know 9S is somewhere in row C.
We know {2C, 4C} are in row D and 8S is in row B.
The remaining unplaced cards for powers of 2 are {2D, 4D, 4H, 8H},
of which row B has a diamond and a heart with a product of 16, ie, {2D, 8H} or {4D, 4H}.
Row H’s powerof2 cards can be, at most, either {4D, 4H} or {2D, 8H}
(ie, whichever pair isn’t in row B);
either way it has (at least) another 2^{3} to account for,
which can only be combined with its factors of 3 and 5 to make values of {6, 10, 10}.
The 10D is in cell EL, so row H contains 10H and 10S.
As no more factors of 2 can be accommodated in the multiples of 3 or 5,
the maximum for powerof2 cards (either {4D, 4H} or {2D, 8H})
is also the minimum (ie, rows B and H together contain all of {2D, 4D, 4H, 8H}).
The 6 can’t be the 6C (no clubs in row H),
6S (both spades in row H are accounted for)
or 6H (in the outstanding deck).
Thus row H consists of {KS, 10H, 10S, 6D, 4D, 4H} or {KS, 10H, 10S, 6D, 2D, 8H} in some order.
As the 10S is now in row H, the row C possibilities with 10S are ruled out
and we’re left with (7H 7D 9S; AS 6S 5D) or (7H 7D 6S; AS 9S 5D),
so we can enter CL = AS and CN = 5D.
In row H, Player 1 wins with a hand starting with KS;
that can’t be a flush (not enough spades),
straight (no queen)
or pair (the two 10s would be listed first),
so it must be a winning highcard hand,
which means Player 2 can’t have a pair of 10s either.
Each player therefore has one 10;
the ordering of hands means that Player 1’s 10 must be in cell HJ
and Player 2’s 10 is in cell HL.
The one in HL can’t be the 10S (column L’s only spade is the AS in cell CL),
so HL = 10H and HJ = 10S.
Column L has three factors of 5;
two of them are accounted for by the 10D and 10H;
the other can’t be a 5 (they’re all spoken for in rows A, C, D),
so it must be the only remaining 10, the 10C, in row F or G (not B, which has no factor of 5).
That accounts for all of column L’s factors apart from 2^{2},
which means cell BL can’t be 8 = 2^{3},
so Player 2 in row B has {4D, 4H, 8S} in some order, not 882.
Cell BL can’t have the 4D (both diamonds in column L are accounted for),
so the hand is (4H 4D 8S) in that order.
That rules out one of the options for row H,
so its three gaps are now {2D, 6D, 8H} in some order.
The remaining cell in column L has an A;
the only suits left for that A and the K in column L are D and H.
In row G, cells GI and GL have the same face value, which has to be higher than a Q,
so it’s an A or K.
The KS is in cell HI and we know KD and KH are in row A,
so there can’t be two Ks in row G.
Thus both cells must have As, the only available ones being AD and AH.
AD can’t be in column I (no diamonds), so cell GI = AH and GL = AD,
which makes AL = KH and the only place left in column L for the 10C is FL.
The remaining heart in row A has to be AI = 9H and the remaining cell is AM = KD.
Cells GK and GN have the same face value;
the only face value still available in more than one suit is 3 (3D and 3S);
both of column N’s spades are accounted for, so GN = 3D and GK = 3S.
In column I, the only remaining factor is a 13,
and the only unplaced K is the KC, so cell FI = KC.
The only remaining suit in the column is S, so EI = JS.
In column K, the remaining factors are 2^{3}.3^{2}.
Cell CK is one of {6S, 9S},
DK is one of {2C, 3C, 4C} (not 5C, as there are no factors of 5 in column K)
and HK is one of {2D, 6D, 8H}.
HK can’t be 8H because then DK and CK would both be powers of 3, which isn’t possible for CK;
so HK is one of {2D, 6D} (definitely a diamond) and the remaining heart is in EK = JH,
which makes EJ = JD (the remaining diamond in row E).
Row H now has suits D and H left; column N has C and D; so cell HN is a diamond and HM is a heart, therefore HM = 8H.
The remaining suit in column N is a C in cell FN.
The only available clubs are 6C and 9C;
the 9 doesn’t fit in FN (there’s only one factor of 3 left for column N), so FN = 6C.
HN is one of {2D, 6D} (from row H) but the factors left for column N are 2^{3},
so HN = 2D and HK = 6D.
The remaining factors of 2^{2} in column N make DN = 4C.
Then the only possible places for the 3C and 2C in row D are DJ = 3C, DK = 2C,
leaving DM = 5C to complete the row.
The remaining factors of 2.3 in column K make CK = 6S,
leaving CM = 9S to complete the row.
Row F is (KC __ JC; 10C __ 6C);
one of the gaps is a Q and the other is the only remaining card, the 9C.
Player 1 can’t have the 9C
because (KC 9C JC) would be a flush with cards listed in the wrong order, so FM = 9C.
Player 2 has (10C 9C 6C), a flush, but Player 1’s hand beats it,
so it must be a higher flush, namely FJ = QC for (KC QC JC).
Row G is (AH Q_ 3S; AD Q_ 3D) and the queens are QD and QH.
Player 2 can’t have (AD QD 3D), a flush,
as that would beat Player 1 but we know the result is a split.
So GM = QH and GJ = QD.
The grid is complete.
The locations of the cards listed in the preamble are
cells FI, GM, CI, AN, AK and OD (the outstanding deck).
Jumbling these coordinates gives A Kind of Magic,
a reference to the setter’s “wizardry” and the title of a song by Queen,
like the puzzle’s title.
