A partial deck of cards is laid out in the 5x5 grid below. No card has a value less than 8. Place the remaining cards given the clues provided. Warning: this is a tough one

You were right, it is hard, but I am determined to get this
Nice puzzle.

Step by step with this one. Every entry is logical, there are just many factors to keep in mind as you solve. Don't give up!

flotsam! I looked at it last night for a pretty long time but I'm not sure I can go back to it.

Step by step is right - until you hit one that can't be right

I'll wait a couple of days to see if anyone can solve it completely. If not, maybe we can try a group step-by-step solve.

I am half way there I think

Or not

Remember that not every card from 8 - A of each suit is used. 25 out of 28 are used.

This one is a puzzler for sure...I almost have it

Kind of close to half of a pinochle deck of cards sans the 8s

Soot did get it and he sent me the answer. I'll leave this up for others to try.

Congrats to soot!

Congrats indeed

This is a puzzle from the 2012 World Puzzle Championship, so of course it's a toughie. Here's an explanation of the solution:



1. In row 3, r3c1 and r3c4 are both hearts, so one must be part of the 4-of-a-kind, and the other is unmatched. In particular, r3c3 and r3c5 must both be one of the set, and r3c2 and r3c3 must be a diamond and a club.
2. In column 5, r3c5 must be the unmatched card, the remaining being 4-of-a-kind.
3. The two 4-of-a-kinds cannot have four 8’s, J’s, Q’s or A’s, as one of each of those sets already appears elsewhere in the grid.
4. Since both r3c3 and r5c5 are part of a royal flush, they must be the 10 and K.
5. The 10 or K of hearts is in r3c1 or r3c4, and the 10 or K of spades is in r3c5. Thus the royal flush must be clubs. (This forces r3c2 to be diamonds.)
6. In row 4, the straight must be 8 through Q, Thus r4c5 must be 10 (and not a K). So column 5 is four 10’s and row 3 is four K’s. r1c5 must be spades.
7. Column 4 is a full house, and whatever appears in r3c4 must appear once or twice more in c4. Since all the K’s are in r3, r3c4 cannot have the K of hearts; so it must be in r3c1.
8. In column 2, we need three-of-a-kind. It can’t be K’s, so there are two options: three 8’s or the remaining three cards are all the same. The latter can’t be true since two of them are clubs. Thus it’s three 8’s in r1c2, r4c2, and r5c2.
9. In column 1, the K cannot be part of a pair. Thus r5c1 must be part of a pair in c1. But it must also be part of a pair in the diagonal (since the remaining four cards are unpaired). The K’s and 10’s are all used, and there is only one 8 unused, so r5c1 must be an A.
10. Similarly, in row 1, the 10 cannot be part of a pair. Thus one pair is 8’s. 8 cannot appear in r1c1, nor in c4 (the full house must be A’s and either J’s or Q’s from r4). Thus r1c3 is an 8.
11. In column 3, r4c3 must pair with one of the other cards. But in r4, it must be a J or Q. Also r4c4 must be a J or Q. Thus the 9 in r4 must be in r4c1.
12. In column 1, the remaining cards to be paired are 9 and A (all K’s are used). r1c1 is part of the royal flush, so these must go in r2c1 and r1c1 respectively.
13. In row 1, r1c4 must be A.
14. In r5c1, the A must be hearts.
15. In column 4, r5c4 can’t be an A (the A of spades just got placed in r1c4), so it must match what is in r4c4, which must be a J or Q. Thus r4c4 and r5c4 are both J’s.
16. In row 4, r4c3 must be Q.
17. In the royal flush diagonal, r2c2 must be Q.
18. In column 4, all the A’s are used, so r3c4 must be J.

I am still determined to get this!

I have faith in you, flotsam!

You can do it flotsam!