# 6 Cards, Top to Bottom

The six cards shown below were turned face down, "shuffled" (i.e.. re-arranged), and put in a pile, one on top of each other. Each card has an obvious numeric value. In addition, the "J", "Q", and "K" have values of 11, 12, and 13 respectively. make use of the information provided in the specifications to determine each card's position in the pile (top to bottom).

1. The seven is somewhere above the "Q".
2. The heart is somewhere above the spade.
3. A three is directly above the six.
4. One red card is directly on top of the other.
5. A black card is on the very top.
6. The bottom card's value is the sum of three other face down cards.

• Why are the images necessary? – Deusovi Jan 11 '17 at 12:17

Order

7 Clubs
3 Clubs
6 Diamonds
Q Hearts
K Clubs

Step 1

See clue #3

Our temporary stack is 3(unknown)6(diamond)

Step 2

See clue #4

Our temporary stack is 3(unknown)6(diamond)Q(Heart)
[only other red card is the Q]

Step 3

See clue #2

The heart(Q) is above the spade(3). So Q cannot be the last card.

See clue #6

The last card has to be the sum of three cards. Only the K and the Q can be the last cards. Since there is a card below Q, it has to be the king.

Step 4

See clue #1

The seven is above the Q. According to clue #2, the 3(spades) is below the Q. But in step two, our stack is fixed with three cards
So effectively (minus the K, which we know is the last card, the other five are)
7 3(Unknown)-6-Q 3(S)
(and obviously the 3 Unknown now is the 3 of clubs)

merging steps 3 and 4

since the last card is the king, we get the order

7-3C-6-Q-3S-K

• I think you did a great job – bill bsl Jan 11 '17 at 7:37

3 and 4 tell us that we have (36Q) in that exact order. 6 tells us K is bottom. 1 and 2 reveal final order. 5 is redundant.

1. 7$\clubsuit$
2. 3$\clubsuit$
3. 6$\diamondsuit$
4. Q$\heartsuit$
5. 3$\spadesuit$
6. K$\clubsuit$

@Karan Atree got the good answer first.

Here is how I found it.

- We know that 7 is before Q. (...7...Q...)
- There is only 1 heart and 1 spade and 1 must be above the other. (...7...Q...3...)
- Only K and Q can be the sum of 3 cards, but the Q cannot be last. (...7...Q...3...K)
- We know that one of the 3s must be on top of the 6 and that the 6 must be in contact with the only other red card (Q). So the 6 must be in front of the Q and the remaining 3 must be in front of the 6. (7 3c 6 Q 3s K)
- The first card is black so that condition is also respected.

• One small issue though, Q can also be the last card since it can be the sum of 3C, 3S and 6D – Karan Atree Jan 11 '17 at 7:29
• aaaaah, you are right. – stack reader Jan 11 '17 at 7:29

Late to the party, but here's what I got:

7C

3C

6D

12H

3S

13C

• Welcome to Puzzling! Can you explain why you think this is the answer? Answers without explanation are usually deleted. – Deusovi Jan 11 '17 at 14:44