Trick with 13 cards of one suit

There's really no magic to the trick; I just used a brute-force method in my analysis. Here is how I figured it out:

I will use b1 to mean the first burn card, b2 for the second, etc.

So the 13 cards at the beginning must be:
A b1 2 b2 3 b3 4 b4 5 b5 6 b6 7

So, you already know where seven of the cards go.
Now, the next cards after you put down the 7 card are the first six burn cards:
b1 b2 b3 b4 b5 b6

You must burn a card after you put down the 7 and that would be b1.

So, the cards you have in your hand after that are (in order):
b2 b3 b4 b5 b6 b1

So, just match up the last six cards since every other one must be burned:

 b2  b3  b4  b5  b6  b1       
 8       9       10            b2=8 then burn b3 then b4=9 then burn b5 then b6=10
     J               Q         then burn b1 then b3=J then burn b5 then b1=Q 
             K                 then b5=K.  

 
Therefore,
b1 = Q
b2 = 8
b3 = J
b4 = 9
b5 = K
b6 = 10

Now, plug those back into the original sequence of cards and you have your answer!

A  b1  2  b2  3  b3  4  b4  5  b5  6  b6  7
A  Q   2   8  3  J   4   9  5  K   6  10  7

I used the same analysis to figure out the answer for the entire deck.
I used H = Hearts, D = Diamonds, C = Clubs, and S = Spades.

The order would be:

A H
A D
2 H
A C

3 H
2 D
4 H
Q C

5 H
3 D
6 H
2 C

7 H
4 D
8 H
8 C

9 H
5 D
10 H
3 C

J H
6 D
Q H
J C

K H
7 D
A S
4 C

2 S
8 D
3 S
9 C

4 S
9 D
5 S
5 C

6 S
10 D
7 S
K C

8 S
J D
9 S
6 C

10 S
Q D
J S
10 C

Q S
K D
K S
7 C

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