Each of A and B is a tridecimal (base 13) positive integer, where A is formed by writing the digit 4 precisely 666 times and, B is formed by writing the digit 6 precisely 666 times.
Determine the distinct digits that appear in the tridecimal representation of the product A*B.
Note: For an extra challenge, solve this problem using only pencil and paper.
list
10 cls
20 for Psn=0 to 665
30 A=A+4*13^Psn
40 next
50 print A
60 for Psn=0 to 665
70 B=B+6*13^Psn
80 next
90 print B
100 C=A*B
105 Rw=40:Cl=79
110 while C>0
120 Dig=C @ 13
130 C=C\13
135 locate Cl,Rw
140 print mid("0123456789abc",Dig+1,1)
145 Cl=Cl1:if Cl<0 then Cl=79:Rw=Rw1
150 wend
155 locate 0,42
160 print
OK
finds the product expressed as tridecimal, is
2222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222222222
222222222222222222222222222222222222222222222222222221aaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab
where a represents ten and b is eleven.
So the distinct digits are 2, 1, A and B.
Partial explanation:
44444...44444
X 66666...66666
In the usual multiplication algorithm, each line of partial product is derived:
4*6 = 24 decimal = 1B tridecimal, so the rightmost digit is B and there's a carry of 1.
The position to the left would also be a B, but the carry makes it a C, and there's still only a carry of 1. So this continues with C's all the way to the left (665 of them), and the final carry gets placed as a solitary 1.
Each successive partial product looks just the same, but offset to the left one position:
1CCCCC...CCCCCB
1CCCCC...CCCCCB
1CCCCC...CCCCCB
1CCCCC...CCCCCB
...
1CCCCC...CCCCCB
1CCCCC...CCCCCB
1CCCCC...CCCCCB
1CCCCC...CCCCCB
On the rhs, B is carried down to the final position. Then C + B = A with a carry of 1. Then C+C+B+1 = A with a carry of 2, etc. In tridecimal the C is similar to the 9 in decimal acting like 1 mod the base, and being offset by the increased carry at each successive position.
The last time this works, though is the leftmost column that consists of all C's and a B  the 666th row, where the B has shifted 665 positions to the left, to be directly under the first C in the first row. The next column to the left then starts with a 1, has the same number of C's as the previous column, and no B. So compared to what we were expecting, the C has been replaced by a 1, and there is no B. Mod our base, replacing a C with a 1 is like going up 2. Eliminating the B is like going up another 2. Going four up from A (10) mod our base (13) brings us to the 1 that shows up in the middle of the product.
This is the point where I lose track of the carries, in explaining the series of 2's to the left.

Posted by Charlie
on 20100807 14:41:52 