To try to simplify the problem, I extracted powers of two from the factors going in to 1C!:
Below is a tally accumulating powers of 2, extracted from the given number at the left. When each reached 16 (decimal) I add 1 to the tally of powers of 16 in the rightmost column and extract (divide) the cumulated power of 2.
what's left
after removing accumulated count of extracted
factor power of 2 power of 2 powers of 16
2 1 2
3 3
4 1 8
5 1
6 3 16 1 1
7 7
8 1 8
9 9
A 5 16 1 2
B B
C 3 4
D D
E 7 8
F F
10 1 128 8 3
11 11
12 9 16 1 4
13 13
14 5 4
15 15
16 B 8
17 17
18 3 64 4 5
19 19
1A D 8
1B 1B
1C 7 32 2 6
When the "accumulated power of 2" column shows two values, the first is the one just after extracting a power of 2, and the other after having converted a factor of 16 onto a new tally in the last column.
The table shows that there should be only six trailing zerosnot the eleven shown in the puzzle. The remaining 2 in the "accumulated power of 2" column indicates that the last nonzero digit should be even, though what that should be would depend on the "what's left" column. It probably would be easy enough, though boring, to multiply these mod 16 to identify this digit. As for the lettered digits and the remaining zeros that shouldn't be zeros, I can't see doing it without mechanical aid.
A verification via computer does validate that there are indeed exactly 6 trailing zeros and that the digit to the left of those is indeed an even nonzero digit. It also shows that the leading digits shown in the puzzle are correct and that the total number of digits is correctly given in the puzzle; it's just that the first five zeros shown after the w should not be zeros.

Posted by Charlie
on 20110618 16:51:21 