Friedman's chopping 2003 puzzle:
(reminder to myself to work on this one :-)

Emrehan Halici's puzzles:
1. 3 x 668
2. 12 x 167
3. 334 x 6 (leaving one of the diagonal bits of the 7 in place to form part
of the x)

Marek Penszko's one-odd-digit digititis:
Since there is no remainder the numbers in the last subtraction must be
identical, x04 with x even. In the second subtraction, we must have 2xx -
20x because subtracting 10x leaves too big a difference. The ones digits are
the same (since they subtract to 0) so we must have a partial product of
202, 204, 206, and 208. 202 and 206 do not factor into a 2-digit number
times a digit, so it must be 204 = 51 x 4 = 68 x 3 = 34 x 6 or 208 = 52 x 4
= 26 x 8. However, the only possible x04 for the 208 cases is 104, which
puts two 1s into the puzzle. So the second (and third) partial products are
204. Then the only valid factoring is 34 x 6 because the other ways make us
use two odd digits (two 3s in the 68 x 3 case). Thus, the second subtraction
is 224 - 204 = 20. The first partial product must not contain any odd digits
and both 34x4 = 136 and 34x8 = 272 do, and 34x2 = 68 is too small, so it
must again be 34x6 = 204.

  666
22644 : 34
204
 224
 204
  204
  204

I guess it *is* diabolic!

Dave Millar's room puzzle:
Is it intended (as in Erich Friedman's puzzles I had not seen before) that
the rooms must be of equal sizes? Otherwise there are many solutions. There
are still multiple solutions with this constraint, but not so many:

xxxx
zzzx
zyyx
zyyx
zyyx
zzyy

xxxx
zzxx
zzyx
zyyx
zyyy
zzyy

December 23 Colonel Sicherman question:
In my original solution to the logical hats problem, I explained why the man
with the largest number on his hat always figures it out first. The
logicians can figure this out in the same way, so when one does so, the
others immediately they do NOT have the largest number, so their numbers are
the difference between the numbers they see.

Dave Millar pentominoes:

ddddc
adccc
aacbb
aaeeb
eeebb

This is rather easy once you notice that the pentominoes must stay aligned
to the grid, and simply number each cell of the pentominoes.

Joseph DeVincentis