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8x8 medium 18th Oct.
http://www.calcudoku.org/forum/viewtopic.php?f=18&t=633
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Author:  sica  [ Sat Oct 18, 2014 3:33 pm ]
Post subject:  8x8 medium 18th Oct.

I think that this puzzle has a mistake. I can't understand how over 200 users solve it.

a3=336x=7x48
b4=21x=7x3
a2=5

So a1=105x=3x5x7, 7 can be only on c1, 5 can be on b1 and 3 can be on a1.

e8=14x=7x2
c8=6x=6x1 or 3x2 but there is already a 2 on this row so c8=6x1
a7=384x=64x6=8x8x6x1 or 8x8x3x2

On row 8 there are already 1, 2, 6 and 7 and we must write an 8 and a (6 or 1) or (3 or 2).
The only one not used on this row is 3.
But we can't use 3 becouse there is one on a1 and anotherone on column b (b4 or b5).

Author:  wjgjr  [ Sat Oct 18, 2014 4:19 pm ]
Post subject:  Re: 8x8 medium 18th Oct.

Puzzle spoilers included:

a7 = 384x = 8x6x4x2 as well, which does not violate any of the constraints you listed.

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