Can anyone help with this? Is there a logic for when there are 3 numbers to go in 3 spaces eg c2 sq 1 (5,8,6) but there are no double pairs?So as example 6 can go in R1 and R3. 5 can go in r2 and r3 and 8 could go in r1,r2,r3
Thanks
IJ wrote:Zeb,
All the 3s in col 7 are in box 3, so are the 7s...
gmc wrote:sorry to confuse things but
3 can't go in r5c9, so must go somewrere in c9r-1-3, so must go somewhere in c7 r4-6, so in the end must go in c8r9
Anonymous wrote:IJ wrote:Zeb,
All the 3s in col 7 are in box 3, so are the 7s...
Hi IJ,
Forgive my ignorance, but I'm not sure I follow what you say. I thought that there couldn't be more than one 3 in one box! The position that I reached suggests that a 3 could go in c7r2, c7r4, c7r6 or c7r8. The 7 in c7 could also go in r7 or r8. I'm still confused........
IJ wrote:Anonymous wrote:IJ wrote:Zeb,
All the 3s in col 7 are in box 3, so are the 7s...
Hi IJ,
Forgive my ignorance, but I'm not sure I follow what you say. I thought that there couldn't be more than one 3 in one box! The position that I reached suggests that a 3 could go in c7r2, c7r4, c7r6 or c7r8. The 7 in c7 could also go in r7 or r8. I'm still confused........
Sorry, I meant col 9 - this probably clarifies things. I am a muppet sometimes! You can probably see that the possible placements for 3 in that column are all in box 3, and likewise for 7s (because r5c9 can't be a 3 or a 7), which means that r2c7 is an 8...