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Katie · Joined: 05 Aug 2005 Posts: 13 Location: London

Given that R8C6 can be only be a 3, 6, or 9.

If =3, then R8C3=1, R8C1=5 leaving R8C7 with no option, as can only = 1 or 5.

If =6, then R3C6=2, R3C5=6, R5 centre = 5,2,9, R9C4=8, leaving R9C6=6, which impossible.

Thus the only option is 9.

No doubt someone will find a quicker and easier solution to this.

bird316 · Guest

So when will w know more [^]

chobans · Joined: 21 Aug 2005 Posts: 39

I don't know if this is easier but I'll make the solution more logical.

r1c4 and r1c6 both only can be 3 or 7. So you can eliminate 3 and 7 from other cells in box 2(r1c4-r3c6) and get 2 or 6 in r3c6.

row 8: r8c1 - {3,5}, r8c3 - {1,3}, r8c7 - {1,5}.

That makes it a naked triple, so you can eliminate 1,3 and 5 from other cells in row 8. So r8c6 becomes {6,9}.

At this point, you can say r7c4 can only be a 3 since that's the only place in box 8 where 3 can go.

column 6: r3c6 - {2,6}, r5c6 - {2,9}, r8c6 - {6,9}

Again, a naked triple. So you can eliminate 2,6 and 9 from other cells in column 6. So r9c6 becomes 8 after elimination. You'll eventually get a 9 at r8c6 after simple elimination now.

So I guess the system should've given you a 3 at r7c4 first. But since you asked as how you can get a 9 at r8c6...