shazbot wrote:...how did you get the 4 in the upper left corner?

r1c1 cannot be a 1,3,5 or 6 because those values are already in box 1.

It cannot be 7,2,or 1 because those values are already in column 1.

It cannot be 8,6,or 9 because those values are already in row 1.

That leaves 4 as the only option. r1c1 MUST be 4.

issac newton wrote:... what comes next, and WHY?????

A possible next move is to note that r4c2 and r4c6 are the only cells on row 4 that contain 1 and 3. Therefore

r4c2: {135} >> {13}

r4c6: {1234} >> {13}

Also note that r2c8 and r2c9 are the only cells in box 3 that contain a 1. Therefore one of those cells MUST be 1. Therefore you can remove pencilmark 1 from 42c4,5,and 6.

By the same logic, noting r5c4 and r5c5 are the only cells in box 5 to contain a 2, r5c8: {2459} >> {459}

Also note that r1c2 and r1c3 now contain {27} each, therefore

r1c4: {1235} >> {135}

r1c6: {123} >> {13}

r1c7: {2357} >> {35}

Now r8c1=3 since it in the only pencilmark 3 in column 1.

[Above edited on 20051014. It said r9c8 before. Don't ask me why]

Note that r1c6={13}=r4c6. Therefore we can remove pencilmarks 1 and 3 from the other cells in column 6. In particular,

r6c6: {134} >> {4} <--- A solution!

After filling that in, r5c5=2 and r6c9=7

Well, try to continue from there and post back if you can't complete it.

Mac