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In column 6, 1 can ONLY go in box 5(r4c4-r6c6) so you can eliminate 1 from other cells in box 5. So r4c4 - {7,8} and r6c4 - {5,8,9}.

If you look at box 5, {1,2} can ONLY go in r4c6 and r6c6. This i ...

In column 3, 8 can only go in box 7(r7c1-r9c3) so you can eliminate 8 in other cells in box 7. So r7c1 - {4,5} and r7c2 - {1,2,4,5}

r7c1 - {4,5}, r7c3 - {1,4,8}, r7c4 - {1,5,8}, r7c8 - {1,5,8}

Nake ...

If you look at box 2(r1c4-r3c6), 2 HAS to go in row 2. The 2 at r8c5 prevents 2 from being able to fit in at r1c5 and r3c5. So you can eliminate 2 from other cells in row 2. So even though r2c9 CAN ...

In column 1, 3 can ONLY go in box 1(r1c1-r3c3) so you can eliminate 3 from other cells in box 1. So r1c3 becomes {8,9}.

In column 1, 9 can ONLY go in box 1 so you can eliminate 9 from other cells in ...

Sudokus are fun... but don't neglect your family to play it! Just get them hooked on it as you are. Then you all can neglect one another. :D

r3c6 should be {1,9}. You have a 5 in there but that's ...

Just do as much as you can and then share with us as where you got stuck at. Then we can help you to get to the next step and go from there. Everybody takes different routes on getting to a certain ...

r2c5 only has possibilities of {2,8}. r6c5 also only has possibilities of {2,8}. So you can eliminate 2 and 8 from other cells in column 5.

So even though r8c5 CAN be {2,5}, you can eliminate the 2 ...

Solution #1 has two 2s in column 5 and two 7s in column 6.

r3c8 and r3c9 can only be {2,5} so you can eliminate 2 and 5 from other cells in box 3(r1c7-r3c9). So you get {3,4} at r1c9 and {3,4,6} at r2c9.

r1c9 - {3,4}, r2c9 - {3,4,6}, r6c9 - {3,4,6,9}, r9c9 ...

I'll try to explain it with elimination method.

At your position, these are the logical steps I took.

In column 1, 6 has to be in box 1(r1c1-r3c3) so you can eliminate 6 from other cells in box 1. ...

2 in row 9 can ONLY go in box 8(r7c4-r9c6) so you can eliminate 2 from other cells in box 8. So r8c4 becomes 4 or 9.

This creates a pair at r7c4 and r8c4 of {4,9}. So you can eliminate 4 and 9 from ...

Yeah, you have two 4s at column 4. Your solution was almost way off from the final answer. That column 4 probably messed you up.

I have found the cells referred to by Chobans: 2,8 ( R2,C6 ); 1,8 ( R8,C7 ); 1,8 ( R8,C8 ) and 3,8,9 ( R2,C7 ) but still cannot understand what this tells me. Please explain!!

Actually it's r8c6 a ...

Perhaps someone else can elaborate the rule?
Two questions:
Is my elimination of 7 from r1c2 and r3c2 (described above) correct?
What cells are involved in the XY-wing?

Well, 6 is a possibility at r9 ...

Click on the "Draw!" button at the bottom.

http://www.simes.clara.co.uk/programs/sudokutechnique6.htm

That site has some great methods with easy to understand explanation. I possibly can't explain better than he did.

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 r3c ...

You need X-Wing for this.

By looking at the grid, 8 has a X-Wing at r2c6,r2c7,r8c6 and r8c7. So you can eliminate 8 at r1c6, r1c7, r4c7 and r9c6.

See if you can solve it after those eliminations.

17th March

xxx 91x xxx
3xx x2x x1x
175 xx8 629

x52 xxx 496
6xx 5x9 237
xxx x6x 185

5x6 8xx 94x
x8x x5x xx2
xxx x94 xxx

the hint is 1 in r5c3.Why?

I had to think a bit about this one.

In row 7, 7 ...

Second problem is more recent 7th August

8x1 xx4 7xx
xx5 7xx x38
6x7 9x8 45x

5x9 xxx xxx
x7x x9x x15
xxx xx5 3x9

x52 1x7 xx4
78x xx9 xxx
xx6 8xx 5xx
the hint is 1 in r4c6.

In box 1(r1c1-r3c3), 3 c ...