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geoff h · Joined: 07 Aug 2005 Posts: 58 Location: Sydney

Need help with today's puzzle - I got as far as shown below. However, the site gave me a hint as Nr 7 in r1c5 and I just can't figure out how they arrived at this. I had Nrs 1 and 7 as possibilities for this cell and can't see how they eliminated the Nr 1?? I have Nrs 1,2 and 7 as possibilities in r1c3 and r1 c9. Plus I've got Nrs 1 and 7 all over the place in column 5, so can't see how they eliminate Nr 1 in r1c5??? It's driving me crazy!

3 9 - 8 - 5 6 4 -
- - - - - - - - 5
4 5 8 - 9 - 3 - -

- 8 4 3 - 2 - - -
2 3 9 - 5 - 1 - 4
- - - 9 - 4 - 2 3

9 - - - 2 - 7 5 8
7 - - 5 - - - - -
8 - 5 - - 9 - 3 6

Thanks

Guest · Guest

Hello Geoff,

I'm stuck on this too but I haven't progressed as far as you. If you don't mind me picking your brain, how did you arrive at the '3' in the centre of the middle left 3x3? (Apologies for not using row and column notation as I am not familiar with it. I think it is the Nr 3 in row5 column2 I am asking about.)

Thanks for any assistance.

Guest · Posted: Sat Aug 20, 2005 2:46 pm Post subject:

r1c5 can only be 1,7

But in the middle square (r4c4 - r6c6), 1 can ONLY be in r4c5 or r6c5. So we can eliminate 1 from the rest of the cells in c5.

So r1c5 ONLY can be a 7.

After that, the grid solves itself.

August 20 puzzle · Guest

Dear Guest No.2,

Ah yes, very good. Thanks for your help - that slipped by me!!

To Guest No. 1,

Yes, your notation is correct. You are enquiring about placing the Nr 3 in r5c2? This is because r5c2 is the only cell in row 5 in which 3 can appear.

It can not appear in 5c4 or r5c6 because there is already a 3 in that square ( r4c4 ). It also can not appear in r5c8 because there is already a 3 in r9c8.

Hope this helps.

sar3000 · Joined: 23 Aug 2005 Posts: 1

3 9 - 8 - 5 6 - -
- - - - - - - - 5
4 5 8 - 9 - - - -

- 8 4 3 - 2 - - -
2 3 9 - 5 - 1 - 4
- - - 9 - 4 - 2 -

9 - - - 2 - 7 5 8
7 - - 5 - - - - -
8 - 5 - - 9 - 3 6

I can not figure how to place the 3 in 6c7 or 6c9. What am i missing?

Thanks

chobans · Joined: 21 Aug 2005 Posts: 39

In box 4 (r4c1-r6c3), the 7 can ONLY go in row 6 (r6c2 or r6c3). So we can eliminate 7 in the other columns in row 6. Since the ONLY possibility in r6c9 was 3 and 7, we can eliminate 7 and get 3 in r6c9.

Guest · Guest

I see the logic of the responses once at the point described in the first post. I have had trouble getting to that point, though. I have the same grid, except for the 4 in r1c8, and the 3 in r3c7, neither of which I've been able to get.

How did you get to a 4 in r1c8?

How did you get to a 3 in r3c7?

Thanks,

Brien

chobans · Joined: 21 Aug 2005 Posts: 39

In row 2, if you noticed, 8 and 9 only shows up in column 7 and 8. So, eventhough, r2c7 can be 2,3,4,8,9 and r2c8 can be 1,4,7,8,9 it can really only be 8 and 9 for both cells.

So after eliminating other numbers from r2c7 and r2c8, 3 in box 3 (r1c7-r3c9) only shows up in r3c7. So r3c7 can ONLY be 3.

Same with 4. In Box 3, 4 only shows up in r1c8. So r1c8 can ONLY be 4.

Guest · Posted: Fri Aug 26, 2005 10:00 pm Post subject:

Thank you chobans, I now see what I was missing. The rest is cake.

Thanks,

Brien