dailysudoku.com

[lolumens]

I got this far:
--7 -25 ---
-58 -73 4--
--3 894 571
--5 2-8 -17
7X- 9-6 -58
38- 517 6--
53- 76- ---
--6 38- 795
87- 45- 1--

The hint was a 1 where I have an X
I don't understand the logic, can someone help me?

[geoff h]

Sorry if my reply is a bit convoluted, but here goes;

- You can eliminate Nr9 in r6c3 because it is already in r7c3 and r9c3 ( must be a 9 in the 3 x 3 square ).
- This leads to a pair of 6,9 in r4c1 and r4c2. Therefore, r4c7 must be Nr 3 and r4c5 must be Nr 4
- This leads to a pair of 1,4 in r5c2 and r5c3.
- Eventually, leads to a pair of 2,4 in r8c1 and r8c2 which means r8c3 must be Nr1.

You can solve it from there.

Cheers.

[geoff h]

Sorry, typo - I meant r7c3 mst be Nr 1 and therefore Nr 1 is eliminated from r5c3.

[lolumens]

Your explanation was very helpful. Thank you.