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Guest
Nov 14, 2022

@Guest#f2b1 what was your pencilmarking to get the r8c9 as the only remaining possible place for 4?

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Guest
Jan 27, 2023
Replying to

I too would like to know what their pencil marking was to eliminate 4 from c9 in box 3 if r2c1 is 7...

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Guest
Nov 14, 2022

congratulations on missing the point, david :)

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Guest
Nov 14, 2022

I was able to solve the puzzle without using uniqueness.


After pencil marking,


Step 1. R8c9 is 4 since whether r2c1 is 4 or 7 it eliminates 4 from c9 in box 3. R8c9 is the only remaining possible place for 4.


This places 8 at r8c7 and 7 at r7d7.


Step 2. 67 must be on the diagonal at r9c1, and c4r6. 8 can only be on the diagonal at r6c4. The 9 must be on the diagonal in box 3. And, r5c8 is 8 by sudoku.


step 3, r5cs is 24, it is 245 by sudoku, but five breaks the puzzle.


step 4, By sudoku 9 is on row 2 of box 2, and, 4 at r2c5 breaks…


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