Red wrote:Looks like I have the solution now though. The approach I used in the first row, box 2, 3 and box 6 of row 1, all shared the digits 235. From here I eliminated the 2 and 3 as options in column 2, box 3, leaving just the 1 as the only number to fill that spot. From here I solved the rest of the puzzle relatively easily.
First of all, what you are calling "box" is more commonly known as cell or square. Cells are commonly referred to by their row (r) and column (c) numbers, e.g., r1c2, r1c3, and r1c7 for the 235 naked triplet you identified.
Your elimination of 23 from r3c2 was a "happy accident" -- eliminations that turned out to be correct even though your logic was faulty. When a naked triple exists in a row, you can only make eliminations of those digits in other cells of that same row.
how do I determine the locked candidate in box 7 that you spotted? Am I just missing it?
I'd be willing to bet you didn't spot it because of your interpretation of "box". Sorry, I didn't say "3x3 box" to be clearer.
My 'locked candidate' hint was concerning the digit 2 in box 7, the lower left 3x3 box. Since all the candidate 2s are in the intersection of box 7 and col 3, a 2 cannot exist anywhere else in col 3.
Solving Sudoku by Angus Johnson has a good tutorial of the basic techniques.
Techniques at the Sadman Sudoku site is also recommended a lot.