LinkedIn Mini Sudoku #272 Answer & Analysis

Top-left box (R1-2, C1-3) has three initial givens (including a 5 at R1C3 and 3 and 4 at R2C2/R2C3). Column 1 already contains a given '1' at R3C1. That blocks any '1' in the box positions that lie in column 1 (R1C1 and R2C1). The only remaining cell in that box that can hold the digit '1' is R1C2, so R1C2 = 1.

After placing R1C2 = 1, check column 3. The column already contains 5 (R1C3), 4 (R2C3) and 2 (R5C3) as initial givens. The rows that already contain a '1' (R1 and R3 and R4 — R3 and R4 have initial '1's in other columns) eliminate those rows as candidates for '1' in column 3. The only remaining row that can take '1' in column 3 is R6, so R6C3 = 1.

With a couple of forced placements established, re-scan each 2x3 box. For example, middle-right box (R3-4,C4-6) already has a 5 and a 1 and a 3 among its givens; only three digits remain to place there. Because column and row exclusions block some of those positions, the remaining candidates fall into place quickly by elimination.

After the earlier forced numbers, many cells become single-candidate cells. Place those digits and keep iterating: each new placement removes candidates from its row, column and box. Continue this scan/eliminate/place cycle until every cell is filled.

Once the grid is filled, verify each row, column and 2x3 box contains digits 1 through 6 exactly once. If any duplicate appears, backtrack to the last non-ambiguous step and re-check candidates. In this puzzle the normal elimination sequence produces a unique completed grid.