Sudoku-YFoundations
Killer combinations: a set is not an order
A two-cell total of 11 does not automatically mean 5 and 6. Keep every combination that still fits.
Use the total and the cage size together
On a 9×9 board, two distinct digits totaling 11 can be {2,9}, {3,8}, {4,7} or {5,6}. There is no 10, so {1,10} is unavailable. Repetition is forbidden inside a cage: a two-cell total of 8 cannot use {4,4}, even across different rows.
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Our example: filter, then arrange
Let A1 and A2 form a cage totaling 11. Suppose neither cell can contain 3 or 4 after checking placed digits. This removes {3,8} and {4,7}, leaving {2,9} and {5,6}. Neither surviving set alone tells us the answer.
Later, suppose A1 has {2,5} and A2 has {6,8}. Test positions as well as sets: 2+6=8, 2+8=10, 5+6=11 and 5+8=13. Only A1=5 with A2=6 fits this partial scenario.
One combination can still have two orders
If both cells instead have {5,6}, the set is fixed but its order is not. Either 5–6 or 6–5 could fit. Use a row, column or box restriction to decide which cell gets which digit; do not promote a combination into two placements without that extra step.
Check simple bounds first
Three distinct 9×9 digits total at least 1+2+3=6 and at most 7+8+9=24. If you read a three-cell cage as 5, revisit the boundary or printed clue. These bounds catch reading mistakes, but passing them does not make every arrangement legal.
Try before looking
Try before revealing
Two cells totaling 10 both have {4,6}. Can you enter 4 in the left cell?
Show explanation
Not yet. The set is fixed, but its two orders remain possible until another constraint separates them.
References and further reading
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