Kakuro
Tutorial -- Challenging
27 in 6 has a limited number of remaining valid permutations. With the first two cells being 2 and 1, 3 and 1, or 3 in 2, we know that the remaining cells will be limited to 5, 6, 7, 8, and 9. The right-most cell is part of 12 in 4, and 5 and 6 are the only valid values for that. Given that, the remaining two cells are limited to 7, 8, and 9.
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