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January 2015

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This month's challenge was (far) too easy. Any uniquely solvable 2x2 Kakuro will do; for example,

3 A1 A2
4 B1 B2
3 A1 B1
4 A2 B2

For a bonus answer, a possible solution is:

3 A2 A3
15 B1 B2 B3
17 C1 C2
16 B1 C1
15 A2 B2 C2
4 A3 B3

If you want a harder version of this question, see February's challenge.

If you have any problems you think we might enjoy, please send them in. All replies should be sent to: