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Thanks for the feedback guys!

My intended path was (Spoiler, I guess): If 15 was a sum (456), you cannot obtain 10 in any way, so it is 1x3x5, then 10 is surely 4+6, so you can place the 2 in that column, this prevents the 4 from being 2x1x2 or 1+2+1, only 1x4x1 remains. Now 16 is found out to be a sum, as its 4th row cell can only be 5 or 6, and it surely contains a 6 in one of its 3rd row cells. From there I think it goes nicely.

Also I like how 6 in three digits is 123 regardless of the operation, it helps too.

The only (kinda) sad part is 11, I was trying to make it so that no operation is obvious at the first sight. Oh well!

Thank you both very much!

While I don't generally do that, I inserted some uniqueness logic in this one (it was possible to solve without using it, harder though.). Glad you liked it!