This is the 4th entry from the Pencils puzzle week. The next puzzles will get bigger (and maybe harder ?), but for this one you should not have any surprises after solving the last 3.
Have you noticed ? For a pencil of size N, you have 2N+1 squares occupied by that pencil. You can decompose the number of white cells into sums of odd digits (1 is impossible) : If you have 7 leftover squares, you must have a length 3 pencil; is you have 10 squares, it's either a 1 and a 3, or two 2s. You don't need this tip for now.
Check the oldest post for the rules.
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