The complete set of all ominoes from 1 to 5 will cover 1 + 2 + 6 + 20 + 60 = 89 squares and 89 = 64 + 25 = 8·8 + 5·5 so it may be possible to cover squares 8×8 and 5×5 using the complete set. This is done in the following examples, where the two squares are joined to form a single polygon and the aim is to have as few or as many cuts of the edges as possible.

H. D. Benjamin (*FCR* Jun/Aug 1937 ¶2777-8) the first shows minimum cuts of the edge (9), the other shows maximum cuts of the edge (26).

H. D. Benjamin (*FCR* Aug/Oct 1937 ¶2854-6). The same problem as above but with the squares differently aligned. The second diagram is Benjamin's solution showing maximum cuts (26), but the minimum cuts (9) was achieved by solvers (H. Fahlander, G. Fuhlendorf, S. H. Hall, H. Perkins, W. H. Reilly and T. R. Dawson) and it is not stated whose solution the diagram is.

W. H. Reilly and G. Fuhlendorf (*FCR* Oct/Dec 1937 ¶2925-6), showing 9 and 27 cuts respectively.

The final example (which I have rotated 90 degrees) is by H. D. Benjamin (*FCR* Aug/Oct 1937 ¶2857) and shows 29 cuts.

W. H. Reilly (FCR Apr/Jun 1938 ¶3164), dedicated to G. Fuhlendorf.

These examples by W. E. Lester (*PFCS*/*FCR* June/August 1936 ¶2297-8) ask for the chessboard to be covered by the maximum number of different pieces (16) of sizes 1 to 5 with the 2×2 on the edge and showing the minimum (7) and maximum (20) number of cuts of the edge The pieces used are all those of sizes 1, 2, 3, 4 (9 pieces in all) plus 7 of size 5. The same selection of 5-square pieces is used in each solution.

T. R. Dawson (FCR Dec/Feb 1946/7 ¶7090). Hollow rectangles. 8×9 minus 2×3, 7×10 minus 1×4, 6×11 minus 0×5 (degenerate case).

H. D. Benjamin (FCR Aug/Oct 1948 ¶7793). Half-hollow rectangles, width 3.

These two versions of the Cross of Lorraine by H. D. Benjamin (*FCR* Jun/Aug 1944 ¶6010, Aug/Oct 1944 ¶6087) are formed with two sets of [5]s and two dominoes. If the dominoes are removed the patterns close up to outline two rectangles 2×4 and 2×6.