If a figure can be tiled by N 2x3 rectangles it is trivial to place the same number of U pentominoes in that figure.
Find the smallest N such that at least N+1 U pentominos can be placed in a figure tiled by N 2x3 rectangles.
It's possible lo place 10 ps in 9 2x3 rectangles.
Suppose a grid 10x8
I give the square of the grid in which the simmetric or central element of the pentomino is placed.
If the higher dimension of the pentomino is in vertical position I add a V, if the legs are up a U, down a D, right a R, left a L
38D
26U
25D
33U
67D
55U
53VR
72VL
85D
93U
The left superior square of rectangles is
18V
15V
38V
35V
57H
55V
75V
95V
52H
with H Horizontal and V Vertical for the rectangle

Posted by armando
on 20160309 17:47:22 