# Smallest square that can pack thin digits

If we draw the digits 0 to 9, segmented into squares, across a rectangle of 2x5 (except the 1) they use up 81 total squares.

Is it possible to pack them all into a 9x9 grid.

What is the smallest n by n grid that can pack them all?

Reflection and rotation are allowed.

Update

The above picture is a grid 9x11 and has 18 gaps.

After the answer below in a grid of size 9x10 leaving 9 gaps. I've discovered a solution in a grid of 8x11 with only 7 gaps

Is 8x11 smallest size rectangle with the fewest gaps?

• do they need to be listed in any particular order? Feb 6, 2023 at 14:10
• Not at all. Rotations and mirroring is also allowed
– Maff
Feb 6, 2023 at 15:25
• No smaller rectangles than 8x11 are possible. The only candidates are 7x12, 6x14, 5x17, 4x21, 3x29, and 2x43 but none of them can take all the pieces. Feb 6, 2023 at 21:49

The smallest square that can contain the digits is

10x10.
Putting them in a 9x9 square is not possible unfortunately. By hand it does quickly become obvious that not all nooks and crannies of the digits can be filled, I don't know of a simple way to prove it impossible. A computer search quickly exhausts all possibilities.

The pieces do fit in a 10x9 rectangle, even without rotating/mirroring any pieces. For example:

The 7 is sandwiched between the 2 and 4, but all the other pieces can be fairly freely permuted.

I believe I can prove that 9x9 is

Impossible

The structure of this proof is:

I will place the 3 without loss of generality, then prove that no set of figures can fill both of its gaps.

Argument to follow:

$$1$$. Place the 3 anywhere.
$$2$$. Rotate the grid so that the 3 is oriented like a numeral 3.
$$3$$. If the 3 is in the top two rows, flip the grid vertically. (The center of the 3 is now not above the center of the grid)
$$4$$. Observe that if 2, 5, 6, or 9 are used to fill the bottom gap, they must either cover the top gap (disallowed) or 3 spaces below the 3 (which would be out of bounds, by line $$3$$).
$$5$$. This means that the bottom gap must be filled by 1, 4, or 7, which must be oriented horizontally.
$$6$$. Now attempt to fill the top gap with any available numeral. Observe that it creates a new gap to the right of the middle of the 3, which is flush with the numeral from step $$5$$.
$$7$$. This new gap also can only be filled by 1, 4, or 7.
$$8$$. Now consider the space below the numeral from step $$5$$. This space can only be filled by the numeral 4, because any other option would leave an un-fillable space under the 3.
$$9$$. The 1, 4, and 7 are now all spoken for (steps $$5$$, $$7$$ and $$8$$). The upper gap of the 3 must be filled by 2, 5, 6, or 9. There is now a 3x1 gap above the 3's long vertical line.
$$10$$. Observe that no remaining numeral can fit in this gap when oriented vertically.
$$11$$. Observe that if any remaining numeral is placed horizontally in the top of the gap, it creates a 1x1 gap with a 1x2 filled space below it, and no remaining numeral could fill that space.
$$12$$. Observe that if any remaining numeral is placed horizontally in the bottom of the gap, no remaining numeral can be placed above it.

Therefore: There is no possible position of the 3 in a packed 9x9 square.