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This Python program runs in 47ms.

Solution:(a) The slab cannot be completed forn=6; (b) Forn=7 the second row is a descending sequence.In general for

n> 6 there is a solution with the bottom row of the form:with (

n– 7) 1’s in the middle.As can be seen all but 4 of the numbers in the bottom row are 1, hence there are

n– 3 1’s overall (as required).There are two 2’s in the bottom row, hence 3 2’s overall.

There is one 3 in the bottom row, hence 2 3’s overall.

Whatever

n– 3 is there is only one of it in the bottom row, hence 2 occurrences overall.The remaining numbers don’t appear in the bottom row, so only appear once in the overall grid.

In case it’s of interest, I found no slabs for n < 4, two for n = 4 (2, 3, 2, 1 and 3, 1, 3, 1), and one for n = 5 (3, 2, 3, 1, 1).