Arrange nine digits in a 3x3 array such that each different digit that appears appears as many times as its cardinality, so for example, if the digit 3 appears at all, it appears 3 times. Also the eight 3-digit numbers formed from the rows, columns (top to bottom) and two diagonals (also top to bottom) must all be different.

Also, the sum of the nine digits used cannot itself be written using the digits you've included in the array.

What is that sum of these nine digits?

There are eight different ways to draw digits as specified (9 81 72 612 63 513 54 423); seven of these sums cannot be written using the digits (only 1x1,3x3,5x5, sum 35 could be so written). Starting with the 2x2, 3x3, 4x4 option (sum 29) the following arrangement gives eight different three-different row/col/diag numbers: ROWS: 442, 234, 343; COLS; 423, 434, 243; DIAG: 433, 233.

The sum of these nine digits is 29. (Since I just started with what seemed the most likely, I do not know if this is unique.)