Various comments so far:

1.Antirice asked if I can repeat the numbers: Answer is yes: ie.. [ [1,2],[4,2]] is allowed.

2. pzbagel: In mentioning of next if $n =~ /0\d+/;. \d+ was necessary to because I just don't wanted the formed-numbers that begins with 0. Anywhere else 0 in the number is fine. For 2x2 matrix this should work fine.

3. I like Rhose's appraoch as initiated by CountZero.

4. Other apporach:, Start forming matrix starting from the highest divisor.

4. Few other related challanges are

• Find a matrix for each divisor possible with lowest possible total of numbers in the matrix. ex.. [[1 2] [3,0]]:Total = 6
• Origianl matrix can have numbers with K digits. K > 1. Example: K= 2  [ [12, 34],[56,78]]
• It could be NXNXN Matrix : Example: 3 x 3 x 3 Matrix: If we take K= 1 here, according to Rhose's approach we will have 1000/27 = 35 as maximum divisor.

Update: tall_man's answer at Re: Re: Matrix Formation is a good but rather cumborsum attempt. The best answer for 3x3 matrix -> Highest divisor is '44'. I am sure coming up with 4x4 or 15x15 matrix answers would be quite challanging.

artist