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 formednumbers 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
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