note artist Various comments so far:<p> 1.[Antirice] asked if I can repeat the numbers: Answer is yes: ie.. <code>[ [1,2],[4,2]]</code> is allowed.<p> 2. [pzbagel]: In mentioning of <code>next if \$n =~ /0\d+/;</code>. \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. <p> 3. I like [Rhose]'s appraoch as initiated by [CountZero]. <p> 4. Other apporach:, Start forming matrix starting from the highest divisor.<p> 4. Few other related challanges are <ul> <li> Find a matrix for each divisor possible with lowest possible total of numbers in the matrix. ex.. <code>[[1 2] [3,0]]: </code>Total = 6 <li> Origianl matrix can have numbers with K digits. K > 1. Example: K= 2 <code> [ [12, 34],[56,78]] </code> <li> 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.<p> </ul> <b>Update:</b> [tall_man]'s answer at [id://266681] 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.<p> [artist] 266572 266572