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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]
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