Moments on Perl or other Programming Issues

a b c
d e f
g h i

Statement: The center term, "$e" , is 5.
Proof:
Given the sum of each "segment" is 15.

Consider
  sum of 2 outer rows - the middle column - sum of 2 diagonals 
= (2-1-2)*15

Then we have
(a+b+c + g+h+i) - (b+e+h) - (a+e+i + c+e+g) = (2-1-2)*15
-3*e = -15
e = 5  □


Then loop the values of $a and $b over 1 to 9

$i = 10-$a
$c = 15-$a-$b
$h = 10-$b
$g = 10-$c
$d = 15-$a-$g
$f = 15-$c-$i

first (top) layer
a b c 
d e f 
g h i
 
second (middle) layer
j k l 
m X n 
o p q 

third (bottom) layer
r s t 
u v w 
x y z 


1+2+3+..+27 = 27*28/2 = 27*14
There are exactly 9 pillars composing the cube. 
Since the sums of each pillar are the same, we have 

             27*14/9 = 42 ,

as the sum of each "segment".


Statement: The center term, "$X", is 14.
Proof:

 Consider:

 (sum of 4 corner pillars)
 - (sum of 2 outer columns of the middle layer) + (m+X+n)
 - (sum of 2 outer columns of the middle layer)
 - (sum of 4 space diagonals)


  (4+2+1-2-4)*42 = a+j+r + c+l+t + i+q+z + x+o+g
                - (j+m+o) -(l+n+q) + m+X+n
                - (a+X+z) - (c+X+x) - (i+X+r) - (g+X+t) 
= -3X


(A bit worry about making errors during cancelling terms? 
Use a computer algebra system.)

-42 = -3X
X = 14  □


Then loop the values of $e,$a,$b,$g from 1 to 27

$c = 42 - $a - $b
$z = 28 - $a
$x = 28 - $c
$y = 42 - $x - $z
$d = 42 - $a - $g
$t = 28 - $g
$w = 42 - $z - $t
$h = 42 - $e - $b
$p = 42 - $y - $h
$f = 42 - $d - $e
$i = 42 - $c - $f
$r = 28 - $i
$s = 42 - $t - $r
$v = 42 - $y - $s
$j = 42 - $a - $r
$l = 42 - $c - $t
$k = 42 - $j - $l
$q = 42 - $z - $i
$n = 42 - $l - $q
$m = 28 - $n
$o = 42 - $j - $m
$u = 42 - $v - $w

 8 10 24 
 22 9 11
 12 23 7

 13 27 2 
 3 14 25
 26 1 15

 21 5 16 
 17 19 6
 4 18 20