## CY's Brute-Force Take on Task 2 of PWC#089

*If you want to challenge yourself on programming, especially on Perl and/or Raku, go to https://perlweeklychallenge.org, code the latest challenge, submit codes on-time (via GitHub or email).*

(In a rush, sorry for the "raw" style this week.)

a b c d e f g h iStatement: 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

Then check whether the values of variables ~~match the 8 given equations and~~ use every integer from 1 to 9 .

I got 8 magic squares.

I got 8 magic squares.

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The definition of "magic cube" as recorded on Wikipedia, early December 2020, is : "the sums of the numbers on each row [9 rows], on each column [9 columns], on each pillar [9 pillars] and on each of the four main space diagonals [4 space diagonals] are equal to the same number."
And I think we may explore 3x3x3 cube in a similar fashion.

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

I got 192 different magic cubes.

One of them is

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

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*link for codes:*

`ch-1.pl (Task 1 on gcd sum)`,`ch-2.pl (Task 2; the 3x3x3 magic cube code is on POD of the bottom of the file)`*Do tell or correct me, if you have oppositions, want to discuss or give me advice!*

Stay healthy! □

(Note after 6-hour published: should write more caution or analysis on this task (Task 2 #089) and Sudoku (Task 2 #086) in next blogpost! )

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