Fill the Squares

a(j,j) ∈ {1,2,3,4,5,6,7,8,9}; i,j ∈ {1,2,3}

it's convenient to start from multiplications and large total resultates;

• row 2: a(2.1)xa(2,2) – a(2,3) = 29;

• col. 2: a(1,2) + a(2,2)xa(3,2) = 9;

the possible triplets for row 2 are (9,4,7), (4,9,7);

but eq. for col 2 implies a(2,2) = 4; so: a(2,1) = 9; a(2,3) = 7;

• col. 2: a(1,2) + 4a(3,2) = 9; choice in {1,2,3,5,6,8};

the triplet that satisy eq. col. 2 is: (5,4,1);

• row 3: a(3,1) – a(3,3) = -1; choice in {2,3,6,8};

the triplet that satisfy eq. row 3 is: (2,1,3), with:

a(3,1) = 2; a(3,2) = 1; a(3,3) = 3;

• col. 1: a(1,1) – 9 – 2 = - 5; choice in {6,8};

the triplet that satisfy eq. col. 1, so that:

a(1,1) = 9 + 2 – 5 = 6, is: (6,9,2);

• row 1: 6 + 5 + a(1,3) = 3;

• col. 3: a(1,3) + 7 + 3 = 12;

choice of a(1,3) in {8}, and a(1,3) = 8 satisy eqs row 1 and col. 3;

Sd1 = a(1,1) + a(2,2) + a(3,3) = 6 + 4 + 3 = 13;

(sum of the elements of the first diagonal);

Sd2 = a(1,3) + a(2,2) + a(3,1) = 8 + 4 + 2 = 14;

(sum of the elements of the second diagonal);

so the total sum of the diagonal elements is Sdt:

Sdt = Sd1 + Sd2 = 13 + 14 = 27. $\square$

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And the rows are: (6,5,8); (9,4,7); (2,1,3).

As a reference to empty squares, I will mark them with $ sign and letters from " a " to " i " and call them cells; starting from first row; left to right. Also I will name given results (3, 29, -1, -5, 9, 12) with word "TOTAL"

We will start with second row, TOTAL=29 . For now; possible multiplications required to arrive to TOTAL=29 are 9x4 , 8x4 , 7x5 and 6x5 . Having in mind this factors we conclude that cell $e can only be one of these integers $e'={4, 5, 6, 7, 8, 9} .

We conclude that $d and $e ≠ 5 or 6 , because if $d = 5 , in first column we would need $a and $g to be |$a|=|$g| in order to arrive to TOTAL=-5 , and this is not allowed since we can't have „double integers“ in matrix. Also to assume $e = 5 , implicates $f and $h = 1 , so again we would have to have double integers, what leads us to conclusion that $e ≠ 5 . If $d and $e ≠ 5 , we cant have 5x6 multiplication in second row, so conlclusion is that $d and $e ≠ 5 or 6 .

We conclude that $h and $e ≠ 8 or 9 , because; in case of integer 8, we would have to have double integers in second column when arriving to TOTAL=9 , and in case of integer 9 we would not be able to arrive to mentioned TOTAL.

We conclude; since $d and $e ≠ 5 ; that:

1.) e$ ≠7 , because 7x5 multiplication therefore is not allowed;

2.) that only possible $e value is $e = 4 (from $e' we excluded integers 5, 6, 8, 9 earlier and now integer 7.).

Than we come to conclusion that $d = 8 or 9 .

Furthermore, in third column; to be able to arrive to TOTAL=12 respecting given operators; we must follow $f ≠ 1 or 2 or 3 .

Since $e = 4 , if $d would be $d = 8 , then $f would need to be $f = 3 in wich case we would not follow our $f ≠ 1 or 2 or 3 rule. So we conclude $d ≠ 8 and sequentialy $d = 9 .

Since $d=9 ; than $f is $f = 7 .

Also;

$h = 1 or 2 .

$g = 2 (all other integers excluded because they require use of double integers in matrix to arrive to TOTAL).

Then sequentialy we conclude that;

$h = 1 .

$b = 5 .

$a = 6 .

$c = 8 .

And finaly $i = 3 .

To sum it up we have:

6 5 8

9 4 7

2 1 3

6,5,8: 9,4,7: 2,1,3

I know this isn't the most efficient solution, but just in case somebody is curious to the possible combination of numbers to be put in the box. haha.

a(j,j) ∈ {1,2,3,4,5,6,7,8,9}; i,j ∈ {1,2,3} it's convenient to start from multiplications and large total resultates; row 2: a(2.1)xa(2,2) – a(2,3) = 29; col. 2: a(1,2) + a(2,2)xa(3,2) = 9; the possible triplets for row 2 are (9,4,7), (4,9,7); but eq. for col 2 implies a(2,2) = 4; so: a(2,1) = 9; a(2,3) = 7; col. 2: a(1,2) + 4a(3,2) = 9; choice in {1,2,3,5,6,8}; the triplet that satisy eq. col. 2 is: (5,4,1); row 3: a(3,1) – a(3,3) = -1; choice in {2,3,6,8}; the triplet that satisfy eq. row 3 is: (2,1,3), with: a(3,1) = 2; a(3,2) = 1; a(3,3) = 3; col. 1: a(1,1) – 9 – 2 = - 5; choice in {6,8}; the triplet that satisfy eq. col. 1, so that: a(1,1) = 9 + 2 – 5 = 6, is: (6,9,2); row 1: 6 + 5 + a(1,3) = 3; col. 3: a(1,3) + 7 + 3 = 12; choice of a(1,3) in {8}, and a(1,3) = 8 satisy eqs row 1 and col. 3; Sd1 = a(1,1) + a(2,2) + a(3,3) = 6 + 4 + 3 = 13; (sum of the elements of the first diagonal); Sd2 = a(1,3) + a(2,2) + a(3,1) = 8 + 4 + 2 = 14; (sum of the elements of the second diagonal); so the total sum of the diagonal elements is Sdt: Sdt = Sd1 + Sd2 = 13 + 14 = 27.