B is for Brilliant

Let us consider

$\boxed{A1}\boxed{A2}\boxed{A3}\boxed{A4}\boxed{A5}\boxed{A6}\boxed{A7}$

$\boxed{B1}\boxed{B2}\boxed{B3}\boxed{B4}\boxed{B5}\boxed{B6}\boxed{B7}$

$\boxed{C1}\boxed{C2}\boxed{C3}\boxed{C4}\boxed{C5}\boxed{C6}\boxed{C7}$

$\boxed{D1}\boxed{D2}\boxed{D3}\boxed{D4}\boxed{D5}\boxed{D6}\boxed{D7}$

$\boxed{E1}\boxed{E2}\boxed{E3}\boxed{E4}\boxed{E5}\boxed{E6}\boxed{E7}$

$\boxed{F1}\boxed{F2}\boxed{F3}\boxed{F4}\boxed{F5}\boxed{F6}\boxed{F7}$

$\boxed{G1}\boxed{G2}\boxed{G3}\boxed{G4}\boxed{G5}\boxed{G6}\boxed{G7}$

Now, A7, D7, G7 are white so we will try to minimize them and allocate them 1, 2,3. To maximize the blue B7, C7, E7, F7, we will allocate them 7,6,5,4

Now, to maximize the blue B1, C1, E1, F1, we will allocate them 4,5,6,7.

Now there are many possible ways to fill up the rest of the table,

We just have to allocate the Ath row(1st) with, 3,4,2,7,6,5,1. 4th row (Dth row) with 1,3,7,6,5,4,2. 7th row (Gth row) with 2,7,6,5,4,1,3.

Adding the values we get $\boxed{122}$

The matrix I came up with is

$\boxed{3}\boxed{4}\boxed{2}\boxed{7}\boxed{6}\boxed{5}\boxed{1}$

$\boxed{4}\boxed{5}\boxed{3}\boxed{1}\boxed{2}\boxed{6}\boxed{7}$

$\boxed{5}\boxed{1}\boxed{4}\boxed{2}\boxed{3}\boxed{7}\boxed{6}$

$\boxed{1}\boxed{3}\boxed{7}\boxed{6}\boxed{5}\boxed{4}\boxed{2}$

$\boxed{6}\boxed{2}\boxed{1}\boxed{4}\boxed{7}\boxed{3}\boxed{5}$

$\boxed{7}\boxed{6}\boxed{5}\boxed{3}\boxed{1}\boxed{2}\boxed{4}$

$\boxed{2}\boxed{7}\boxed{6}\boxed{5}\boxed{4}\boxed{1}\boxed{3}$

"The table can be filled up in many ways, but you will get the same answer."

there are 5 line: horizontal up, horizontal middle, horizontal bottom, vertical left, vertical right. each because the line containing permutation 1 to 7, each line sum is (1+2+3+4+5+7) there are 3 intersection 1 in vertical left, and 2 in vertical right (that why there some hole in right). if we want to maximize the sum, the intersection must be as small as possible, each intersection has 3 points: up,middle, and bottom, so the intersection sum is (1+2+3).

answer is: 5x(1+2+3+4+5+6+7)-3x(1+2+3)=5x35-3x6=140-18=122

For this problem, I am going to refer to the rows as $A \rightarrow G$ and the columns as $1 \rightarrow 7$ . A group of cells will be notated like in Microsoft Excel; $R_1C_1$ : $R_2C_2$ refers to all of the cells bounded by rows $R_1$ and $R_2$ and columns $C_1$ and $C_2$ , inclusive.

There are five major regions of cells that need to be maximized. They are $A1$ : $G1$ , $A2$ : $A6$ , $D2$ : $D6$ , $G2$ : $G6$ , and $B7$ : $C7 \cup E7$ : $F7$ .

Obviously, $A1$ : $G1$ will contain every digit, but the digits can be arranged to maximize $A2$ : $A6$ , $D2$ : $D6$ , and $G2$ : $G6$ . If $A1$ , $D1$ , and $G1$ are minimized, then their rows' blue cells' sums will be maximized.

Looking at the white entries in column $7$ shows that a maximum sum is achieved when the white entries are $1$ , $2$ , and $3$ in some order (similarly, $A1$ , $D1$ , and $G1$ are $1$ , $2$ , and $3$ in a different order). The blue cells in column $7$ are $4$ , $5$ , $6$ , and $7$ in some order.

Putting in numbers as large as possible to fill rows $A$ , $D$ , and $G$ yields that the maximum sums for $A2$ : $A6$ , $D2$ : $D6$ , and $G2$ : $G6$ are $\textbf{25}$ , $\textbf{24}$ , and $\textbf{23}$ . The sum of $A1$ : $G1$ is $\textbf{28}$ . The sum of the blue cells in column $7$ is $\textbf{22}$ .

The sum of the entries in the blue entries is $25+24+23+28+22=\boxed{122}$ .

An example of a possible board is $\downarrow$

$2654371\\ 7143526\\ 6425137\\ 3761452\\ 4237615\\ 5312764\\ 1576243$

Think that there are 5 rows or columns blued in total :: (Summing to a total of 28*5 == 140)

Now we are counting the 2 blued corners twice and the 3 non blue squares on the column also twice

We need to substract them and the lowest possible value would be a combination of 1,2,3 with the conditions satisfied..

By this we get the sum as :: (140-(1+2+2)3) == 122

Sum of the numbers in the first blue column is 28.In the first,middle and last rows only one number is missing,let's say 1,2,3 in any order is missing. Individual sums 25,26,27. In the four remaining squares only 4,5,6,7 (in any order) can be there.Total sum $28+27+26+25+22=128$ .Now to the three common squares,for the sum to be the greatest those 3 should contain 1,2,3 (in any order). Greatest sum= $128-(3+2+1)=122$