Mind Your P, Q, R and S

Solution 1:

The sum of all of the numbers in the grid is

$P +Q+R+S+10+16+22+28+4=P +Q+R+S+80$

But the sum of the three numbers in the second column is $4+16+28 = 48$ ,

so the sum of the three numbers in each column is $48$ .

Thus, the sum of the numbers in the second row is $10+16+22 = 48$ ,so the sum of the numbers in any row, column or diagonal is $48$ .

From the first row, $P + 4 + Q = 48 \ so P + Q = 44$ .

From the first column, $P + 10 + R = 48 \ so P + R = 38$ , total of the nine numbers in the grid is $3(48) = 144$ ,

so $P + Q + R + S + 80 = 144$ ,

Or $P+Q+R+S=64$ .

Solution 2:

The sum of the numbers in the second row is $10+16+22 = 48$ , so the sum of the numbers in any row, column or diagonal is 48.

In the first row, $P + 4 + Q = 48 \ so P + Q = 44$ .

In the third row, $R + 28 + S = 48 \ so R + S = 20$ .

Therefore, $P + Q + R + S = 44 + 20 = 64$ .

The sum has to be 48 on every row, column and diagonal. Since P+Q+4 = 48 and R+S+ 28 =48 it follows that P+Q = 44 and R+S = 20. Hence the total sum is 64.

From the picture,

P + S + 16 = 28 + 4 + 6

Q + R + 16 = 28 + 4 + 6

Similarly, we get P + S = Q + R,

P + S = 32

P + S + Q + R

2(P + S)

64