6 possibilities,
i) R - B - R - B  
ii) R - R - B - B 
iii) B - R - R - B 
iv) B - R - B - R  
v) B - B - R - R  
vi) R - B - B - R
(R- RED, B- BLUE)

There are 5 horizontal lines and 5 vertical lines.
Total 10 lines minimum needed.

As we see in the figure in the middle 1, 2, 3 and 4 are 4 small squares.
Also we have 9 and 10 => 2 big squares. Outer squares 5, 6, 7, & 8 => 4.
Total = 4 + 4 + 1 + 1 = 10 squares.

Smaller squares 4 \(\times\) 4 = 16 (As numbered).
As a whole bigger = 4.
Total squares = 20.

The given star can be viewed as two magical triangular as.

Now the required arrangement is

Some other arrangements are

Here there are even number of terms.
Also we know that 1 + 6 = 7, 2 + 5 = 7, 3 + 4 = 7;
so placing 7 at the centre, and the pairs (1, 6) (2, 5) and (3, 4) at the opposite ends we get, the answer.

The odd numbers between 1 to 17 are 1, 3, 5, 7, 9, 11, 13, 15, 17.
Step 1: Place the smaller numbers 1, 3 : 5 on the comers.

Step 2: Arrange another set of smaller numbers 7, 9 and 11 on each side.

Step 3: Arrange the remaining numbers 13, 15, 17 to give the total 30.

Magic sum = 30.

 

 

 

 

 

Step 1: Complete the corners with smaller numbers 1, 2 and 3.

Step 2: The side having smallest numbers 1 & 2 are to be filled with the greatest number 6, the second smallest 1 & 3 side to be filled with the second largest 5 at the middle and so on.

The magic sum is 1 + 6 + 2 = 2 + 4 + 3 = 3 + 5 + 1 = 9. Some other ways are given below

The magic sum = 1 + 6 + 3 = 3 + 2 + 5 = 5 + 4 + 1 = 10

The magic sum 6 + 1 + 4 = 4 + 5 + 2= 2 T 3 + 6 = 11.

The magic sum 4 + 3 + 5 = 5 + 1 + 6= 6 + 2 + 4 = 12.