Almost A Magic Square

There are four lines in the diagram, each producing the sum $S$ . Adding these four sums together, we add together the numbers in all nine boxes, but the middle number $m$ is counted four times. Thus: $4S = 4m + \sum \text{other boxes} \\ 4S = 3m + \sum \text{all boxes} \\ 4S = 3m + \sum \text{numbers from 1 to 9} \\ 4S = 3m + 45.$

Therefore the sum is greatest if $m$ is greatest. Also, because $S$ is an integer, $3m + 45$ must be a multiple of four. This is only possible if $m$ is one more than a multiple of four; in other words, $m = 1, 5, 9$ .

It makes sense first to try $m = 9$ . This means $S = 18$ , so that the numbers not in the middle should add up to 9 in pairs. It is easy to do this: $9 = 1+8 = 2+7 = 3+6 = 4+5.$ Thus we place 1 across from 8, 2 across from 7, and so on. It does not matter where! This proves that the maximum sum is $\boxed{18}$ , and that this can actually be done.

By using Gauss' number paring Trick we can easily solve the problem without any algebraic equations.

The answer is $18$ but I did it by trial and error. I placed the numbers in the boxes until I got the maximum sum.

If I am looking for the largest value and I must use all numbers then I start with the smallest value. Which is 1. So what is the largest value I can achieve if 1 is one of the numbers. That is 1+8+9=18. No larger sum can be achieved. Placing 9 in the center with 1 on one side and 8 on the other. It becomes easy to fill in the remaining boxes to add each to 18.

I thought you always put the highest in the middle. The next number (8) is offset by (1) so on and so forth, no advanced math needed for this one

We can start by writing sum of all digits i.e 45. If we assign variables a,b,c,d,e,f,g,h,i. Putting e at the centre box. As condition is the sum along the lines is same but actually the centre box doesn't bothers the sum, i.e the sum of boxes other than the central box is also the same. a+b+c+d+e+f+g+h+i= 45
a+i=b+h=d+f= g+c=∆-e (∆=sum along lines)
4(a+i)+e=45. So 4(∆-e)+e=45
4∆-3e=45. So. ∆=45+3e\4 So ∆is max. When e is max. But ∆ is integer only when e=5,9 But we want max. So e=9. So ∆= [18]

$Let$ $the$ $highest$ $number$ $go$ $in$ $the$ $middle$ $and$ $the$ $highest$ $sum$ $that$ $can$ $be$ $formed$ $between$ $two$ $numbers$ $for$ $all$ $the$ $remaining$ $numbers$ $go$ $in$ $their$ $respective$ $places,$

$9$ $is$ $the$ $highest$ $number$ $and$ $the$ $highest$ $number$ $that$ $can$ $be$ $formed$ $from$ $the$ $remaining$ $numbers$ .

$Hence,$ $9+9=18$

$Therefore,$ $the$ $maximum$ $possible$ $sum$ $that$ $can$ $be$ $achieved$ $is$ $18.$

to maximise each line I needed to put the max number possible in the middle box which is : 9 and then I tought that every two boxes must have same outcome, and instinctly got 9 8+1 7+2 6+3 5+4

To maximize our value - We put 9 in the middle, then we have 1-8 notice that pairing off gives us 1+8 2+7 3+6 4+5 this is 9 for all pairs, then putting this on our line we get 18 as our sum.

Every line shares the central, so for all lines to add up to the same value the outer numbers of every line also have to add up to the same value.

Chosen any central number the biggest of the remaining numbers has to be matched with the smallest, otherwise, the smallest number should be matched with a number bigger than the biggest in order for the pair to have the required sum.

We can, therefore, say that the sum of any line is equal to the sum of the central number and the biggest and smallest of the remaining. The number 9 will always be either the central one or the biggest one. Similarly, the number 1 will always be either the central one or the smallest one. Therefore 9 and 1 will always be in the same line.

We can choose the last number of the line however we want and therefore to maximize the sum we choose 8. So the result is $9+8+1=18$

For each sum of numbers in a line to be a maximum each must contain the number 9, Hence place 9 in the middle.

Now for the surrounding numbers realize that for all numbers in a line to sum to the same each number startng with 1 must be paired with the largest number yet unused.

Hence the pairs (1,8) , (2,7), (3,6) , (4,5)

1+2+3+........+9 = 45
as per magic box total is equal to 60.......[45 + 3 * 5 = 60
now by considering 9 instead of 5 total becomes.....45 + 3 * 9 = 72
72/4 = 18