the question means: how many ''balanced subsets'' does M have?

Case 1: the subset doesn't include 5 Case 1.1: the subset includes 4,6. Now the numbers 4,5,6 don't affect the arithmetic means, so we find the subset M of {1,2,3,7,8,9} with P(M) = 5. We find that there are 8 of them. Case 1.2: the subset doesn't include 4,6. It is the same as case 1.1, because the numbers 4 and 6 doesn't affect the arithmetic means. Case 1.3: the subset includes 4, but not 6. We find that there are 5 of them. Case 1.4: the subset includes 6, but not 4. By symmetry, there are 5 as well. So there are 26 subsets Case 2: the subset includes 5. Since the number 5 doesn't affect the arithmetic mean, it is the same as case 1. But we've counted the empty set, so the actual total number is 51.

I'd look at the numbers not as they are written down now, but at their difference with 5. So: 4, 3, 2, 1, 0, 1, 2, 3, 4 At both the left and the right part, the sum of the differences must be the same. D is the total difference at one side: If D = 0: One possibility at both sides: (empty) 1*1=1 If D = 1: One possibility at both sides: 4, 6 1*1=1 If D = 2: One possibility at both sides: 3, 7 1*1=1 If D = 3: Two possibilities at both sides: 3+4 / 2, 6+7 / 8 2*2=4 If D = 4: Two possibilities at both sides: 2+4 / 1, 6+8 / 9 2*2=4 If D = 5: Two possibilities at both sides: 1+4 / 2+3, 6+9 / 7+8 2*2=4 If D = 6: Two possibilities at both sides: 1+3 / 2+3+4, 7+9 / 6+7+8 2*2=4 If D = 7: Two possibilities at both sides: 1+2 / 1+3+4, 8+9 / 6+7+9 2*2=4 If D = 8: One possibility: no 3, 7 1*1=1 If D=9: One possibility: no 4,6 1*1=1 If D=10: all numbers: 1 Counting gives us 26. Multiplied by 2 for including or excluding 5 gives 52, minus the empty subset gives us 51 as well!

Nice solution! Now try the case $M = \{1,2,3,...,k\}$

I'm going away for a few days, but of course I already thought about this a little. Maybe, induction? If the answer for {1,2,3....k} is A(k): A(2a+1) = A(2a)*2+1 When k = 2a, the number of elemens in a subset must be even, since the mean is a+1/2. A(1) = 1 A(2) = 1 A(3) = 3 A(4) = 3 A(5) = 7 A(6) = 7 A(7) = 15 For these small numbers there is a pattern, but A(9) = 51 does not fit :'(. A(2a) >= A(2a-1) Maybe you can count by the number of elements in the subset you want to create. I'll check this topic later, arrivederci!

Hmm... then it seems I can't solve this

Can someone pls check my solution, I got 39 instead

Help pls Edit:I had a careless again whoops Edit: I just realized I think I have the shortest most elegant solution using complements