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Solution 1

To start with I'll list all possible cases by brute force (this was how I solved it...). Therefore I find the whole layout of the question a bit weird as a) doesn't lead to b) etc. with my method...

Important to keep in mind is the criterion (which means the sum has to be possible to compose in 8 ways with the digits 1-9): a+b+c=d+e+f=g+h+k=a+d+g=b+e+h=c+f+k=a+e+k=c+e+g

SUM

• 24: 789 = 1 way
• 23: 986 = 1 way
• 22: 985, 876 = 2 ways
• 21: 984, 876 = 2 ways
• 20: 983, 974, 965, 875 = 4 ways
• 19: 982, 973, 964, 874, 865 = 5 ways
• 18: 981, 972, 963, 954, 873, 864, 765 = 7 ways
• 17: 971, 962, 953, 872, 854, 863, 764 = 7 ways
• 16: 961, 952, 943, 871, 862, 853, 763, 754 = 8 ways
• 15: 951, 942, 861, 852, 843, 762, 753, 654 = 8 ways
• 14: 941, 932, 851, 842, 761, 752, 743, 653 = 8 ways
• 13: 193, 184, 175, 283, 274, 265, 364 = 7 ways
• 12: 129, 138, 147, 156, 273, 264, 345 = 7 ways
• 11: 128, 137, 146, 236, 245 = 5 ways
• 10: 127, 136, 145, 235 = 4 ways
• 9: 135, 234 = 2 ways
• 8: 125, 134 = 2 ways
• 7: 124 = 1 way
• 6: 123 = 1 way

From this we can see that there are 3 possible ways of getting a sum that has eight ways of being combined. However, for this type of array we are trying to create 'e' will appear in four sums (d+e+f and b+e+h and a+e+k and c+e+g). For the sum 14 the max number of times a digit appears is three (for 1, 2, 3, 4, 5 and 7) and for 15 the max number of times a digit appears is four (for 5), lastly for 16 the max number of times a digit appears is three (for 9, 8, 7, 6, 5 and 3). This means to create an array of this type the combinations that will be present are the ones presented in the emboldened line in the list of possibilities.

Therefore follows the answers to the questions

i) Shown 15 above.

ii) Shown 5 above.

iii) Implicitly shown as we only have two sums with 9 involved when creating the sum 15. This means that the 9 must be located at a position where it only appears in two summations (therefore b, d, f or h). (On a sidenote e appears in four sums as earlier mentioned and a, c, g and k appear in three summations)

iv) Setting b=9 (knowing e=5) gives h=1 and a+c= 4 and 2 (from our table of possibilities) we get two possible arrays (one with first line 492 and one with 294).

v) Calculating these possibilities I think is just to multiply 4 by 2 (because there are four possible positions of the 9, b, d, f and h) to get 8 because once you fix one number all the other will fall out with two possibilities.

Solution by nota bene.

Solution 2

(i) 45 = (a+b+c)+(d+e+f)+(g+h+i) = 3n => n = 15.

(ii) So 45 +3e = (a+b+c+d+e+f+g+h+i) + 3e = (a+e+i)+(c+e+g)+(b+e+h)+(d+e+f) = 4n = 60. So 3e=15, so e = 5.

(iii) If not, then one of a,c,g,i = 9. Wlog, suppose a = 9. Then i = 1 and b+c = 6. b isn't 1 and so c is at most 4. Then as c+f+i = 15 we have f > 9, which is impossible. So one of b,d,f,h = 9.

(iv) b = 9, so a+c = 6. and h = 1. As neither a,c can be 1, one must be 2 and the other must be 4. Suppose a = 2. Then i = 8, so g = 6, so d=7 and f=3. So there's only 1 solution with a=2. By symmetry there's only one solution with c=2. So total of 2 solutions.

(v) There are 4 choices for where the 9 goes (b,d,f,h), and then 2 choices for where the 2 goes (either side of the 9). These choices fix the remaining numbers. So total number of solutions = 8.

Solution by DFranklin.