TRIAL METHOD 1
If we label the given equations in order (1)-(4), we can see that (1) and (2) differ in only one number, as do (2) and (4). We can make use of this fact by deriving values for the first and last numbers in the triads.
253094-151872=101222≡2 (5-3=2), 1≡50611 in the lead position, so 9≡455499;
253094-251573=1521≡3 (6-3=3) in the last position, so 1 in the last position≡507 and 7≡3549.
5 in the lead position is equivalent to 5×50611=253055;
7 in the last position is equivalent to 3549, so 5 in lead + 7 in last is equivalent to 253055+3549=256604.
Therefore, 6 in the central position is equivalent to 303585-256604=46981, and 4 in the central position is equivalent to (4/6)46981=31320⅔, making 9 +4 +7=455499+31320⅔+3549=490368⅔.
However, this looks unlikely and in any case doesn't satisfy all the equations, so this method fails to provide consistent results.
TRIAL METHOD 2
There is a pattern for the first 4 digits of each sum: if A, B, C are the three "addends" and S is the given sum:
S=10000AB+100AC+X=100A(100B+C)+X where X has to be determined. So, 10000(3×5)+100(3×6)+72=151872, 10000(5×5)+100(5×6)+94=253094, 10000(5×6)+100(5×7)+85=303585,10000(5×5)+100(5×3)+73=251573. So the solution to the problem is of the form 366300+X. X has different values for each of the four given sums.
We have to figure out how to get X in terms of A, B, C, that is, work out how 72, 94, 85, 73 have been derived. F(A,B,C), where F is a 3-variable function, has the following values:
F(3,5,6)=72, F(5,5,6)=94, F(5,6,7)=85, F(5,5,3)=73. But what is F(9,4,7)?
Using similar logic to Trial Method 1, taking the deduced values for the lead number and last number, we see that:
LEAD: (94-72)/(5-3)=11, 72-3×11=39=3×13=94-5×11; LAST: (94-73)/(6-3)=7, 94-6×7=52=4×13=73-3×7. There appears to be a pattern (but it might just be coincidence).
The next multiple of 13 is 65, so X=65+9×11=164, making the solution 366300+164=366464. This method is too manipulative so we need a better way of finding X.
TRIAL METHOD 3
More to follow...