IF
5+3+2 = 151022
9+2+4 = 183652
8+6+3 = 482466
5+4+5 = 202541
THEN
7+2+5 = ?
i knew that was going to be higher grade… 
143547
Correct, 143547. Now the real question where the real fun begins: Is 7+2+5 unique? That is, can 143547 be generated by the same scheme but using a different triplet?
'Luthon64
Ok, so I can figure out the first bit, and I can get to the last two numbers, but then I neuk off the bus with the middle two figures. What am I so obviously missing?
If you can get the last two numbers, you must necessarily be able to get the middle two.
'Luthon64
Huh? Wait, lemme go figure my shit out. Dont tell me. I’ll get this figured out.
Flippin basic maths getting me down - sorry about the hijack, but this is what I’ve got:
72 = 14
75 = 35
14+35-2 = 47
;D
Heh, std 7 maths in action!! (and I failed that got a grand 18% for it)
Throw some more! I’m enjoying this!
Yes. No. (Answered in order.)
Correct. Final question: Can you prove the general case that a given valid result from applying this scheme uniquely fixes the triplet?
(For the interested reader, the original problem belongs to a branch of mathematics called “group theory.”)
Later today I’ll see if I can dig up a few more number problems, preferably ones that Google knows nothing of. >:D
'Luthon64
Not formally (IANAM), but let’s give it a bash. BTW, I’m not doing your homework, am I? Taking the last triplet as an example, we have
[tex]$xy=14$[/tex],
[tex]$xz=35$[/tex], and
[tex]$xy+xz-y=47$[/tex]
All of these are linear equations–we have three variables, three equations with a single solution for each variable. Therefore a valid result can only be arrived at by one triplet?
Again correct in essence – and, no, you’re not doing my homework. 
In the general case, we have the relation
[b][i]a[/i][/b] [s]+[/s] [b][i]b[/i][/b] [s]+[/s] [b][i]c[/i][/b] = [b][i]r[/i][/b]<sub>1</sub>[b][i]r[/i][/b]<sub>2</sub>[b][i]r[/i][/b]<sub>3</sub>[b][i]r[/i][/b]<sub>4</sub>[b][i]r[/i][/b]<sub>5</sub>[b][i]r[/i][/b]<sub>6</sub>.
(Note the use of “+” vs. “+” to distinguish the operation from usual arithmetical addition.)
Given r1r2r3r4r5r6, you know that
[b][i]r[/i][/b]<sub>5</sub>[b][i]r[/i][/b]<sub>6</sub> = [b][i]r[/i][/b]<sub>1</sub>[b][i]r[/i][/b]<sub>2</sub> + [b][i]r[/i][/b]<sub>3</sub>[b][i]r[/i][/b]<sub>4</sub> – [b][i]b[/i][/b],
whence
[b][i]b[/i][/b] = [b][i]r[/i][/b]<sub>1</sub>[b][i]r[/i][/b]<sub>2</sub> + [b][i]r[/i][/b]<sub>3</sub>[b][i]r[/i][/b]<sub>4</sub> – [b][i]r[/i][/b]<sub>5</sub>[b][i]r[/i][/b]<sub>6</sub>.
The above clearly fixes the value of b uniquely.
Once b is known, a is uniquely determined by a × b = r1r2. Finally, knowing a, c is uniquely determined by a × c = r3r4. Thus, a, b and c are unique to a given r1r2r3r4r5r6, and the desired result is proved.
'Luthon64
If
2 ○ 6 ○ 9 = 248
3 ○ 2 ○ 7 = 641
4 ○ 8 ○ 3 = 242
6 ○ 7 ○ 3 = 218
8 ○ 3 ○ 6 = 488
then
5 ○ 5 ○ 5 = ?
'Luthon64
641 is a prime, hmmm interesting, curious facts about 641
Answer: 555
Correct. Just to be sure, what’s 7 ○ 4 ○ 9?
Anyone else worked it out?
'Luthon64
863
Ag nee bliksem … this last one is killing me!
Okay, that clinches it. You’ve got it.
Perhaps you’d like to try your hand at one that’s rather trickier:
If
1 ○ 3 ○ 7 → 2195
2 ○ 6 ○ 9 → 7841
4 ○ 8 ○ 3 → 6129
6 ○ 7 ○ 4 → 6433
7 ○ 4 ○ 9 → 7514
then
2 ○ 4 ○ 6 → ?
'Luthon64
Please explain previous one … PM (or spoiler tag) is fine … I’m beginning to pull my hair out