The previous one in Reply #18 might well give you sleepless nights. If so, try this instead for relief:
If
1 ○ 3 ○ 7 → 353
2 ○ 9 ○ 8 → 595
4 ○ 8 ○ 5 → 193
6 ○ 7 ○ 4 → 119
7 ○ 4 ○ 6 → 239
then
3 ○ 5 ○ 7 → ?
'Luthon64
I don’t think the forum software supports spoiler tags, but I may be wrong. I sent you a PM just in case.
'Luthon64
If 1 ○ 3 ○ 7 → 353 2 ○ 9 ○ 8 → 595 4 ○ 8 ○ 5 → 193 6 ○ 7 ○ 4 → 119 7 ○ 4 ○ 6 → 239 then 3 ○ 5 ○ 7 → ?
3 ○ 5 ○ 7 → 371
Correct. Well done for getting it so quickly!
'Luthon64
Bloody nora, Lurkie! I was planning on working on it over Easter Weekend so as to impress the socks off everyone first thing in May. Respect!
Mintaka
Well, you can do that anyway if you like. For everyone but Lurkie, what does 6 ○ 1 ○ 9 yield in the scheme of Reply #20?
'Luthon64
Bloody nora, Lurkie! I was planning on working on it over Easter Weekend so as to impress the socks off everyone first thing in May. Respect!
Thanks Mintaka! 
Do the puzzle anyway. It is mathematically elegant. I’m never mad about problems that involve stuff like modular arithmetic - don’t like throwing parts of numbers away.
1 ○ 3 ○ 7 → 2195 2 ○ 6 ○ 9 → 7841 4 ○ 8 ○ 3 → 6129 6 ○ 7 ○ 4 → 6433 7 ○ 4 ○ 9 → 7514 then 2 ○ 4 ○ 6 → ?
Mefiante, this puzzle was a SOB. I tried the usual combinations of addition, subtraction, multiplication, inverses then graduated to powers, factorials, Laplace transforms and Lagragian multipliers (ok, not the last two!)
Finally,
2 ○ 4 ○ 6 → 4148
Didn’t like having to throw some numbers away …
I sound like I’m complaining, but I’m not. Thoroughly enjoyed the chase, despite some head-bashing moments!
Correct again, well done! Anyone else? 
'Luthon64
Nope … I have now finally embraced the new and somewhat less agonizing passtime of sticking roasting forks in my legs. 
ETA: … but just for interest, is there an elegant way of solving these chestnuts, or is inspection and a keen eye the only hope? They look a bit like simultaneous equations, and its almost as if they beg to be solved in a similar manner … but, of course, they are not equations at all.
Mintaka
ETA: .... but just for interest, is there an elegant way of solving these chestnuts, or is inspection and a keen eye the only hope? They look a bit like simultaneous equations, and its almost as if they beg to be solved in a similar manner ... but, of course, they are not equations at all.
I’d love to learn an elegant way to solve these rotters. Roasting fork moments almost drove me to write some code to implement a brute force search to test lots and lots of possible combinations of operators for the given number sequence. Only problem is that there are gazillions of combinations. But this will still miss the problems where only a demented human brain will pick up a non-mathematical pattern.
Enough of these brilliant minds being wasted on games. Here is a real life number problem: My bank balance need to grow! Fast!
No problem, if
$1 ○ $3 ○ $7 → $353
$2 ○ $9 ○ $8 → $595
$4 ○ $8 ○ $5 → $193
$6 ○ $7 ○ $4 → $119
$7 ○ $4 ○ $6 → $239
then
$3 ○ $5 ○ $7 → ?
No problem, if$1 ○ $3 ○ $7 → $353 $2 ○ $9 ○ $8 → $595 $4 ○ $8 ○ $5 → $193 $6 ○ $7 ○ $4 → $119 $7 ○ $4 ○ $6 → $239then
$3 ○ $5 ○ $7 → ?
Hmm. Not good if the operators divide the numbers resulting in an infinitely diminishing balance. I suggest a new operator where ○=$
Mathematically speaking, these problems aren’t especially interesting or edifying. At best, they can be used to illustrate some core mathematical ideas whose application is mostly cryptographic. And no, there is no fixed approach to them because the variety of possible schemes is virtually limitless.
That said, for the insatiably curious, frustrated and bagshot among us, here’s…
[spoiler=Solution to Reply #12]a ○ b ○ c → 100×(a×b [mod 10]) + 10×(b×c [mod 10]) + (c×a [mod 10])
where “[mod 10]” denotes “take the remainder after division by 10.” For example, 1234 [mod 10] = 4 because 1234 divided by 10 leaves a remainder of 4. In our ordinary decimal system of numeration, this is of course always the least significant (right-most) digit of the integer.
Using “4 ○ 8 ○ 3” to illustrate, 4×8 = 32 = 2 [mod 10], 8×3 = 24 = 4 [mod 10] and 3×4 = 12 = 2 [mod 10], giving 100×2 + 10×4 + 2 = 242.
Similarly, 5 ○ 5 ○ 5 → 100×(5×5 [mod 10]) + 10×(5×5 [mod 10]) + (5×5 [mod 10]) = 100×(25 [mod 10]) + 10×(25 [mod 10]) + (25 [mod 10]) = 100×5 + 10×5 + 5 = 555.[/spoiler]
[spoiler=Solution to Reply #18]a ○ b ○ c → (ab + bc + ca) [mod 10,000]
where “[mod 10,000]” denotes “take the remainder after division by 10,000.” For example, 123,456 [mod 10,000] = 3,456 because 123,456 divided by 10,000 leaves a remainder of 3,456. In our ordinary decimal system of numeration, this is of course always the four least significant (right-most) digits of the integer, padding out to the left with leading zeroes if necessary.
This one needs a calculator capable of 10-digit accuracy.
Using “7 ○ 4 ○ 9” to illustrate, 74 = 2,401, 49 = 262,144 and 97 = 4,782,969, giving 2,401 + 262,144 + 4,782,969 = 5,047,514 [mod 10,000] = 7,514.
Similarly, 2 ○ 4 ○ 6 → 24 + 46 + 62 [mod 10,000] = 16 + 4,096 + 36 [mod 10,000] = 4,148 [mod 10,000] = 4,148 .[/spoiler]
[spoiler=Solution to Reply #20]a ○ b ○ c → a + b2 + c3
Using “6 ○ 7 ○ 4” to illustrate, 6 + 72 + 43 = 6 + 49 + 64 = 119.
Similarly, 3 ○ 5 ○ 7 → 3 + 52 + 73 = 3 + 25 + 343 = 371.[/spoiler]
'Luthon64
Elsewhere, fun with numbers suggests that religion may soon become extinct in nine developed countries.
'Luthon64
Got this in a mail today…
THIS IS THE ONLY TIME WE WILL SEE AND LIVE THIS EVENTCalendar for July 2011. JulySun Mon Tue Wed Thu Fri Sat
1 2
3 4 5 6 7 8 9
10 11 12 13 14 15 16
17 18 19 20 21 22 23
24 25 26 27 28 29 30
31This year, July has 5 Fridays, 5 Saturdays and 5 Sundays. This happens
once every 823 years.Kinda interesting - read on!!!
This year we’re going to experience four unusual dates.
1/1/11, 1/11/11, 11/1/11, 11/11/11 and that’s not all…
Take the last two digits of the year in which you were born - now add
the age you will be this year,The results will be 111 for everyone in whole world…