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…

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.

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 .

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.

'Luthon64