C addition using modulus

With little-endian ints (and assuming ASCII text, and 8-bit bytes, and all the other assumptions the code requires), and ignoring all the technically-wrong-in-modern-C stuff in the code, your "What I understand so far" is correct.

gets(&n) will store the ASCII values of A, space, and B into the first 3 bytes of n. It'll also store a null terminator into the 4th byte. Storing those ASCII values into those bytes of n results in n taking the value B*256*256 + space*256 + A, where B, space, and A represent the corresponding ASCII values.

256 mod 85 is 1, so by the properties of modular arithmetic,

(B*256*256 + space*256 + A) % 85 = (B + space + A) % 85

Incidentally, with 4-byte big-endian ints, we get

(A*256*256*256 + space*256*256 + B*256) % 85 = (B + space + A) % 85

so endianness doesn't matter, as long as we have 4-byte ints. (Bigger or smaller ints could be a problem; for example, with 8-byte ints, we'd have to worry about what's in the bytes of n that gets didn't set.)

Space is ASCII 32, and the ASCII value for a digit character is 48 + the value of the digit. Defining a and b as the numeric values of the digits entered (rather than the ASCII values of the digit characters), we have

(B + space + A) % 85 = (b + 48 + 32 + a + 48) % 85
                     = (a + b + 128) % 85
                     = (a + b + 43) % 85

(B + space + A) % 85 - 43 = (a + b + 43) % 85 - 43
                          = (a + b) % 85
                          = a + b

where the last two equivalences rely on the fact that a and b take values from 0 to 9.