HHC 2019 programming contest entries

[Albert Chan]

(10-10-2019 10:24 PM)dramsey Wrote:  An anonymous note left at my place when I was away reads "Anonymous note to judge: A smart programmer can cut search space in half by requiring first two digits to have the same even-odd parity." I have no idea if this is true but wish the submitter had included a program...

Proof:

Let N = 100 a + 10 b + c

Rephrase the problem to search (a^3 - 100 a) + (b^3 - 10 b) + (c^3 - c) = 0

Mod 2: (a^3 - 100 a) + (b^3 - 10 b) + (c^3 - c) ≡ a + b + 0 ≡ 0

→ a ≡ b (mod 2)