The Code - Part 2

For the last two weeks, I am addicted to The Code. Those who successfully passed pre-selection, were offered to download a PDF file (you need the passwords from the pre-selection to open it) with even more puzzles.

Here is another one I liked:

A Mathematician's Apology
ABC = A³ + B³ + C³
CDE = C³ + D³ + E³
CDA = C³ + D³ + A³
FED = F³ + E³ + D³
A⋅100 + A⋅10 + D = ?

Now, if you Google for "A Mathematician's Apology", you will come to the following link, which is a book with the same title by Hardy. Section 15 reveals all the solutions for the equation:

x⋅100 + y⋅10 + z = x³ + y³ + z³
where x∈{1..9}, y,z∈{0..9}. They are (1,5,3), (3,7,0), (3,7,1) and (4,0,7).

The pattern is easy to spot now:

(A,B,C) = (1,5,3)
(C,D,E) = (3,7,0)
(C,D,A) = (3,7,1)
(F,E,D) = (4,0,7)
and A⋅100 + A⋅10 + D = 117

However, if (like me) "Google search" isn't (always) an obvious option, the following Java code:

public class Code {
  public static int func(int a, int b, int c) {
    return a*100 + b*10 + c - a*a*a - b*b*b - c*c*c;		
  }

  public static void main(String[] args) {
    for (int a = 1; a < 10; a++) {
      for (int b = 0; b < 10; b++) {
        for (int c = 0; c < 10; c++) {
          if (func(a,b,c) == 0) {
            System.out.println("("+a+","+b+","+c+")");
          }
        }
      }
    }
  }
}

will also return (1,5,3), (3,7,0), (3,7,1) and (4,0,7).