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The Code, Part 2

, 8 Aug 2012
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Solving a puzzle.

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).

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

About the Author

rtybase
Software Developer (Senior) Snappli Ltd.
United Kingdom United Kingdom
My name is Ruslan Ciurca. Currently I am a Software Developer at snappli.com.

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