Why does this code using Math.pow print "HELLO WORLD"?

Question

I discovered the following code. I know, it looks less weird/exciting than this one using seemingly random numbers, but it seems to be more complex than this one using bit shifts on a large number:

long[] c = {130636800L, -5080148640L, 13802573088L, -14974335980L, 8683908340L,
           -3006955245L, 651448014L, -89047770L, 7457160L, -349165L, 6998L};

for (int x = 0; x < 11; x++) {
    long s = 0;
    for (int i = 0; i < 11; i++)
        s += c[i] * Math.pow(x, i);

    System.out.print((char)(s / 1814400));
}

Code on Ideone

Output:

HELLO WORLD

How does it work? Is it some form of encryption or did anyone get mad constructing it?

Answer 1

Let's get into some math:

Solve the below equations and you get your answers. These equations have one unique solution as the number of equations equals the number of unknown variables.

Let c[0] = 72, which is the ASCII value of 'H'.

For clarity: I've used ^ for raised to convention. Now solve:

1^0 * c[0] + 1^1 * c[1] + 1^2 * c[2] + 1^3 * c[3] + 1^4 * c[4] + 1^5 * c[5] + 1^6 * c[6] + 1^7 * c[7] + 1^8 * c[8] + 1^9 * c[9] + 1^10 * c[10] = 69
2^0 * c[0] + 2^1 * c[1] + 2^2 * c[2] + 2^3 * c[3] + 2^4 * c[4] + 2^5 * c[5] + 2^6 * c[6] + 2^7 * c[7] + 2^8 * c[8] + 2^9 * c[9] + 2^10 * c[10] = 76
3^0 * c[0] + 3^1 * c[1] + 3^2 * c[2] + 3^3 * c[3] + 3^4 * c[4] + 3^5 * c[5] + 3^6 * c[6] + 3^7 * c[7] + 3^8 * c[8] + 3^9 * c[9] + 3^10 * c[10] = 76
4^0 * c[0] + 4^1 * c[1] + 4^2 * c[2] + 4^3 * c[3] + 4^4 * c[4] + 4^5 * c[5] + 4^6 * c[6] + 4^7 * c[7] + 4^8 * c[8] + 4^9 * c[9] + 4^10 * c[10] = 79
5^0 * c[0] + 5^1 * c[1] + 5^2 * c[2] + 5^3 * c[3] + 5^4 * c[4] + 5^5 * c[5] + 5^6 * c[6] + 5^7 * c[7] + 5^8 * c[8] + 5^9 * c[9] + 5^10 * c[10] = 32
6^0 * c[0] + 6^1 * c[1] + 6^2 * c[2] + 6^3 * c[3] + 6^4 * c[4] + 6^5 * c[5] + 6^6 * c[6] + 6^7 * c[7] + 6^8 * c[8] + 6^9 * c[9] + 6^10 * c[10] = 87  
7^0 * c[0] + 7^1 * c[1] + 7^2 * c[2] + 7^3 * c[3] + 7^4 * c[4] + 7^5 * c[5] + 7^6 * c[6] + 7^7 * c[7] + 7^8 * c[8] + 7^9 * c[9] + 7^10 * c[10] = 79  
8^0 * c[0] + 8^1 * c[1] + 8^2 * c[2] + 8^3 * c[3] + 8^4 * c[4] + 8^5 * c[5] + 8^6 * c[6] + 8^7 * c[7] + 8^8 * c[8] + 8^9 * c[9] + 8^10 * c[10] = 82  
9^0 * c[0] + 9^1 * c[1] + 9^2 * c[2] + 9^3 * c[3] + 9^4 * c[4] + 9^5 * c[5] + 9^6 * c[6] + 9^7 * c[7] + 9^8 * c[8] + 9^9 * c[9] + 9^10 * c[10] = 76
10^0 * c[0] + 10^1 * c[1] + 10^2 * c[2] + 10^3 * c[3] + 10^4 * c[4] + 10^5 * c[5] + 10^6 * c[6] + 10^7 * c[7] + 10^8 * c[8] + 10^9 * c[9] + 10^10 * c[10] = 68

Note that the number of unknowns are c[1] to c[10], so 10. We know that c[0] = 72, so it is not an unknown and the number of equations is 10.

Now we just multiply all numbers with 1814400, divide by the same in the answer, so it doesn't change anything or probably the answer found by solving the equations would not be whole numbers, so multiply by 1814400 to get whole numbers.

You can solve these equations by using this code for solving simultaneous equations of any order.