Interesting properties of numbers

Python 2.7:

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def digit_sum(n):
    return sum([int(c) for c in str(n)])
def goal(n):
    return digit_sum(n) > digit_sum(n**3)
n = 0
while not goal(n):
    n += 1
print "Answer:", n

I had Bookmarked this site a long time ago , and now it's coming in handy .

For those who don't read the comments below and just want to look at the solution , let me present a solution that my friend Kishlaya has come up with .

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class prog  {

    public static void main (String[] args)
    {
        for(int x=1;;x++) {
            if(sum(x)>sum(x*x*x)) {
                System.out.println(x);
                break;
            }
        }
    }

    static int sum(int n) {
        int sum=0;
        while(n>0) {
            sum+=n%10;
            n=n/10;
        }
        return sum;
    }
}

returns 587

My python 2.7 solution:
for i in range(5000):
a=str(i)
s1=0
for j in a:
s1+=int(j)
s2=0
b=str(i**3)
for k in b:
s2+=int(k)
if s1>s2:
print a

Mine is a bit convoluted but here is what I did:

My C# code (Uses System.Numeric.dll library for BigInteger)

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static void Main(string[] args)
        {
            BigInteger n = 1;
            while (n > 0)
            {
                BigInteger DSn = DigitSum(n);
                BigInteger DSn3 = DigitSum(n * n * n);
                if (DSn>DSn3)
                {
                    Console.WriteLine(n);
                    Console.ReadLine();
                }
                n++;
            }
        }
        public static BigInteger DigitSum(BigInteger n)
        {
            BigInteger num = 0, r, sum = 0;
            num = n;
            while (num > 0)
            {
                r = num % 10;
                num = num / 10;
                sum = sum + r;
            }
            return sum;
        }

$\quad$

Output = 587

Using javascript. With google chrome, you can easily run this code.

for (x = 1; x <= 1000; x++) { var f1 = String(x), f2 = String(x x x); var n1 = 0, n2 = 0; for (var i in f1) n1 += Number(f1[i]); for (var j in f2) n2 += Number(f2[j]); if (n1 > n2) {alert(x); break;} }

Infinite Loop hehe

C language :

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#include <stdio.h>
int ds(int i)    
{
    int s=0;
    while(i>=1)
    {
        s+=i%10;
        i=i/10;
    }
    return s;
}
int main(void)
{
      int j;
      for(j=1;j>0;j++)
      {
          if(ds(j)>ds(j*j*j))
          {
              printf("%d",j);
              break;
          }
      }
}

Java solution -

public static void main(String[] args){
    for(int i = 1;;i++)
        if(sum(i) > sum(i * i * i))
            System.out.println(i);
}

private static int sum(int n){
    int sum = 0;
    while (n != 0){
        sum += n % 10;
        n /= 10;
    }
    return sum;
}

Execution time - 0.000456635 seconds