What is the sum of the digits of the number 2^1000?

Question

This is a problem from Project Euler, and this question includes some source code, so consider this your spoiler alert, in case you are interested in solving it yourself. It

Answer 1

Try using BigInteger type , 2^100 will end up to a a very large number for even double to handle.

Answer 2

For calculating the values of such big numbers you not only need to be a good programmer but also a good mathematician. Here is a hint for you, there's familiar formula a^x = e^{x ln a} , or if you prefer, a^x = 10^{x log a}.

More specific to your problem 2¹⁰⁰⁰ Find the common (base 10) log of 2, and multiply it by 1000; this is the power of 10. If you get something like 10^53.142 (53.142 = log 2 value * 1000) - which you most likely will - then that is 10⁵³ x 10^0.142; just evaluate 10^0.142 and you will get a number between 1 and 10; and multiply that by 10⁵³, But this 10⁵³ will not be useful as 53 zero sum will be zero only.

For log calculation in C#

Math.Log(num, base);

For more accuracy you can use, Log and Pow function of Big Integer.

Now rest programming help I believe you can have from your side.

Answer 3

 main()
 {
   char c[60];
  int k=0;
     while(k<=59)
      {
    c[k]='0';
   k++;

    }
       c[59]='2';
       int n=1;
     while(n<=999)
       {
       k=0;
     while(k<=59)
      {
        c[k]=(c[k]*2)-48;
        k++;
      } 
    k=0;
     while(k<=59)
        {
        if(c[k]>57){ c[k-1]+=1;c[k]-=10;   }
       k++;
         }
       if(c[0]>57)
        {
         k=0;
         while(k<=59)
           {
         c[k]=c[k]/2;
          k++;
           }
           printf("%s",c);
             exit(0);
           }

            n++;
            }
          printf("%s",c);
              } 

Answer 4

Python makes it very simple to compute this with an oneliner:

print sum(int(digit) for digit in str(2**1000))

or alternatively with map:

print sum(map(int,str(2**1000)))