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

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.