HINT for summing digits of a large power

Question

I recently started working through the Project Euler challenges, but I've got stuck on #16 (http://projecteuler.net/problem=16)

$2^{15} = 32768$ and the sum of its digits is $3 + 2 + 7 + 6 + 8 = 26$. What is the sum of the digits of the number $2^{1000}$?

(since I'm a big fan of generality, my interpretation is to find a solution to the sum of digits of $a^b$ in base $c$, and obviously I'm trying to solve it without resorting to "cheats" like arbitrary-precision numbers).

I guess this is simpler than I'm making it, but I've got no interest in being told the answer so I haven't been able to do a lot of internet searching (too many places just give these things away). So I'd appreciate a hint in the right direction.

I know that $2^{1000} = 2^{2*2*2*5*5*5} = (((((2^2)^2)^2)^5)^5)^5$, and that the repeated sum of digits of powers of 2 follows the pattern $2, 4, 8, 7, 5, 1$, and that the last digit can be determined by an efficient pow-mod algorithm (which I already have from an earlier challenge), but I haven't been able to get further than that… (and I'm not even sure that those are relevant).

Answer 1

In this case, I'm afraid you just have to go ahead and calculate $2^{1000}$. There are various clever ways to do this, but for numbers this small a simple algorithm is fast enough.

The very simplest is to work in base 10 from the start. Faster is to work in binary, and convert to base 10 at the end, but this conversion is not completely trivial. A compromise (on a 32-bit machine) would be to work in base 1000000000, so that you can double a number without overflow. Then at the end the conversion to base 10 is much simpler.

Answer 2

Here are 2 hints that really helped me along to get to the solution:

1)

"We can use an array, or hash map, to store the digits of any big number. Every time we multiply each position (the values) of the array/hash map by two, we need to start from the ‘One’ position to carry over the digits.

With this mechanism, we can multiply 1000 times 2 quite quickly. We don’t even need 1000 arrays – one array is enough when we sum the result directly."

Source: https://www.educative.io/edpresso/what-is-the-power-digit-sum

You may want to stop reading that page before looking at the implementation code towards the bottom of that page.

2.

"Consider an array of ints with the digits, and repeatedly multiply by 2."

Source: https://www.hackerrank.com/contests/projecteuler/challenges/euler016/forum/comments/189057

This one isn't extremely insightful, but it did help light a lightbulb in my head. To go a bit further (and without revealing everything), how can you cache results as you "repeatedly multiply by 2"?

Edit: Note that my comment #2 is a rewording of Sil's answer (https://math.stackexchange.com/a/2879132/162331) above, particularly "Try 2⋅2, 4⋅2, 8⋅2, … by hand, notice how digits are carried over. Try to formalize this into an algorithm that uses array for digits.".