Here’s the problem:

—–

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?

—–

## My Solution

Here’s the solution implementing the large number as an array of digits:

```
#include <stdio.h>
#define MAX 400
int main(){
int vet[MAX]={0};
int i,x,y,v,f,sum;
vet[0]=1;
y=1000;
while (y>0){
v = 0;
for (i=0;i<MAX;i++){
x = vet[i]*2 + v;
v = 0;
if (x > 9){
f = x % 10;
v = x / 10;
}
else
f = x;
vet[i] = f;
}
y = y - 1;
}
sum=0;
for (i=MAX-1;i>=0;i--)
sum=sum+vet[i];
printf("%dn",sum);
return 0;
}
```

And using the bignum library:

```
#include <stdio.h>
#include <math.h>
#include <gmp.h>
int main(){
mpz_t x,y;
int i,sum;
sum=0;
mpz_init(x);
mpz_init(y);
mpz_ui_pow_ui(x,2,1000);
while (mpz_sgn(x)){
sum += mpz_mod_ui(y,x,10);
mpz_fdiv_q_ui(x,x,10);
}
printf("%dn",sum);
return 0;
}
```