# Solution to Problem 30 on Project Euler

Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:

1634 = 14 + 64 + 34 + 44
8208 = 84 + 24 + 04 + 84
9474 = 94 + 44 + 74 + 44

As 1 = 14 is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

## My Solution

``````#include <stdio.h>
#include <math.h>

int main(){
int i,a,b,c,d,e,f,n,sum;
double number;
sum =0;

for (i=10;i<100;i++){
a = i%10;
b = i/10;
number=pow(a,5)+pow(b,5);
if ((int)number==i)
sum+=(int)number;
}

for (i=100;i<1000;i++){
n=i;
a = n%10;
n /= 10;
b = n%10;
n /= 10;
c = n;
number=pow(a,5)+pow(b,5)+pow(c,5);
if ((int)number==i)
sum+=(int)number;
}

for (i=1000;i<10000;i++){
n=i;
a = n % 10;
n /= 10;
b = n % 10;
n /= 10;
c = n % 10;
n /= 10;
d = n % 10;
number=pow(a,5)+pow(b,5)+pow(c,5)+pow(d,5);
if ((int)number==i)
sum+=(int)number;
}

for (i=10000;i<100000;i++){
n=i;
a = n%10;
n /= 10;
b = n%10;
n /= 10;
c = n%10;
n /= 10;
d = n%10;
n /= 10;
e = n;
number=pow(a,5)+pow(b,5)+pow(c,5)+pow(d,5)+pow(e,5);
if ((int)number==i)
sum+=(int)number;
}

for (i=100000;i<1000000;i++){
n=i;
a = n%10;
n /= 10;
b = n%10;
n /= 10;
c = n%10;
n /= 10;
d = n%10;
n /= 10;
e = n%10;
n/=10;
f=n;
number=pow(a,5)+pow(b,5)+pow(c,5)+pow(d,5)+pow(e,5)+pow(f,5);
if ((int)number==i)
sum+=(int)number;
}

printf("%d\n",sum);

return 0;
}
``````