Sum of Fifth power of digits

Computer Science Level 3

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

$1634 = 1^{4} + 6^{4} + 3^{4} + 4^{4}$

$8208 = 8^{4} + 2^{4} + 0^{4} + 8^{4}$

$9474 = 9^{4} + 4^{4} + 7^{4} + 4^{4}$

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

Find the sum of all the numbers (less than 1 million) that can be written as the sum of fifth powers of their digits.

Note: $1^5 = 1$ is not considered a sum and it is not included.

Try other problems Here

The answer is 443839.

3 solutions

Kunal Gupta
Sep 19, 2017

Small Python Script :

def X(n):
  x = n
  val = 0
  while x>0:
    val+=(x%10)**5
    x/=10
  if val==n:
    return 1
  return 0
ans= 0
for i in range(2,1000000):
  if X(i)==1:
    ans+=i
print ans

Rishabh Deep Singh
May 31, 2017

import java.lang.Math;

public class main {
    public static int last_digit(int n){
        n=n%10;
        return n;
    }

    public static void main(String[] args){
        int n,temp,temp_sum=0;
        for (int i=2;i<=9999999;i++) {
            n=i;
            temp=i;
            int sum=0;
            while (n > 0) {
                int temp_last = last_digit(n);
                int temp_power = (int) Math.pow(temp_last,5);
                n = (n - (last_digit(n))) / 10;
                sum += temp_power;
            }
            if(temp==sum){
                temp_sum+=temp;
                System.out.println("required number"+temp);
            }
        }
        System.out.println("Sum of Numbers = "+temp_sum);
    }
}

Steven Chase
May 31, 2017

Here's some primitive code (Python)

msum = 0

for j in range(1,10**6):

    num = j

    ####################################

    p6 = num / (10**6)

    num = num % (10**6)

    ####################################

    p5 = num / (10**5)

    num = num % (10**5)

    ####################################

    p4 = num / (10**4)

    num = num % (10**4)

    ####################################

    p3 = num / (10**3)

    num = num % (10**3)

    ####################################

    p2 = num / (10**2)

    num = num % (10**2)

    ####################################

    p1 = num / (10**1)

    num = num % (10**1)

    ####################################

    p0 = num / (10**0)

    num = num % (10**0)

    ####################################

    q = p6**5 + p5**5 + p4**5 + p3**5 + p2**5 + p1**5 + p0**5

    res = q - j

    if res == 0:

        msum = msum + j
        print j

print ""
print ""
print (msum - 1)


####### Results #######

1
4150
4151
54748
92727
93084
194979


443839

Type your code like this

"```python

Code

```"

Without Quotes, it will show numbers Too and looks beautiful.

BTW nice solution.

Rishabh Deep Singh - 4 years ago

Sum of Fifth power of digits

The answer is 443839.

3 solutions

Code

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