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

\begin{align}

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

\end{align}

As $1 = 1^4$ 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.

### Finding numbers that are equal to the sum of the fifth powers of their digits is a bit computationally intensive because these numbers could technically be as large as 354,294 (since $9^5*6=354,294$; any number larger than this couldn’t possibly be such a number). Therefore, we need to search through all the numbers from 2 up to 354,294.

To calculate the sum of fifth powers for each number we separate digits from the number, raise each digit to the 5th power and sum them. If this sum will be equal to the original number – we have a match.

Doing that, we find there are only four such numbers:

\begin{align}

4150 &= 4^5 + 1^5 + 5^5 + 0^5\\

4151 &= 4^5 + 1^5 + 5^5 + 1^5\\

54748 &= 5^5 + 4^5 + 7^5 + 4^5 + 8^5\\

92727 &= 9^5 + 2^5 + 7^5 + 2^5 + 7^5\\

194979 &= 1^5 + 9^5 + 4^5 + 9^5 + 7^5 + 9^5

\end{align}

Adding these all up gives you the sum of $4150 + 4151 + 54748 + 92727 + 194979 = 443839$. So, 443839 is sum of all the numbers that can be written as the sum of fifth powers of their digits.

##### More Answers:

Self PowersPrime Permutations

Consecutive Prime Sum

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