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Question: Find the sum of digits of the $1^4 +2^4 + ...... + 25^4$

I can't seem to find a way to solve this question without using brute force. Is there a simpler way to solve this question?

Hector Lombard
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    https://math.stackexchange.com/questions/241760/geometric-interpretation-for-sum-of-fourth-powers – lab bhattacharjee Dec 31 '18 at 05:56
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    For reference OP, lab's post presumably stems from the relevant sum discussed there:

    $$\sum_{k=1}^nk^4=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1)$$

    Take $n=25$. While not a direct answer, it should help you on your way.

     – PrincessEev Dec 31 '18 at 05:59
  • Brute force isn't too difficult if you consider $odd^4$ all end with $1$ or $5$. And $even^5$ all end in $6$ or $0$. – fleablood Dec 31 '18 at 06:27

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$$\sum_{n=1}^{25}n^4=\frac{1}{5}\sum_{n=1}^{25}((n+1)^5-n^5-10n^3-10n^2-5n-1)=$$ $$=\frac{1}{5}\left(26^5-1-10\cdot\frac{25^226^2}{4}-10\cdot\frac{25\cdot26\cdot51}{6}-5\cdot\frac{25\cdot26}{2}-25\right)=$$ $$=2153645.$$ I got 26.

I used the telescoping sum: $$\sum_{n=1}^{25}((n+1)^5-n^5)=26^5-1$$ and known identities, which we can got by the same way: $$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4},$$ $$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$$ and $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$

Michael Rozenberg
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