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📅 May 4, 2026 👤 scraface

May 04, 2026

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📝 Recent Posts

\left(\sum_{k=1}^{n} k^3
ight) = \left(rac{n(n+1)}{2}
For $n = 10$:
\sum_{k=1}^{10} k^3 = \left(rac{10 \cdot 11}{2}
ight)^2 = (55)^2 = 3025.
Now compute $3025 \mod 9$. Use the digit sum method:
Digit sum of $3025$: $3 + 0 + 2 + 5 = 10$, then $1 + 0 = 1$, so $3025...
Alternatively, divide: $3025 \div 9 = 336 imes 9 = 3024$, remainder $...
Thus, the remainder is $1$.
Therefore, the answer is $oxed{1}$.

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