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Let the fifth number be \( x \). Then: - Portal da AcÃºstica

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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\sum_{k=1}^{10} k^3 = \left( \frac{10 \times 11}{2} \right)^2 = (55)^2...
Now compute \(3025 \mod 11\). First, reduce 55 modulo 11:
\(55 \div 11 = 5\), so \(55 \equiv 0 \pmod{11}\).
Then \(55^2 \equiv 0^2 = 0 \pmod{11}\).
Thus, \(3025 \equiv 0 \pmod{11}\).
Alternatively, since \(55 \equiv 0 \mod 11\), the square is divisible...
Therefore, the remainder is \(\boxed{0}\).
An emerging technology researcher exploring base systems in quantum co...
To convert \(321_9\) to base ten, expand using powers of 9:
321_9 = 3 \times 9^2 + 2 \times 9^1 + 1 \times 9^0

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