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Question Number 213001 by efronzo1 last updated on 28/Oct/24

$$\underset{―}{\underbrace{⋖}\cancel{}}{14}^3+{15}^3+{16}^3+...+{24}^3+{25}^3$$

Commented by Ghisom last updated on 28/Oct/24

$$12$$

Answered by mehdee7396 last updated on 28/Oct/24

$$1^3+2^3+...+{25}^3={\left(\frac{25\times 26}{2}\right)}^2$$$$1^3+2^3+...+{13}^3={\left(\frac{13\times 14}{2}\right)}^2$$$$s={\left(25\times 13\right)}^2-{\left(7\times 13\right)}^2={13}^2\times {24}^2$$$$\Rightarrow p=13\times 24$$$$\Rightarrow p-300=12$$$$$$

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