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Let \(p(x)\) be defined on \(2 \le x \le 10\) such that where 

\(p(x) = \begin{cases} x + 1 &\quad \lfloor x \rfloor\text{ is prime} \\ p(y) + (x + 1 - \lfloor x \rfloor) &\quad \text{otherwise} \end{cases}\)

 is the greatest prime factor of \(\lfloor x\rfloor\) Express the range of \(p\) in interval notation.

 

I've tried to do this problem and I think it's [3,10) but i'm not sure... can someone comfirm this maybe?

 Feb 26, 2021
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The function p gets bumped up by one at each prime, so the range is [2,4) U [5,6) U [7,8).

 Feb 26, 2021

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