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

{    x + 1                    if $$\left \lfloor{x}\right \rfloor$$ is prime

p(x) =  {    p(y) + (x + 1 - $$\left \lfloor{x}\right \rfloor$$)  otherwise

where Y is the greatest prime factor of $$\left \lfloor{x}\right \rfloor$$. Express the range of $$p$$ in interval notation.

I tried [3, 8] because when x = 2, p(x) = 3, and when x = 7, p(x) = 8. I thought that was the range, but it is apparently not.

Jan 29, 2019
edited by Guest  Jan 29, 2019

#1
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Try [3, 8)   i.e. open interval on the right.

Jan 29, 2019
#2
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I tried that for guest, and I am pretty sure it is incorrect. I think p(x) can equal 8 because if x = 7, p(x) = 7+1 = 8.

PartialMathematician  Jan 30, 2019
edited by PartialMathematician  Jan 30, 2019
#3
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True!

Alan  Jan 30, 2019
#4
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If you only consider the integer values, then the range is [3,8].

Consider a number infinitesimally close to 8. (7.9999999999999999999999...)

p(7.9999999999999999999...) = x + 1 = 8.999999999999999999999... = 9

So the range is actually [3,9].

Feb 1, 2019
#5
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Oh, I see what you are saying. I thought the first condition with x+1 if floor x is prime was saying: If floor x is prime, then p(x) = floor x + 1.

That explains a lot. But again, the number can be infinitely close to 8, such as 7.99999..., but the number can never be 8, so the answer range is $$\boxed{[3,9)}$$.