How Many Integers Meet the Condition {√n - √(23×24)}² < 1?

AI Thread Summary
The discussion focuses on finding the integers that satisfy the inequality {√n - √(23×24)}² < 1. One participant solved the problem using trial and error and seeks a systematic, step-by-step solution. It is noted that the square root of 23×24 can be approximated by the square root of 23 squared for simplicity. The conversation emphasizes the importance of sharing methods to find the solution effectively. Ultimately, the goal is to determine the specific integers that meet the given condition.
pratikaman
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How many integers satisfy {√n-√(23×24)}^2<1


I was able to solved this by trial and error method , but i want to know systematic step-wise solution.
 
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pratikaman said:
How many integers satisfy {√n-√(23×24)}^2<1


I was able to solved this by trial and error method , but i want to know systematic step-wise solution.


Show us what you did; perhaps your method was as good as anything.
 
Notice that the square root of 23x24 is approximately the square root of 23 squared. Since asked about only integers, this approximation suffices
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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