Solving equations with greatest integer function

AI Thread Summary
To solve equations involving the greatest integer function, it's essential to understand how it operates. For example, if you have the equation 99 = [2x+1]/3, you can multiply both sides by 3 to isolate the greatest integer function, leading to [2x+1] = 297. This implies that 297 ≤ 2x + 1 < 298, which can be simplified to find the range for x as 148 ≤ x < 148.5. The discussion emphasizes the importance of recognizing the boundaries set by the greatest integer function to solve such equations effectively. Understanding these principles allows for accurate solutions to similar problems.
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Homework Statement


I can't find a step by step explanation for solving these types of equations

eg.
99 = [2x+1]/3


Homework Equations



eg.
99 = [2x+1]/3

or

48 = 4[2x/3]

How do you handle the multipliers iand constants inside the brackets?
thx
3. The Attempt at a Solution
 
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Think about how the greatest integer function works. For example,

<br /> \lfloor 3.2 \rfloor = \lfloor 3.582 \rfloor = 3<br />

and in fact, if 3 \le x &lt; 4 it is true that

<br /> \lfloor x \rfloor = 3<br />

So, if you know that

<br /> \frac{\lfloor 2x+1\rfloor}{3} = 99<br />

you also know that

<br /> \lfloor 2x+1 \rfloor = 297<br />

(the 3 in the denominator is not in the function). What does the final
statement above tell you about how large 2x + 1 must be?
 
statdad said:
Think about how the greatest integer function works. For example,

<br /> \lfloor 3.2 \rfloor = \lfloor 3.582 \rfloor = 3<br />

and in fact, if 3 \le x &lt; 4 it is true that

<br /> \lfloor x \rfloor = 3<br />

So, if you know that

<br /> \frac{\lfloor 2x+1\rfloor}{3} = 99<br />

you also know that

<br /> \lfloor 2x+1 \rfloor = 297<br />

(the 3 in the denominator is not in the function). What does the final
statement above tell you about how large 2x + 1 must be?

--------------------
so 297 <= 2x+1 < 298

296 <=2x and 2x < 297
148 <=x and x < 148.5

Did I get it?
 
Yup.
 
statdad said:
Yup.


Thanks!
 
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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