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anemone
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Find the values of integer $x$ such that $\left\lfloor{\dfrac{x-1}{2}}\right\rfloor+\left\lfloor{\dfrac{x^2-x}{3}}\right\rfloor=x$.
anemone said:Find the values of integer $x$ such that $\left\lfloor{\dfrac{x-1}{2}}\right\rfloor+\left\lfloor{\dfrac{x^2-x}{3}}\right\rfloor=x$.
kaliprasad said:we take 2 separate cases
case 1) x is odd say 2m+1
then we get
$m+\lfloor\dfrac{2m(2m+1)}{3}\rfloor= 2m+ 1$
or $\lfloor\dfrac{2m(2m+1)}{3}\rfloor= m+ 1$
or $m+1\le \dfrac{2m(2m+1)}{3}\lt m+ 2$
or $3m+3\le 2m(2m+1) \lt m+ 6$
or $ 3 \le 4m^2 -m \lt 6$
m = 1 or = -1 giving x = 3 or -1case 2) x is even say 2mthen we get
$m-1 +\lfloor\dfrac{2m(2m-1)}{3}\rfloor= 2m $
or $\lfloor\dfrac{2m(2m-1)}{3}\rfloor= m+1 $
or $m+1\le \dfrac{2m(2m-1)}{3}\lt m+2 $
or $3\le 4m^2-5m \lt 6$
giving m = 2 or x = 4so solution x = -1 or 3 or 4
anemone said:Hi kaliprasad,
Thanks for participating but one of the solutions doesn't satisfy the original equation...(Mmm)
anemone said:Okay, and thanks, kaliprasad, for participating!
My solution:
Whereas for $x≤0$, all the three terms are negative.
kaliprasad said:for x = 0 the 2nd term is zero
johng said:Here's another solution.
The value of x cannot be determined without more information. The equation is a floor function equation, meaning that it has infinite solutions. The possible solutions can be found by trial and error.
As mentioned before, the equation has infinite solutions and cannot be solved algebraically. The best way to find solutions is by trial and error. You can start by plugging in different values for x and checking if the equation holds true.
The equation has infinite solutions because it is a floor function equation. The floor function rounds the input to the nearest integer that is less than or equal to the input. Therefore, the equation is satisfied for any value of x that rounds to the same integer when plugged into both floor functions.
No, there is no specific method for finding solutions to floor function equations. As mentioned before, the best approach is to trial and error by plugging in different values for x and checking if the equation holds true.
No, this equation cannot be solved using calculus. Calculus deals with continuous functions, whereas the floor function is a discrete function. Therefore, the techniques used in calculus cannot be applied to this equation.