2023 AMC 10B Problems/Problem 22

Problem

How many distinct values of 𝑥 satisfy $\lfloor{x}\rfloor^2-3x=0.$ where $\lfloor{x}\rfloor$ denotes the largest integer less than or equal to 𝑥?

Solution (Quick)

A quadratic equation can have up to 2 real solutions. With the $\lfloor{x}\rfloor$ , it could also help generate another pair. We have to verify that the solutions are real and distinct.

First, we get the trivial solution by ignoring the floor. $(x-2)(x-1) = 0$ , we get $(2,1)$ as our first pair of solutions.

Up to this point, we can rule out A,E.

Next, we see that $\lfloor{x}\rfloor^2-3x=0.$ This implies that $-3x$ must be an integer. We can guess and check $x$ as $\dfrac{k}{3}$ which yields $(\dfrac{2}{3},\dfrac{11}{3}).$

So we got 4 in total $(\dfrac{2}{3},1,2,\dfrac{11}{3}).$

~Technodoggo

Solution

First, $x=2,1$ are trivial solutions

We assume from the shape of a parabola and the nature of the floor function that any additional roots will be near 2 and 1

We can now test values for $\lfloor{x}\rfloor$ :

$\lfloor{x}\rfloor=0$

We have $0-3x+2=0$ . Solving, we have $x=\frac{2}{3}$ . We see that $\lfloor{\frac{2}{3}}\rfloor=0$ , so this solution is valid

$\lfloor{x}\rfloor=-1$

We have $1-3x+2=0$ . Solving, we have $x=1$ . $\lfloor{1}\rfloor\neq-1$ , so this is not valid. We assume there are no more solutions in the negative direction and move on to $\lfloor{x}\rfloor=3$