2023 AMC 10B Problem 22

P21-25~5 minPrint

Number Properties Absolute Value & Inequalities Bound the quantity above/below or estimate to pin it down Substitute to simplify (u = x+1/x, shifting, scaling)

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

Show solution without answering

Key insight

Call the floor k. Then x = (k squared + 2)/3, and this x must really have floor k. Only k = 0, 1, 2, 3 pass that test.

Solution

$\lfloor x \rfloor$ is always a whole number, so give it a name and try values.

1. Name the floor. Let $k = \lfloor x \rfloor$ , a whole number (possibly $0$ or negative). The equation says $k^2 - 3x + 2 = 0$ , so
$x = \frac{k^2 + 2}{3}.$
2. The catch. This $x$ only counts if its floor really is $k$ . That means $x$ must be at least $k$ but less than $k + 1$ .
3. Try values of $k$ .

$k$	$-1$	$0$	$1$	$2$	$3$	$4$
$x = \frac{k^2+2}{3}$	$1$	$\frac{2}{3}$	$1$	$2$	$\frac{11}{3}$	$6$
$\lfloor x \rfloor$	$1$	$0$	$1$	$2$	$3$	$6$
equals $k$ ?	no	yes	yes	yes	yes	no

4. Why no other $k$ works. The condition $x < k + 1$ means $k^2 + 2 < 3k + 3$ , that is, $k^2 < 3k + 1$ .
- For $k = 4$ : $16 < 13$ is false. For bigger $k$ , $k^2$ grows much faster than $3k + 1$ , so it stays false.
- For $k = -1$ : $1 < -2$ is false. For more negative $k$ , $k^2$ is positive and $3k + 1$ is negative, so it stays false.
(The other condition, $x \ge k$ , turns into $k^2 - 3k + 2 \ge 0$ , or $(k - 1)(k - 2) \ge 0$ , which is true for every whole number $k$ .)
5. Count. The solutions are $x = \frac{2}{3}, 1, 2, \frac{11}{3}$ . Check one: $x = \frac{11}{3}$ has floor $3$ , and $9 - 11 + 2 = 0$ . That is $4$ values.

The answer is $\boxed{\textbf{(B)}\ 4}$ .

Why this works

An equation with $\lfloor x \rfloor$ is really many ordinary equations, one for each whole-number value of the floor. Solve for $x$ in terms of that whole number $k$ , then keep only the answers that land between $k$ and $k + 1$ . That condition allows only a few values of $k$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let $k = \lfloor x \rfloor$ , an integer. The equation gives $x$ in terms of $k$ :
$x = \frac{k^2 + 2}{3}.$
For this to be consistent we need $\lfloor x \rfloor = k$ , i.e. $k \le \frac{k^2 + 2}{3} < k + 1$ . Multiply by $3$ :

Left inequality: $3k \le k^2 + 2 \iff k^2 - 3k + 2 \ge 0 \iff (k-1)(k-2) \ge 0$ , so $k \le 1$ or $k \ge 2$ : every integer satisfies this.
Right inequality: $k^2 + 2 < 3k + 3 \iff k^2 - 3k - 1 < 0$ . The roots of $k^2 - 3k - 1$ are $\frac{3 \pm \sqrt{13}}{2} \approx -0.30$ and $3.30$ , so the integers allowed are $k = 0, 1, 2, 3$ .

Each $k$ gives one $x$ : $k = 0 \Rightarrow x = \frac{2}{3}$ ; $k = 1 \Rightarrow x = 1$ ; $k = 2 \Rightarrow x = 2$ ; $k = 3 \Rightarrow x = \frac{11}{3}$ . Quick check: $\lfloor 2/3 \rfloor = 0$ and $\lfloor 11/3 \rfloor = 3$ , as required. That is $4$ distinct solutions.

The answer is $\boxed{\textbf{(B)}\ 4}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)