2023 AMC 10B Problem 22

$\lfloor x \rfloor$ 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 $k$.

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 $\boxed{\textbf{(B)}\ 4}$.

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 $k$.

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 $\boxed{\textbf{(B)}\ 4}$.