2023 AMC 10B Problem 22

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

A floor equation is really a family of ordinary equations, one for each integer value of the floor, each valid only on the interval where that floor is correct. Solving for $x$ in terms of $k$ and then imposing $k \le x < k+1$ converts the problem into a quadratic inequality in the integer $k$, which has only finitely many solutions.

Since $3x = k^2 + 2$ with $k \le x$, we get $k^2 + 2 \ge 3k$, always true, and from $x < k + 1$, $k^2 + 2 < 3k + 3$. Test integers directly: $k = -1$ gives $3 < 0$, false; $k = 0, 1, 2, 3$ give $2 < 3$, $3 < 6$, $6 < 9$, $11 < 12$, all true; $k = 4$ gives $18 < 15$, false. Four solutions.

• Replacing floor(x) by x and solving x^2 - 3x + 2 = 0, which finds only x = 1 and x = 2 and misses x = 2/3 and x = 11/3.
• Finding the four candidate values but not verifying the floor condition, or wrongly rejecting $x = 11/3$ because it is not an integer.