AMC 10 · 2023 · #22

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

floor-functioncaseworklinear-equations-one-varbound-inequality-then-enumerate caseworksystematic-enumerationbound-inequality-then-enumerate ↑ Prerequisites: floor-functionlinear-equations-one-var

📏 Long solution 💡 3 insights

Problem

Find how many real numbers x satisfy ⌊ x ⌋² - 3x + 2 = 0, where ⌊ x ⌋ is the greatest integer ≤ x.

Pick an answer.

(A)

$text{an infinite number}$

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

⌊ x ⌋ being an integer is the key opening. Set n = ⌊ x ⌋ (tool #9 — replace the floor function with the easier integer variable n). For each integer n, the equation forces one candidate x = (n²+2)/3. Now tool #6 (Guess and Check) plus tool #2 (Systematic List) sweep small integers n = …, -1, 0, 1, 2, 3, 4, … and check whether ⌊ (n²+2)/3 ⌋ = n. The candidates grow quadratically in n while the valid range grows linearly, so the search has only a few cases.

1STEP 1

Let n = ⌊ x ⌋, so n² - 3x + 2 = 0 gives x = (n²+2)/3 — valid only if n ≤ x ＜ n+1.

x = (n² + 2)/3, n ≤ x ＜ n + 1

2STEP 2

n = 0: x = (0+2)/3 = $\frac{2}{3}$ ≈ 0.667, and 0 ≤ 0.667 ＜ 1, so x = $\frac{2}{3}$ is valid.

n=0: x =

\frac{2}{3}

, ⌊

\frac{2}{3}

⌋ = 0 ✓

3STEP 3

n = 1: x = (1+2)/3 = 1, and 1 ≤ 1 ＜ 2, so x = 1 works.

n=1: x = 1, ⌊ 1 ⌋ = 1 ✓

4STEP 4

n = 2: x = (4+2)/3 = 2, and 2 ≤ 2 ＜ 3, so x = 2 works.

n=2: x = 2, ⌊ 2 ⌋ = 2 ✓

5STEP 5

n = 3: x = (9+2)/3 = $\frac{11}{3}$ ≈ 3.667, and 3 ≤ 3.667 ＜ 4, so x = $\frac{11}{3}$ works.

n=3: x =

\frac{11}{3}

≈ 3.667, ⌊

\frac{11}{3}

⌋ = 3 ✓

6STEP 6

n = 4: x = (16+2)/3 = 6, but 6 ≥ 5 so ⌊ 6 ⌋ ≠ 4 — reject; for all n ≥ 4 the candidate overshoots [n, n+1).

n=4: x = 6 ≥ 5, floor=6 ≠ 4 ✗

7STEP 7

Negative n: n = -1 gives x = 1 ∉ [-1, 0); every n ≤ -1 makes a positive candidate that can't fit a non-positive window — all rejected.

n=-1: x = 1 ∉ [-1, 0) ✗

8STEP 8

Valid n ∈ {0, 1, 2, 3} give four distinct solutions x ∈ { $\frac{2}{3}$ , 1, 2, $\frac{11}{3}$ }, so the answer is choice (B).

x ∈ {

\frac{2}{3}

, 1, 2,

\frac{11}{3}

} → 4 solutions → (B)

Answer

Spot-check by plugging back: x =

\frac{2}{3}

gives 0² - 3 ·

\frac{2}{3}

+ 2 = 0 - 2 + 2 = 0 ✓. x = 1: 1 - 3 + 2 = 0 ✓. x = 2: 4 - 6 + 2 = 0 ✓. x =

\frac{11}{3}

: 9 - 11 + 2 = 0 ✓. Four solutions found. The boundary check n = 4 failed cleanly (x = 6 way outside [4, 5)), so no more solutions hide above. Floor of any negative window cannot equal a positive x, so negative n are ruled out. Answer (B) = 4 matches.

💡Key takeaway

This AMC 10 problem only needs Grade 6 expression and number-line reasoning you already know — replace the scary floor function ⌊ x ⌋ with an integer name n, get x = (n²+2)/3, and check whether the candidate truly lives in the window [n, n+1). Sweeping n = 0, 1, 2, 3, 4, … gives exactly four valid candidates, so the answer is (B) = 4.

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