AMC 10 · 2023 · #22

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

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

📏 Long solution 💡 3 insights

Problem

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 $x$ ?

$\textbf{(A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0$

Pick an answer.

(A)

$text{an infinite number}$

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Find how many real numbers $x$ satisfy $\lfloor x \rfloor^{2} - 3x + 2 = 0$, where $\lfloor x \rfloor$ is the greatest integer $\le x$.

Givens: $\lfloor x \rfloor$ is the floor of $x$ (an integer $n$ with $n \le x < n+1$); The equation is $\lfloor x \rfloor^{2} - 3x + 2 = 0$; Answer choices: (A) infinitely many, (B) $4$, (C) $2$, (D) $3$, (E) $0$

Unknowns: The number of distinct real solutions $x$

Understand

Restated: Find how many real numbers $x$ satisfy $\lfloor x \rfloor^{2} - 3x + 2 = 0$, where $\lfloor x \rfloor$ is the greatest integer $\le x$.

Plan

Primary tool: #6 Guess and Check

Secondary: #2 Make a Systematic List, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities

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

Execute — Answer: B

#9 Solve an Easier Related Problem 6.EE.B.6 Step 1

Let $n = \lfloor x \rfloor$.
The equation becomes $n^{2} - 3x + 2 = 0$, so $x = \tfrac{n^{2} + 2}{3}$.
For this $x$ to be a real solution of the original equation, the integer $n$ must actually equal the floor of this $x$ — that is, $n \le \tfrac{n^{2}+2}{3} < n + 1$.

$$x = \dfrac{n^{2} + 2}{3},\quad n \le x < n + 1$$

💡 Grade 6 'use a variable to record an unknown' — name $n$ for the floor and the floor function disappears.

#6 Guess and Check 6.NS.C.6 Step 2

Sweep $n$ starting at $0$ and check each.
$n = 0$: $x = \tfrac{0 + 2}{3} = \tfrac{2}{3} \approx 0.667$.
Need $0 \le 0.667 < 1$?
Yes.
So $x = \tfrac{2}{3}$ is a valid solution.

$$n=0:\;x = \tfrac{2}{3},\;\lfloor \tfrac{2}{3} \rfloor = 0\;\checkmark$$

💡 Grade 6 'rational numbers on the number line' — locate $\tfrac{2}{3}$ between $0$ and $1$.

#6 Guess and Check 6.EE.B.5 Step 3

$n = 1$: $x = \tfrac{1 + 2}{3} = 1$.
Need $1 \le 1 < 2$?
Yes.
So $x = 1$ works.

$$n=1:\;x = 1,\;\lfloor 1 \rfloor = 1\;\checkmark$$

💡 Grade 6 'solving an equation' — plug in and check both inequalities.

#6 Guess and Check 6.EE.B.5 Step 4

$n = 2$: $x = \tfrac{4 + 2}{3} = 2$.
Need $2 \le 2 < 3$?
Yes.
So $x = 2$ works.

$$n=2:\;x = 2,\;\lfloor 2 \rfloor = 2\;\checkmark$$

💡 Grade 6 — check the candidate's floor matches the assumed $n$.

#6 Guess and Check 5.NF.B.3 Step 5

$n = 3$: $x = \tfrac{9 + 2}{3} = \tfrac{11}{3} \approx 3.667$.
Need $3 \le 3.667 < 4$?
Yes.
So $x = \tfrac{11}{3}$ works.

$$n=3:\;x = \tfrac{11}{3} \approx 3.667,\;\lfloor \tfrac{11}{3} \rfloor = 3\;\checkmark$$

💡 Grade 5 'fraction as division' — $11 \div 3 = 3$ remainder $2$, so the floor is $3$.

#2 Make a Systematic List 6.EE.B.8 Step 6

$n = 4$: $x = \tfrac{16 + 2}{3} = 6$.
Need $4 \le 6 < 5$?
But $6 \ge 5$, so NO.
Reject.
Once $n$ grows, $\tfrac{n^{2}+2}{3}$ shoots past $n + 1$.
Check $n = 5$: $x = \tfrac{27}{3} = 9$, way above $6$.
All $n \ge 4$ fail because $n^{2}/3$ already exceeds $n + 1$ when $n \ge 4$.

$$n=4:\;x = 6 \ge 5,\;\text{floor}=6 \ne 4\;\text{✗}$$

💡 Grade 6 'inequality on a number line' — for $n \ge 4$ the candidate overshoots the window $[n, n+1)$.

#2 Make a Systematic List 6.NS.C.7 Step 7

Negative $n$: try $n = -1$.
$x = \tfrac{1 + 2}{3} = 1$.
Need $-1 \le 1 < 0$?
But $1 \not< 0$, so NO.
For $n \le -1$, the candidate $\tfrac{n^{2}+2}{3} > 0$ but the window $[n, n+1)$ is at most $0$.
So all negative $n$ fail.

$$n=-1:\;x = 1 \not\in [-1, 0)\;\text{✗}$$

💡 Grade 6 'order rational numbers' — positive candidate can't sit in a non-positive window.

#3 Eliminate Possibilities 2.OA.C.4 Step 8

Tally: valid integers $n \in \{0, 1, 2, 3\}$ give exactly four distinct solutions $x \in \{\tfrac{2}{3},\,1,\,2,\,\tfrac{11}{3}\}$.
The four values are all different.
Answer: $\boxed{4}$, choice (B).

$$x \in \Big\{\tfrac{2}{3},\,1,\,2,\,\tfrac{11}{3}\Big\}\;\Rightarrow\;4 \text{ solutions} \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 2 'count objects' — four valid values, four solutions.

[1] #9 6.EE.B.6 Let $n = \lfloor x \rfloor$. The equation becomes $n^{2} - 3x + 2 = 0$, so $x =

[2] #6 6.NS.C.6 Sweep $n$ starting at $0$ and check each. $n = 0$: $x = \tfrac{0 + 2}{3} = \tfra

[3] #6 6.EE.B.5 $n = 1$: $x = \tfrac{1 + 2}{3} = 1$. Need $1 \le 1 < 2$? Yes. So $x = 1$ works.

[4] #6 6.EE.B.5 $n = 2$: $x = \tfrac{4 + 2}{3} = 2$. Need $2 \le 2 < 3$? Yes. So $x = 2$ works.

[5] #6 5.NF.B.3 $n = 3$: $x = \tfrac{9 + 2}{3} = \tfrac{11}{3} \approx 3.667$. Need $3 \le 3.667

[6] #2 6.EE.B.8 $n = 4$: $x = \tfrac{16 + 2}{3} = 6$. Need $4 \le 6 < 5$? But $6 \ge 5$, so NO.

[7] #2 6.NS.C.7 Negative $n$: try $n = -1$. $x = \tfrac{1 + 2}{3} = 1$. Need $-1 \le 1 < 0$? But

[8] #3 2.OA.C.4 Tally: valid integers $n \in \{0, 1, 2, 3\}$ give exactly four distinct solution

Review

Reasonableness: Spot-check by plugging back: $x = \tfrac{2}{3}$ gives $0^{2} - 3 \cdot \tfrac{2}{3} + 2 = 0 - 2 + 2 = 0$ ✓. $x = 1$: $1 - 3 + 2 = 0$ ✓. $x = 2$: $4 - 6 + 2 = 0$ ✓. $x = \tfrac{11}{3}$: $9 - 11 + 2 = 0$ ✓. 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.

Alternative: Tool #1 (Draw a Diagram): plot $y = \lfloor x \rfloor^{2}$ as a step function and $y = 3x - 2$ as a line. Intersections occur on flat steps where $3x - 2 = n^{2}$ on the interval $[n, n+1)$. Visually one sees one intersection on each of the steps $n = 0, 1, 2, 3$ and none for $n \ge 4$ (line slope $3$ is much slower than the step jumps $2n + 1$).

CCSS standards used (min grade 6)

2.OA.C.4 Use addition to find the total number of objects in rectangular arrays (Counting the four valid integers $n$ to report the solution count.)
5.NF.B.3 Interpret a fraction as division of the numerator by the denominator (Computing $\tfrac{11}{3}$ as $3$ remainder $2$ to read off its floor.)
6.EE.B.5 Understand solving an equation as a process of answering: which values make the equation true? (Substituting $n = 0, 1, 2, 3, 4, \ldots$ and checking whether the resulting $x$ satisfies both the equation and the floor-window condition.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Introducing $n = \lfloor x \rfloor$ to convert the floor-function equation into the clean integer-parameter equation $x = (n^{2} + 2)/3$.)
6.EE.B.8 Write an inequality of the form $x > c$ or $x < c$ and graph on a number line (Recording the condition $n \le x < n+1$ as the membership test of the candidate in its floor-window.)
6.NS.C.6 Understand a rational number as a point on the number line (Placing the candidates $\tfrac{2}{3}, 1, 2, \tfrac{11}{3}, 6, \ldots$ on the number line to inspect their floors.)
6.NS.C.7 Understand ordering and absolute value of rational numbers (Ruling out negative $n$ because the corresponding candidate is positive, so it cannot sit in a non-positive window.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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