2023 AMC 10 B Problem 22

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To help you master 2023 AMC 10 B Problem 22, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2023 AMC10 B Problem 22

Solution

Solution 1

Algebraic Bounding Method

By expressing $x$ in terms of its integer part $I=[x]$, we can substitute $x$ into the defining inequality $I\le x<I+1$. This creates a system of inequalities that restricts $I$ to a few integer values.

The fundamental inequality of the floor function states that for any real number $x$, its floor $[x]$ satisfies $[x]\le x<[x]+1$. This is the most reliable way to convert an equation with a floor function into an algebraic inequality.

Step 1: Express $x$ in terms of $I$

Let $I=[x]$. Substitute $I$ into the original equation and solve for $x$.

$$I^2-3x+2=0$$

$$3x=I^2+2$$

$$x=\frac{I^2+2}{3}$$

Step 2: Set up the bounding inequality

By the definition of the floor function, the integer part $I$ must be less than or equal to $x$, and $x$ must be strictly less than $I+1$. Substitute our expression for $x$ into this fundamental inequality.

$$I\le x<I+1$$

$$I\le \frac{I^2+2}{3}<I+1$$

Step 3: Solve the left side of the inequality

Multiply the entire inequality by 3 to clear the denominator, then solve the left part of the compound inequality. The result $(I-1)(I-2)\ge 0$ is true for $I\le 1$ and $I\ge 2$. Since there are no integers strictly between 1 and 2, this holds for all integers $I$.

$$3I\le I^2+2$$

$$0\le I^2-3I+2$$

$$0\le (I-1)(I-2)$$

Step 4: Solve the right side of the inequality

Now solve the right part of the compound inequality. We find the roots of the corresponding quadratic equation using the quadratic formula. Since $\sqrt{13}$ is between 3 and 4, the roots are approximately $-0.3$ and $3.3$. The parabola opens upwards, so the expression is negative between the roots.

$$I^2+2<3I+3$$

$$I^2-3I-1<0$$

$$I=\frac{3\pm \sqrt{9-4(1)(-1)}}{2}=\frac{3\pm \sqrt{13}}{2}$$

$$-0.3<I<3.3$$

Step 5: Test the integer candidates and count the solutions

The only integers satisfying $-0.3<I<3.3$ are $0,1,2,$ and $3$. For each valid $I$, calculate the corresponding $x$ and verify that its floor is indeed $I$. All four candidates produce valid solutions.

$$I=0⟹x=\frac{0^2+2}{3}=\frac{2}{3}⟹\left[\frac{2}{3}\right]=0$$

$$I=1⟹x=\frac{1^2+2}{3}=1⟹[1]=1$$

$$I=2⟹x=\frac{2^2+2}{3}=2⟹[2]=2$$

$$I=3⟹x=\frac{3^2+2}{3}=\frac{11}{3}⟹\left[\frac{11}{3}\right]=3$$

Correct answer: B. $4$
There are exactly 4 distinct values of $x$: $\frac{2}{3},1,2,$ and $\frac{11}{3}$. The correct answer choice is B. We verified each solution by substituting it back into the original equation, confirming that they all evaluate to 0.

Solution 2

Interval Testing Method

Instead of solving quadratic inequalities, we can use the equation to show that $x$ must be positive, test small integer values for $[x]$, and prove that the quadratic term $[x{]}^2$ grows too fast for any solutions to exist when $[x]\ge 4$.

Growth Rates of Functions: A quadratic function (like $x^2$) will always eventually outgrow a linear function (like $3x$). When solving equations mixing these types, solutions are confined to a bounded region, allowing us to test a finite number of cases.

Step 1: Establish that $x$ must be positive

Rearrange the equation to isolate $3x$. Since $[x{]}^2\ge 0$ and $2>0$, the right side is strictly positive. This means $3x>0$, so $x$ must be positive. Therefore, $[x]$ must be a non-negative integer.

$$3x=[x{]}^2+2$$

$$[x{]}^2+2>0⟹3x>0⟹x>0$$

$$[x]\in \{0,1,2,3,\dots \}$$

Step 2: Test small non-negative integer values for $[x]$

Test the first few non-negative integer values for $[x]$. For each case, calculate $x$ and check if the floor of that $x$ matches the assumed value of $[x]$. We find four valid solutions.

$$[x]=0⟹3x=2⟹x=\frac{2}{3}\in [0,1)$$

$$[x]=1⟹3x=3⟹x=1\in [1,2)$$

$$[x]=2⟹3x=6⟹x=2\in [2,3)$$

$$[x]=3⟹3x=11⟹x=\frac{11}{3}\in [3,4)$$

Step 3: Prove that no solutions exist for $[x]\ge 4$

We must ensure no solutions exist for $[x]\ge 4$. By definition, $x<[x]+1$, so $3x<3[x]+3$. Substitute $3x=[x{]}^2+2$ into this inequality. For $[x]\ge 4$, the resulting quadratic expression is strictly positive, which contradicts the requirement that it must be less than zero. Thus, the quadratic term grows too fast, and no larger solutions exist.

$$x<[x]+1⟹3x<3[x]+3$$

$$[x{]}^2+2<3[x]+3$$

$$[x{]}^2-3[x]-1<0$$

$$[x]([x]-3)-1<0$$

$$[x]\ge 4⟹[x]([x]-3)-1\ge 4(1)-1=3>0$$

Step 4: Conclude the total number of solutions

Since $[x]$ cannot be negative and cannot be 4 or greater, the only possible values for $[x]$ are $0,1,2,$ and $3$. Each of these generated exactly one valid $x$, giving a total of 4 distinct solutions.

Correct answer: B. $4$
We found 4 valid solutions and proved that no other integer values for $[x]$ can produce a solution. The correct answer choice is B.

Similar Problems in Real Exams

The following five AMC 10 B problems are closely related to AMC 10 B 2023 Question 22. These real AMC 10 B problems test the same key concept—Number Theory. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2025 AMC 10 B Problem 18

2020 AMC 10 B Problem 24

2025 AMC 10 B Problem 2

2025 AMC 10 B Problem 4

2025 AMC 10 B Problem 8

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Fundamental Inequality of the Floor Function:

$x$ is any real number, and $[x]$ is the greatest integer less than or equal to $x$. It allows us to convert an equation involving a floor function into a system of inequalities involving only integers.

• Comparing Growth Rates:

Applies when an equation sets a quadratic expression equal to a linear expression. It guarantees that for sufficiently large values of $[x]$, the quadratic side $[x{]}^2+2$ will strictly exceed the linear side $3[x]+3$, meaning no large solutions can exist.

Key Takeaway: Remember This Strategy for Next Time

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