2023 AMC 10B Problem 22

Question

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?

an infinite number

4

2

3

0

Set

n = \lfloor x \rfloor;

then

x = \frac{n^2 + 2}{3}

Require

n \le \frac{n^2 + 2}{3} \lt n + 1

and find the integers

n

that work

Set

n = \lfloor x \rfloor.

Then

n^2 - 3x + 2 = 0

gives

x = \frac{n^2 + 2}{3}.

For this to be consistent we need

n \le \frac{n^2 + 2}{3} \lt n + 1.

The left side,

n^2 - 3n + 2 \ge 0,

holds for every integer

n.

The right side,

n^2 - 3n - 1 \lt 0,

holds only for

n \in \{0, 1, 2, 3\}.

Those give

x = \frac{2}{3}, 1, 2, \frac{11}{3},

so there are

4

distinct values. Therefore, the answer is B.

Problem 22 in other years