2024 AMC 10B Problem 19

The table has three rows (slope $0$, a nonzero fraction slope, an irrational slope) and four columns ($0$, exactly $1$, exactly $2$, more than $2$ lattice points). Two facts settle almost everything.

Fact 1: two lattice points give a fraction slope. If a line passes through $(1, 2)$ and $(4, 7)$, its slope is $\frac{7 - 2}{4 - 1} = \frac53$, a fraction. The same happens for any two lattice points, because the slope is one whole number divided by another. So a line with an irrational slope (like $\sqrt2$) passes through at most one lattice point.

Fact 2: a fraction slope repeats. Take the line through $(0, 0)$ with slope $\frac23$. Going $3$ right and $2$ up lands on $(3, 2)$, then $(6, 4)$, then $(9, 6)$, and so on forever. In general, if a line has slope $\frac{p}{q}$ and passes through one lattice point, moving $q$ right and $p$ up lands on another lattice point every time. So such a line has either no lattice points or infinitely many.

Now go row by row.

1. Slope $0$ (lines $y = c$). The line $y = 1.5$ has no lattice points, and $y = 1$ has infinitely many. Exactly $1$ or exactly $2$ is impossible by Fact 2. Possible: $0$ and more than $2$.
2. Nonzero fraction slope. The line $y = 2x$ has infinitely many lattice points. The line $y = x + \frac12$ has none: for any whole number $x$, $y$ ends in $.5$. Exactly $1$ or $2$ is impossible by Fact 2. Possible: $0$ and more than $2$.
3. Irrational slope. The line $y = \sqrt2\,x$ passes through $(0, 0)$ and no other lattice point (by Fact 1). The line $y = \sqrt2\,x + \frac12$ passes through none, since at $x = 0$ it gives $y = \frac12$ and at any other whole $x$, $y$ is not a whole number. Two or more is impossible by Fact 1. Possible: $0$ and exactly $1$.

Each row has $2$ "Possible" entries, so there are $3 \cdot 2 = 6$.

The answer is $\boxed{\textbf{(C)}\ 6}$.

Whether a line's slope is a fraction decides how it meets lattice points. A fraction slope makes the lattice points repeat at a steady step, so there are none or infinitely many. An irrational slope allows at most one. To fill a "possible or not" table, give one example for each "Possible" and a general reason for each "Not Possible".

The rows are slope $0$, nonzero rational slope, irrational slope; the columns are $0$, exactly $1$, exactly $2$, more than $2$ lattice points.

Two general facts do most of the work.

1. If a line contains two lattice points $(a, b)$ and $(c, d)$, its slope $\frac{d - b}{c - a}$ is rational (or the line is vertical). So an irrational-slope line has at most one lattice point.
2. If a line with rational slope $\frac{p}{q}$ (lowest terms) contains a lattice point $(a, b)$, it also contains $(a + q, b + p)$, $(a + 2q, b + 2p)$, and so on. So a rational-slope line with one lattice point has infinitely many.

Slope $0$: $y = c$. If $c$ is not an integer, there are $0$ lattice points; if $c$ is an integer, infinitely many. Exactly $1$ or $2$ is impossible by fact 2. Possible: $0$ and more than $2$.

Nonzero rational slope: $y = \frac23 x + \frac12$ has no lattice points (the right side is never an integer for integer $x$: $\frac{4x + 3}{6}$ has odd numerator); $y = 2x$ has infinitely many. Exactly $1$ or $2$ is impossible by fact 2. Possible: $0$ and more than $2$.

Irrational slope: $y = \sqrt2\, x$ passes through exactly one lattice point, the origin; $y = \sqrt2\, x + \frac12$ passes through none. Two or more is impossible by fact 1. Possible: $0$ and exactly $1$.

Each row has $2$ possible entries, for $3 \cdot 2 = 6$ in all.

The answer is $\boxed{\textbf{(C)}\ 6}$.