2024 AMC 10B Problem 19

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 $\boxed{\textbf{(C)}\ 6}$.

Lattice points on a line are controlled by the slope's rationality: rational slopes make lattice points periodic (none or infinitely many), irrational slopes cap them at one. Filling a possibility table means proving each "Not Possible" with a general argument and each "Possible" with one concrete example; keep both tools ready.

• Believing a nonzero rational slope line can pass through exactly one or two lattice points; once it hits one, stepping by the reduced (q, p) gives infinitely many.
• Marking "exactly one" as impossible for irrational slope, forgetting that a lattice point on the line is allowed, just not two.