# Difference between revisions of "2011 AMC 12B Problems/Problem 19"

## Problem

A lattice point in an $xy$-coordinate system is any point $(x, y)$ where both $x$ and $y$ are integers. The graph of $y = mx + 2$ passes through no lattice point with $0 < x \leq 100$ for all $m$ such that $\frac{1}{2} < m < a$. What is the maximum possible value of $a$?

$\textbf{(A)}\ \frac{51}{101} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{51}{100} \qquad \textbf{(D)}\ \frac{52}{101} \qquad \textbf{(E)}\ \frac{13}{25}$

## Solution

It is very easy to see that the $+2$ in the graph does not impact whether it passes through lattice.

We need to make sure that $m$ cannot be in the form of $\frac{a}{b}$ for $1\le b\le 100$, otherwise the graph $y= mx$ passes through lattice point at $x = b$. We only need to worry about $\frac{a}{b}$ very close to $\frac{1}{2}$, $\frac{m+1}{2m+1}$, $\frac{m+1}{2m}$ will be the only case we need to worry about and we want the minimum of those, clearly for $1\le b\le 100$, the smallest is $\frac{50}{99}$, so answer is $\boxed{\frac{50}{99} \textbf{(B)}}$

## Solution 2

Like in the first solution, we note that the $+2$ does not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line $y = mx$.

A line with a slope of $\frac{1}{2}$ passes through (among other lattice points) the lattice point $(100,50)$. As the slope of the line increases from $\frac{1}{2}$, the first lattice point it hits is at $(99, 50)$, the slope of that line being $\frac{50}{99}$ . So the answer is $\boxed{\textbf{(B)}}$