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 $\boxed{\frac{50}{99} \textbf{(B)}}$

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

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2011 AMC 12B Problems/Problem 19

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