2011 AMC 10B Problems/Problem 24

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 $0 < x \le 100$ 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

We see that for the graph of $y=mx+2$ to not pass through any lattice points, the denominator of $m$ must be greater than $100$ , or else it would be canceled by some $0<x\le100$ which would make $y$ an integer. By using common denominators, we find that the order of the fractions from smallest to largest is $(A), (B), (C), (D), (E)$ . We can see that when $x=\frac{50}{99}$ , $y$ would be an integer, so therefore any fraction greater than $\frac{50}{99}$ would not work, as substituting our fraction $\frac{50}{99}$ for $m$ would produce an integer for $y$ . So now we are left with only $\frac{51}{101}$ and $\frac{50}{99}$ . But since $\frac{51}{101}=\frac{5049}{9999}$ and $\frac{50}{99}=\frac{5050}{9999}$ , we can be absolutely certain that there isn't a number between $\frac{51}{101}$ and $\frac{50}{99}$ that can reduce to a fraction whose denominator is less than or equal to $100$ . Since we are looking for the maximum value of $a$ , we take the larger of $\frac{51}{101}$ and $\frac{50}{99}$ , which is $\boxed{\textbf{(B)}\frac{50}{99}}$ .

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2011 AMC 10B Problems/Problem 24

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